JEE Purification & Characterisation Questions

From the reaction sequence given in Q-17, the molecular formula of compound $$M$$ is already known (see the previous answer) and it contains exactly one sulphur atom per molecule. Its molar mass is

$$M_M = 379\;\text{g mol}^{-1}$$

In the Carius method the entire sulphur present in the sample is oxidised to sulphate ion and is quantitatively precipitated as $$\mathrm{BaSO_4}$$. For every mole of sulphur, one mole of $$\mathrm{BaSO_4}$$ is obtained:

$$\text{S (in sample)} \;\longrightarrow\; \mathrm{SO_4^{2-}} \;\longrightarrow\; \mathrm{BaSO_4}$$

1. Moles of $$M$$ taken:

$$n_M = \frac{3.79}{379} = 0.010\;\text{mol}$$

2. Since each molecule of $$M$$ contains one sulphur atom, the moles of sulphur are also 0.010 mol.

3. Hence, moles of $$\mathrm{BaSO_4}$$ formed:

$$n_{\mathrm{BaSO_4}} = 0.010\;\text{mol}$$

4. Mass of $$\mathrm{BaSO_4}$$ precipitated (molar mass $$= 137+32+64 = 233\;\text{g mol}^{-1}$$):

$$m_{\mathrm{BaSO_4}} = n_{\mathrm{BaSO_4}}\times 233 = 0.010 \times 233 = 2.33\;\text{g}$$

Therefore, from 3.79 g of compound $$M$$ the mass of $$\mathrm{BaSO_4}$$ obtained is

2.33 g.

Reaction of 2-methylpropene with dilute sulphuric acid gives alcohol by hydration.

Reaction:

$$CH_2=C\left(CH_3\right)_2+H_2O\ +\ dil.\ H_2SO_4\longrightarrow\ \left(CH_3\right)COH$$

Major Product Formed:

X= tert-butyl alcohol(2-methylpropan-2-ol)

In TLC:

$$R_f=\frac{\text{Distance travelled by compound}}{\text{Distance travelled by solvent}}$$

More polar compounds interact more strongly with polar stationary phase and therefore move less.

Hence:

More polar compound $$=>$$Lower $$R_f$$ value

2-methylpropene is non-polar.

tert-butyl alcohol contains: $$-OH$$ group and is more polar.

Therefore, product X will have lower $$R_f$$​ value than 0.42.

Hence,

$$R_f\text{ value of }X<0.42$$

Therefore, Option D is correct

The Carius method involves converting the chlorine in an organic compound to silver chloride (AgCl). The mass of chlorine in the AgCl precipitate comes from the organic compound.

Given: - Mass of organic compound = 0.2425 g - Mass of AgCl obtained = 0.5253 g

The molar mass of AgCl is calculated using atomic masses: Ag = 108 g/mol, Cl = 35.5 g/mol. Thus, molar mass of AgCl = 108 + 35.5 = 143.5 g/mol.

In 143.5 g of AgCl, the mass of chlorine is 35.5 g. Therefore, the mass of chlorine in 0.5253 g of AgCl is found using proportion:

Mass of chlorine = $$\left( \frac{35.5}{143.5} \right) \times 0.5253$$

First, compute the ratio: $$\frac{35.5}{143.5} = \frac{355}{1435} = \frac{71}{287} \approx 0.2473867595818815$$

Then, multiply by the mass of AgCl: $$0.2473867595818815 \times 0.5253 = 0.1299522648083624 \text{ g}$$

This mass of chlorine comes from 0.2425 g of the organic compound. The percentage of chlorine is:

Percentage of chlorine = $$\left( \frac{\text{mass of chlorine}}{\text{mass of compound}} \right) \times 100 = \left( \frac{0.1299522648083624}{0.2425} \right) \times 100$$

Compute the division: $$\frac{0.1299522648083624}{0.2425} = 0.5358762886597938$$

Multiply by 100: $$0.5358762886597938 \times 100 = 53.58762886597938\%$$

Rounding to two decimal places gives 53.59%. However, comparing with the options, 53.58% is the closest match.

The options are: - A. 37.57% - B. 53.58% - C. 87.65% - D. 34.79%

Thus, the correct answer is option B, 53.58%.

The sulphur in the organic compound is converted to barium sulphate (BaSO₄) during the experiment. Therefore, the mass of sulphur in the barium sulphate precipitate comes entirely from the organic compound.

The molar mass of BaSO₄ is 233 g/mol, and the molar mass of sulphur (S) is 32 g/mol. The mass of sulphur in the barium sulphate precipitate can be calculated using the proportion:

$$\text{Mass of sulphur} = \left( \frac{\text{Molar mass of S}}{\text{Molar mass of BaSO}_4} \right) \times \text{Mass of BaSO}_4$$

Substituting the given values:

$$\text{Mass of sulphur} = \left( \frac{32}{233} \right) \times 0.4813$$

First, compute the fraction:

$$\frac{32}{233} \approx 0.13733905579399142$$

Now multiply by the mass of BaSO₄:

$$0.13733905579399142 \times 0.4813 \approx 0.06610128$$

So, the mass of sulphur is approximately 0.06610128 g.

This sulphur originated from the organic compound with a mass of 0.314 g. The percentage of sulphur in the compound is:

$$\text{Percentage of sulphur} = \left( \frac{\text{Mass of sulphur}}{\text{Mass of compound}} \right) \times 100$$

Substituting the values:

$$\text{Percentage} = \left( \frac{0.06610128}{0.314} \right) \times 100$$

First, divide the masses:

$$\frac{0.06610128}{0.314} \approx 0.210514$$

Now multiply by 100:

$$0.210514 \times 100 = 21.0514\%$$

Rounding to two decimal places, the percentage is 21.05%.

Comparing with the options:

A. 48.24%
B. 63.15%
C. 42.10%
D. 21.05%

The correct answer is D. 21.05%.

Mass of compound $$(X) = 2.0\;g$$. 

Under Carius halogen analysis bromine is converted to $$AgBr$$, 

so the moles of $$AgBr$$ produced equal the moles of $$Br$$ present.

Molar mass of $$AgBr = 108 + 80 = 188\;g\,mol^{-1}$$.

Moles of $$AgBr$$ obtained $$= \frac{3.36}{188} = 0.017872\;mol$$.

Since,

the ratio $$AgBr : Br$$ is $$1:1$$, moles of bromine in the sample are also $$0.017872\;mol$$.

Mass of bromine in the sample $$= 0.017872 \times 80 = 1.4298\;g$$.

Percentage of bromine $$= \frac{1.4298}{2.0}\times100 = 71.49\%$$.

Percentage of carbon is given as $$26.7\%$$, 

so 

mass of carbon in the sample $$= 0.267 \times 2.0 = 0.534\;g$$.

Moles of carbon $$= \frac{0.534}{12} = 0.0445\;mol$$.

The remaining mass is hydrogen: $$2.0 - 0.534 - 1.4298 = 0.0362\;g$$.

Moles of hydrogen $$= 0.0362\;mol$$ (atomic mass of $$H = 1$$).

Now write the mole ratio:

$$C : H : Br = 0.0445 : 0.0362 : 0.017872$$.

Divide by the smallest value $$0.017872$$:

$$C : H : Br = 2.49 : 2.03 : 1$$.

These are very close to $$2.5 : 2 : 1$$. 

Multiply each by $$2$$ to clear the fractional $$0.5$$ and obtain the simplest integer ratio:

$$C : H : Br = 5 : 4 : 2$$.

Therefore the empirical formula of the bromo-hydrocarbon is $$C_5H_4Br_2$$.

Hence, the total number of carbon atoms present in the empirical formula is 5.

We need to find the percentage of halogen in compound Y.

First, the halogen was identified by treating the sodium fusion extract with $$CHCl_3$$ and chlorine water, which gives a violet color to the $$CHCl_3$$ layer. This is the test for iodine, since iodine dissolves in $$CHCl_3$$ giving a violet/purple color. Hence the halogen is iodine (I), with molar mass 127 g/mol.

Next, using the Carius method data for iodine, the precipitate formed is AgI (molar mass = 108 + 127 = 235 g/mol). The mass of the compound taken was 0.15 g and the mass of AgI obtained was 0.12 g.

Therefore, the mass of iodine can be calculated as $$\text{Mass of I} = \frac{127}{235} \times 0.12 = \frac{15.24}{235} = 0.06485 \text{ g}$$.

From this, the percentage of iodine in the compound is $$\% \text{ of I} = \frac{0.06485}{0.15} \times 100 = 43.23\%$$.

The percentage of halogen is approximately $$43\%$$ to the nearest integer.

Therefore, the answer is 43.

The qualitative detection of extra-elements (nitrogen, sulphur, halogen and phosphorus) in an organic compound is carried out with the help of Lassaigne’s test, also called the sodium-fusion test.

Case 1 : Statement (I)
Statement (I) says that nitrogen, sulphur, halogen and phosphorus present in an organic compound are detected by Lassaigne’s test. During the test the compound is fused with metallic sodium. The fusion converts the covalently bound elements into ionic salts such as $$NaCN$$, $$Na_2S$$, $$NaX$$ (where $$X = Cl, Br, I$$) and $$Na_3PO_4$$, which can then be identified by suitable confirmatory tests. Hence Statement (I) is correct.

Case 2 : Statement (II)
Statement (II) claims that the elements are converted from covalent form into ionic form by fusing the compound with magnesium in Lassaigne’s test. However, the metal actually used in Lassaigne’s test is sodium, not magnesium. Therefore the statement describing fusion with magnesium is incorrect. Hence Statement (II) is false.

Thus Statement (I) is true while Statement (II) is false. The option that matches this combination is Option D.

Final answer : Option D

In titrimetric (volumetric) analysis a primary standard is the substance that is weighed accurately, dissolved, and its solution is used to standardise another reagent. For a compound to serve as a primary standard it must satisfy all the conditions below:
  • Very high purity and known formula
  • Stable in air (non-hygroscopic, non-deliquescent, does not react with $$CO_2$$ or $$O_2$$)
  • Reasonably high molar/​equivalent mass (reduces weighing error)
  • Soluble in the chosen solvent and gives a stoichiometric reaction
  • Non-toxic and easily available

Let us test each compound given in the question against the above requirements.

Case A:

$$Na_2Cr_2O_7$$ (sodium dichromate)
This salt is strongly hygroscopic and absorbs moisture from air, so its weighed mass does not give a fixed number of moles. Hence it fails the “stable in air” criterion and should not be used as a primary standard.

Case B:

Oxalic acid, generally used as the dihydrate $$H_2C_2O_4\cdot 2H_2O$$, is available in high purity, has a fixed water of crystallisation, is not hygroscopic, and has a suitable equivalent mass. Therefore it is a good primary standard and is widely used for standardising $$KMnO_4$$.

Case C:

NaOH (sodium hydroxide) readily absorbs both $$H_2O$$ (moisture) and $$CO_2$$ from the atmosphere, forming $$Na_2CO_3$$. Because of this, its composition changes while weighing. Thus NaOH cannot be used as a primary standard.

Case D:

$$FeSO_4\cdot 6H_2O$$ (ferrous sulphate hexahydrate) slowly oxidises to $$Fe^{3+}$$ in air and also effloresces (loses water). Hence its exact formula weight changes with time. It is therefore unsuitable as a primary standard.

Case E:

Sodium tetraborate (borax), $$Na_2B_4O_7\cdot 10H_2O$$, is available in very pure form, is not appreciably hygroscopic, and has a high equivalent mass. Because it meets all the criteria, it is an excellent primary standard for acid-base titrations.

Summarising the suitability:

• Not suitable → A ($$Na_2Cr_2O_7$$), C (NaOH), D ($$FeSO_4\cdot 6H_2O$$)
• Suitable → B (oxalic acid), E (sodium tetraborate)

The compounds that should NOT be used as primary standards are therefore A, C and D only.

Option D is correct.

Let us analyze both statements:

Statement I: In Lassaigne's test, the covalent organic molecules are transformed into ionic compounds.

In Lassaigne's test, the organic compound is fused with sodium metal. This converts covalent compounds containing N, S, or halogens into ionic compounds like NaCN, Na₂S, NaX (where X is halogen). This statement is true.

Statement II: The sodium fusion extract of an organic compound having N and S gives prussian blue colour with $$FeSO_4$$ and $$Na_4[Fe(CN)_6]$$.

When an organic compound contains both N and S, the fusion with sodium produces NaSCN (sodium thiocyanate) instead of NaCN and Na₂S separately. With $$FeSO_4$$, NaSCN would give a blood-red colour (ferric thiocyanate), not prussian blue.

For prussian blue test: $$Na_4[Fe(CN)_6]$$ reacts with $$Fe^{3+}$$ to give prussian blue $$Fe_4[Fe(CN)_6]_3$$. But in the presence of S, the CN⁻ ions are converted to SCN⁻, so prussian blue is not obtained directly. This statement is false.

Additionally, the reagent combination mentioned is incorrect. For the prussian blue test, we use $$FeSO_4$$ followed by $$FeCl_3$$, not $$Na_4[Fe(CN)_6]$$. Statement II is false.

The correct answer is Option 1: Statement I is true but Statement II is false.

In separation techniques we always match the physical property of the mixture with the most suitable method:

Case A: Aniline-water mixture
Aniline boils around $$184^{\circ}\text{C}$$ and is immiscible with water. When such a liquid is subjected to steam, the total vapour pressure is the sum of the individual vapour pressures, so the mixture boils below $$100^{\circ}\text{C}$$ and aniline distils along with steam. Hence the correct technique is steam distillation.

Case B: Glycerol from spent-lye in the soap industry
Pure glycerol decomposes near its normal boiling point $$\left(\approx 290^{\circ}\text{C}\right)$$. To avoid decomposition it is distilled under low pressure, so that it boils at a much lower temperature. Therefore we need distillation at reduced (vacuum) pressure.

Case C: Different fractions of crude oil
Crude petroleum contains components whose boiling ranges overlap over fairly narrow intervals. Separation is done in a fractionating column where rising vapours repeatedly condense and revapourise. This is classical fractional distillation.

Case D: Chloroform-aniline mixture
Chloroform boils at $$61^{\circ}\text{C}$$ while aniline boils at $$184^{\circ}\text{C}$$. Such a large difference means the lower-boiling component can be collected first without elaborate apparatus. Hence we use simple distillation.

Summarising the matches:

(A) Aniline / water  →  IV Steam distillation
(B) Glycerol / spent-lye  →  III Distillation at reduced pressure
(C) Crude-oil fractions  →  II Fractional distillation
(D) Chloroform / aniline  →  I Simple distillation

Therefore the correct choice is Option A.

In some purification techniques, the solid sample is first converted directly into its vapour phase and then condensed back to the solid state, leaving behind the non-volatile impurities. This direct solid ⇌ vapour ⇌ solid change, without passing through the liquid phase, is called sublimation.

Only compounds that possess high vapour pressure at temperatures well below their melting points show this behaviour. Typical examples are naphthalene, camphor, iodine, benzoic acid, etc. Non-sublimable impurities (sand, salts, metal oxides) remain as residue.

Hence, the purification method that utilises the physical transformation “solid → vapour → solid” is sublimation.

Option C which is: Sublimation

In Dumas’ method the nitrogen gas collected over water is always moist, so we must first obtain the volume of dry nitrogen using Dalton’s law of partial pressures.

Total pressure of the gas mixture, $$P_{\text{total}} = 715 \text{ mm Hg}$$
Aqueous tension of water at $$300 \text{ K},\, P_{\text{H}_2\text{O}} = 15 \text{ mm Hg}$$

Hence, pressure of dry nitrogen
$$P_{\text{N}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} = 715 - 15 = 700 \text{ mm Hg}$$

Convert this pressure into atmospheres because the gas constant $$R$$ in SI units involves atmospheres:
$$1 \text{ atm} = 760 \text{ mm Hg}$$, therefore
$$P = \frac{700}{760} \text{ atm} = 0.921 \text{ atm}$$ (rounded to three significant figures)

The measured volume of nitrogen is
$$V = 60 \text{ mL} = 0.060 \text{ L}$$

Temperature is
$$T = 300 \text{ K}$$

Use the ideal-gas equation $$PV = nRT$$ to find the number of moles of nitrogen.
Gas constant, $$R = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}$$

Substituting the values:
$$n = \frac{PV}{RT} = \frac{0.921 \times 0.060}{0.0821 \times 300}$$

Calculate the numerator:
$$0.921 \times 0.060 = 0.0553$$

Calculate the denominator:
$$0.0821 \times 300 = 24.63$$

Hence,
$$n = \frac{0.0553}{24.63} = 0.00224 \text{ mol}$$

Mass of nitrogen present:
Molar mass of $$\text{N}_2 = 28 \text{ g mol}^{-1}$$, so
$$m_{\text{N}_2} = n \times 28 = 0.00224 \times 28 = 0.0628 \text{ g}$$

The original mass of the organic compound taken was $$0.400 \text{ g}$$. Therefore, percentage of nitrogen in the compound is
$$\%\text{N} = \frac{0.0628}{0.400} \times 100 = 15.7\%$$ (to three significant figures)

Thus the percentage composition of nitrogen in the compound is $$\mathbf{15.71\%}$$.

Option A is correct.

Case A : Simple distillation
Simple (ordinary) distillation is used when the two liquids are completely miscible and their boiling points differ by more than about $$25^{\circ}\text{C}$$. Chloroform boils at $$61^{\circ}\text{C}$$ while aniline boils at $$184^{\circ}\text{C}$$, a difference of $$123^{\circ}\text{C}$$. Therefore the pair chloroform + aniline can be separated by ordinary distillation. Hence (A) → (III).

Case B : Fractional distillation
Fractional distillation is chosen for mixtures whose components are miscible and have boiling points that are close to each other (difference less than about $$25^{\circ}\text{C}$$). Petrol and diesel are overlapping fractions of petroleum whose components possess closely spaced boiling ranges; they are separated industrially in a fractionating tower. Hence (B) → (I).

Case C : Distillation under reduced pressure (vacuum distillation)
When a liquid has a very high boiling point and tends to decompose at its normal boiling point, it is distilled at a lowered external pressure so that it boils at a lower temperature. Glycerol (bp $$\approx 563\ \text{K}$$) is separated from spent-lye (the residual liquor in soap manufacture) by vacuum distillation to avoid its decomposition. Hence (C) → (IV).

Case D : Steam distillation
Steam distillation is applicable to compounds that are volatile in steam, immiscible with water, and possess high boiling points. Aniline is immiscible with water and forms a separate layer; it distils along with steam at a temperature much lower than its normal boiling point, allowing its separation from water. Hence (D) → (II).

Collecting all matches:
(A) - (III), (B) - (I), (C) - (IV), (D) - (II).

This correspondence is provided by Option D.

Lead(II) ions are identified in classical qualitative analysis by passing the precipitate successively through sulphide, sulphate and chromate tests. The colours obtained at every stage fix the formulae of the precipitates.

Case 1 : Treatment of the original lead(II) salt with $$H_2S$$ in acidic medium
$$Pb^{2+} + S^{2-} \rightarrow PbS\downarrow$$

Lead(II) sulphide forms as a black precipitate, so
$$A = PbS$$

Case 2 : Treating the black precipitate with dilute $$H_2SO_4$$
$$PbS + 2\,H_2SO_4 \rightarrow PbSO_4\downarrow + H_2S\uparrow + SO_2\uparrow + 2\,H_2O$$

Lead(II) sulphate is sparingly soluble and appears as a white precipitate, hence
$$B = PbSO_4$$

Case 3 : Adding an alkaline $$K_2CrO_4$$ (or $$Na_2CrO_4$$) solution to the white precipitate
$$Pb^{2+}_{(aq)} + CrO_4^{2-} \rightarrow PbCrO_4\downarrow$$

The insoluble bright-yellow solid obtained is lead(II) chromate, therefore
$$C = PbCrO_4$$

Thus the correct identification is:
A - $$PbS$$, B - $$PbSO_4$$, C - $$PbCrO_4$$

The only option that lists this sequence is
Option D which is: $$PbS,\,PbSO_{4},\,PbCrO_{4}$$

Statement I: In partition chromatography, the stationary phase is a thin film of liquid present on an inert support. This is correct - partition chromatography is based on the partitioning of components between a liquid stationary phase held on a solid support and a mobile phase.

Statement II: In paper chromatography, the material of paper acts as the stationary phase. This is incorrect. In paper chromatography, the water trapped in the cellulose fibers of the paper acts as the stationary phase (liquid), not the paper material itself. Paper chromatography is actually a type of partition chromatography.

The correct answer is Option 1: Statement I is true but Statement II is false.

To determine the volume of $$N_2$$ gas liberated at STP using Dumas' method, we first analyze the molecular structure and composition of piperazine, $$C_4H_{10}N_2$$.

The molar mass of piperazine is calculated as follows:

• Carbon: $$4 \times 12 = 48\text{ g mol}^{-1}$$
• Hydrogen: $$10 \times 1 = 10\text{ g mol}^{-1}$$
• Nitrogen: $$2 \times 14 = 28\text{ g mol}^{-1}$$

Hence, the molar mass of piperazine is

$$48 + 10 + 28 = 86\text{ g mol}^{-1}.$$

For a sample mass of $$0.42\text{ g}$$, the number of moles of piperazine is

$$\frac{0.42}{86} \approx 0.0048837\text{ mol}.$$

Each molecule of piperazine contains two nitrogen atoms, which together correspond to one molecule of $$N_2$$. Therefore, one mole of piperazine liberates one mole of $$N_2$$ gas.

Hence,

$$\text{Moles of }N_2 = 0.0048837\text{ mol}.$$

If the molar volume of an ideal gas at STP is taken as $$22,400\text{ mL mol}^{-1}$$, then the volume of nitrogen liberated is

$$0.0048837 \times 22,400 \approx 109.39\text{ mL}.$$

Rounding to the nearest integer,

$$\boxed{109\text{ mL}}.$$

However, if the standard molar volume at STP is taken as $$22,700\text{ mL mol}^{-1}$$ (corresponding to STP defined at $$1$$ bar), then

$$0.0048837 \times 22,700 \approx 110.86\text{ mL},$$

which rounds to

$$\boxed{111\text{ mL}}.$$

Thus, the slight difference in the final answer arises solely from the convention adopted for the molar volume of an ideal gas at STP. Using $$22.4\text{ L mol}^{-1}$$ gives $$109\text{ mL}$$, whereas using $$22.7\text{ L mol}^{-1}$$ gives $$111\text{ mL}$$.

In Dumas’ method, nitrogen is collected over water, so the measured pressure includes the vapour pressure of water (aqueous tension).
To obtain the pressure of dry nitrogen, subtract the aqueous tension from the total pressure.

Total pressure of the gas mixture at $$300\ \text{K} = 715\ \text{mm Hg}$$.
Aqueous tension of water at $$300\ \text{K} = 15\ \text{mm Hg}$$.

Pressure of dry $$N_2$$:
$$P_{N_2}=715\ \text{mm Hg}-15\ \text{mm Hg}=700\ \text{mm Hg}$$.

Convert this pressure to atmospheres (because the gas constant $$R$$ we shall use is in $$\text{L atm mol}^{-1}\text{K}^{-1}$$):
$$P=\frac{700}{760}\ \text{atm}=0.921\ \text{atm}$$.

Volume of nitrogen collected:
$$V =60\ \text{mL}=0.06\ \text{L}$$.

Temperature:
$$T =300\ \text{K}$$.

Ideal gas equation: $$PV = nRT$$.
Gas constant: $$R =0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$$.

Calculate moles of nitrogen:
$$n=\frac{PV}{RT}= \frac{0.921 \times 0.06}{0.0821 \times 300}$$.

Numerator: $$0.921 \times 0.06 = 0.05526$$.
Denominator: $$0.0821 \times 300 = 24.63$$.
Thus,
$$n = \frac{0.05526}{24.63}=0.002244\ \text{mol}$$.

Molar mass of $$N_2 =28\ \text{g mol}^{-1}$$.
Mass of nitrogen present:
$$m = n \times 28 = 0.002244 \times 28 = 0.06283\ \text{g}$$.

Mass of the organic sample taken: $$0.5\ \text{g}$$.

Percentage of nitrogen in the compound:
$$\%N = \frac{0.06283}{0.5} \times 100 = 12.57\%$$.

The percentage composition of nitrogen in the compound is $$\mathbf{12.57\%}$$.
Hence, the correct choice is Option D.

The percentage of an element in an unknown organic compound can be obtained from its combustion data by first calculating the mass of that element present in the measured amounts of $$CO_2$$ and $$H_2O$$, then relating that mass to the original mass of the compound.

Mass of organic compound taken: $$0.5 \text{ g}$$.

Step 1: Find mass of carbon present
• Every mole of $$CO_2$$ contains one mole of carbon.
• Molar mass of $$CO_2$$ is $$44 \text{ g mol}^{-1}$$, and the mass of carbon in each mole is $$12 \text{ g}$$.

Therefore the mass of carbon obtained from $$1.46 \text{ g}$$ of $$CO_2$$ is
$$\text{mass of C}=1.46\; \frac{12}{44}$$

Simplifying,
$$\text{mass of C}=1.46 \times 0.2727 = 0.398 \text{ g (approximately)}$$.

Step 2: Calculate percentage of carbon
Percentage of carbon $$= \frac{\text{mass of C in product}}{\text{mass of original sample}}\times 100$$

$$\%\,\text{C}= \frac{0.398}{0.5}\times 100 = 79.6$$

Rounding to the nearest integer, the percentage of carbon is $$\mathbf{80}$$.

Hence, the required percentage of carbon in the organic compound is 80 %.

In Dumas’ method the nitrogen gas is collected over water, so the measured pressure is the total (nitrogen + water-vapour) pressure. We must first obtain the partial pressure of dry nitrogen.

Total pressure of the gas mixture, $$P_{\text{total}} = 900 \text{ mm Hg}$$
Aqueous tension at $$300 \text{ K},\; P_{\text{H}_2O} = 15 \text{ mm Hg}$$
Therefore, pressure of dry $$N_2$$ is
$$P_{N_2} = P_{\text{total}} - P_{\text{H}_2O} = 900 - 15 = 885 \text{ mm Hg}$$

Convert this pressure into atmospheres for use in the ideal-gas equation:
$$P_{N_2} = \frac{885}{760} \text{ atm} = 1.164 \text{ atm (approx)}$$

Volume of nitrogen collected,
$$V = 150 \text{ mL} = 0.150 \text{ L}$$

Temperature,
$$T = 300 \text{ K}$$

Ideal-gas constant (in compatible units),
$$R = 0.0821 \text{ L atm mol}^{-1}\text{ K}^{-1}$$

Using the ideal-gas equation $$PV = nRT$$ to find the moles of $$N_2$$:
$$n = \frac{PV}{RT} = \frac{(1.164)\,(0.150)}{(0.0821)\,(300)}$$

Calculate $$n$$:
Numerator: $$1.164 \times 0.150 = 0.1746$$
Denominator: $$0.0821 \times 300 = 24.63$$
$$n = \frac{0.1746}{24.63} = 0.00709 \text{ mol}$$

Mass of nitrogen present (each mole of $$N_2$$ weighs $$28 \text{ g}$$):
$$m_{N} = n \times 28 = 0.00709 \times 28 = 0.1985 \text{ g}$$

The organic sample weighed $$1.00 \text{ g}$$, so the percentage of nitrogen is
$$\%N = \frac{0.1985}{1.00} \times 100 = 19.85\%$$

To the nearest integer, the percentage composition of nitrogen in the compound is **20 %**.

In sulphur estimation, the organic compound is converted to $$BaSO_4$$.

Molar mass of $$BaSO_4 = 137 + 32 + 64 = 233$$ g/mol.

Mass of S in $$BaSO_4$$ = $$\frac{32}{233} \times 0.40 = \frac{12.8}{233}$$ g

Percentage of S = $$\frac{32}{233} \times \frac{0.40}{0.20} \times 100 = \frac{32 \times 200}{233} = \frac{6400}{233} = 27.47\%$$

In the form $$x \times 10^{-1}\%$$: $$27.47 = 274.7 \times 10^{-1}$$, so $$x \approx 275$$.

The answer is 275.

Methods used for purification of organic compounds (such as distillation, crystallization, chromatography, etc.) depend on both the nature of the compound being purified and the type of impurities present.

For example, distillation is used when the compound and impurity have different boiling points, while crystallization depends on solubility differences. The choice of method depends on both factors.

The correct answer is Option 1: nature of compound and presence of impurity.

The purification method based on the principle of solubility in two different solvents is Differential Extraction.

In differential extraction, a substance is partitioned between two immiscible solvents based on its different solubilities in each.

The answer is Option (4): $$\boxed{\text{Differential Extraction}}$$.

We need to identify which purification method uses the principle of adsorption.

Analysis of each method:

Extraction: This method is based on the principle of differential solubility of a compound in two immiscible solvents. It does not use adsorption.

Chromatography: This technique separates components of a mixture based on differential adsorption of the components on an adsorbent (stationary phase). The mixture is passed through a column of adsorbent (such as silica gel or alumina), and different components are adsorbed to different extents, allowing separation. Column chromatography, thin-layer chromatography (TLC), and other forms all rely on this principle.

Distillation: This method separates liquids based on differences in their boiling points. It does not involve adsorption.

Sublimation: This method separates substances based on their ability to sublime (convert directly from solid to vapor). It does not use adsorption.

The correct answer is Option (2): Chromatography.

In chromatography: polar compounds interact more strongly with the polar stationary phase (silica) and move slower, giving smaller Rf values. Non-polar compounds move faster with larger Rf.

The correct answer is Option (3): Rf of a polar compound is smaller than that of a non-polar compound.

We need to determine the correct purification technique used for a substance that is steam volatile and immiscible with water.

Statement (A): Fractional distillation. Fractional distillation is used to separate a mixture of miscible liquids whose boiling points are close to each other (typically a difference of less than $$25 \text{ K}$$). It relies on repeated vaporization and condensation within a fractionating column. This is FALSE.

Statement (B): Fractional distillation under reduced pressure. This technique (also known as vacuum distillation) is used for purifying liquids that have very high boiling points or those that tend to decompose at or below their normal boiling points. This is FALSE.

Statement (C): Distillation. Simple distillation is used to separate a volatile liquid from non-volatile impurities, or to separate two miscible liquids with a large difference in their boiling points (typically greater than $$25 \text{ K}$$). This is FALSE.

Statement (D): Steam distillation. Steam distillation is a specialized separation process specifically designed for purifying liquids that are steam volatile and completely immiscible with water. Passing steam through the mixture lowers the effective boiling temperature of the organic compound below its normal boiling point, protecting it from thermal decomposition. This matches the criteria perfectly. This is TRUE.

The appropriate technique for a steam volatile, water-immiscible substance is steam distillation.

Answer: Option D — Steam distillation

Lassaigne's test involves fusing the organic compound with sodium metal to convert covalent organic compounds into ionic compounds (NaCN for N-containing compounds).

Hydrazine (N₂H₄) does not contain carbon. In Lassaigne's test, N is detected through the formation of NaCN, which requires both N and C. Since hydrazine has no carbon, it cannot form NaCN and hence does not give a positive Lassaigne's test for nitrogen.

The correct answer is Option 4: Hydrazine.

We need to identify which chromatographic techniques are based on the principle of differential adsorption.

A. Column Chromatography:

In column chromatography, the stationary phase is a solid adsorbent (such as silica gel or alumina) packed in a column. Components of the mixture are separated based on their different degrees of adsorption on the solid stationary phase. This is based on the principle of differential adsorption.

B. Thin Layer Chromatography (TLC):

In TLC, a thin layer of adsorbent (silica gel or alumina) is coated on a glass plate. The components separate based on their different adsorption affinities for the stationary solid phase. This is also based on the principle of differential adsorption.

C. Paper Chromatography:

In paper chromatography, water trapped in the cellulose fibres of the paper acts as the stationary phase (liquid), and the mobile phase is an organic solvent. Components separate based on their different partition coefficients between the two liquid phases. This is based on the principle of differential partition, not adsorption.

Therefore, only Column Chromatography (A) and Thin Layer Chromatography (B) are based on differential adsorption.

The correct answer is Option 3: A & B only.

• Silica Gel (A): A highly porous, chemically inert form of silicon dioxide with a strong affinity for polar compounds.
• Alumina (B): Aluminium oxide, another widely used polar stationary phase available in acidic, basic, or neutral forms.
• Quick Lime (C) & Magnesia (D): Not commonly utilized as primary adsorbents in standard adsorption chromatography techniques.

Conclusion:

Silica gel and alumina are the two classic, standard solid adsorbents employed to separate mixtures based on differential adsorption.

Answer: Option D — A and B only

We need to identify the suitable method for extracting essential oils from flowers.

Key Concept: The essential oils are (i) steam-volatile organic compounds, (ii) insoluble in water at room temperature, and (iii) miscible with water vapour in the vapour phase.

Option D: Steam distillation is the correct method. In steam distillation, steam is passed through the plant material. The essential oils, being volatile, vaporize along with the steam at a temperature below their normal boiling point. The mixed vapours of water and oil are then condensed. Since the oils are immiscible with water at room temperature, the condensed mixture separates into two layers, allowing the essential oil to be easily separated. This method is specifically designed for compounds that are steam-volatile but water-insoluble—exactly matching the properties described in the question.

The correct answer is Option D.

We need to determine which statements about purification techniques are correct.

Statement A: "Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point."

Glycerol has a normal boiling point of about 290 degC, but it begins to decompose around 250 degC. Vacuum distillation lowers the boiling point by reducing the pressure, allowing glycerol to be distilled at a lower temperature without decomposition. This statement is correct. ✓

Statement B: "Aniline can be purified by steam distillation as aniline is miscible in water."

While it is true that aniline can be purified by steam distillation, the reason given is incorrect. Steam distillation works for compounds that are immiscible (or sparingly soluble) in water, not miscible. Aniline is only slightly soluble in water (about 3.6 g/100 mL at 20 degC), which is why it can be steam-distilled - the two immiscible liquids distil at a temperature below the boiling point of either component. The statement incorrectly claims aniline is miscible in water. This statement is incorrect. ✗

Statement C: "Ethanol can be separated from ethanol-water mixture by azeotropic distillation because it forms an azeotrope."

Ethanol and water form a minimum boiling azeotrope at 95.6% ethanol (by weight) at 78.1 degC. Simple distillation cannot separate them beyond this composition. Azeotropic distillation (using a third component like benzene or cyclohexane as an entrainer) can break the azeotrope and separate pure ethanol. This statement is correct. ✓

Statement D: "An organic compound is pure, if mixed M.P. is remained same."

The mixed melting point test is a standard purity test: if a sample is mixed with a pure reference compound and the melting point remains unchanged (no depression), then the sample is the same pure compound. If impurities are present, the melting point would be depressed. This statement is correct. ✓

Conclusion: Statements A, C, and D are correct.

The correct answer is Option 2: A, C, D only.

In Kjeldahl's method, we need to identify the role of $$CuSO_4$$.

Kjeldahl's method overview:

This method estimates nitrogen content in organic compounds through three steps: digestion, distillation, and titration.

Role of $$CuSO_4$$ in the digestion step:

During digestion, the organic compound is heated with concentrated $$H_2SO_4$$. The nitrogen in the organic compound is converted to ammonium sulfate $$(NH_4)_2SO_4$$. $$CuSO_4$$ acts as a catalyst in this process:

- The $$Cu^{2+}$$ ions accelerate the oxidative breakdown of the organic compound.

- They help achieve faster and more complete conversion of organic nitrogen to $$NH_4^+$$.

- Without the catalyst, the digestion would be much slower and may be incomplete.

$$K_2SO_4$$ is also added to raise the boiling point of the acid mixture, but $$CuSO_4$$ specifically serves as a catalytic agent.

The correct answer is Option B: Catalytic agent.

In organic compounds, elements such as nitrogen, sulphur, and halogens are covalently bonded to carbon and cannot be detected directly by ordinary inorganic qualitative tests.

Lassaigne's test involves fusing the organic compound with metallic sodium. During fusion, these covalently bonded elements are converted into water-soluble ionic sodium salts, which can then be identified by standard chemical tests.

Nitrogen is converted into sodium cyanide, $$NaCN$$

Sulphur is converted into sodium sulphide, $$Na_2S$$

Halogens $$(Cl,\ Br,\ I)$$ are converted into sodium halides, $$NaX$$

If both nitrogen and sulphur are present in the compound, they combine with sodium and carbon during fusion to form sodium thiocyanate, $$NaSCN$$

Thus, Lassaigne's test is specifically used for the detection of nitrogen, sulphur, and halogens by converting them into ionic sodium salts.

Therefore, the correct answer is $$\text{Nitrogen, Sulphur and Halogens}$$, thus the right option is D.

We need to evaluate both statements about the confirmatory test for sulphide ions.

Statement-I Analysis:

When a sulphide salt is warmed with dilute $$H_2SO_4$$, hydrogen sulphide gas ($$H_2S$$) is liberated:

$$Na_2S + H_2SO_4 \rightarrow Na_2SO_4 + H_2S \uparrow$$

When this $$H_2S$$ gas comes in contact with paper dipped in lead acetate solution, it forms lead sulphide ($$PbS$$), which is black in colour:

$$(CH_3COO)_2Pb + H_2S \rightarrow PbS \downarrow (\text{black}) + 2CH_3COOH$$

This is indeed a standard confirmatory test for sulphide ions. So Statement-I is true.

Statement-II Analysis:

Statement-II claims the black colour is due to the formation of lead sulphite ($$PbSO_3$$). This is incorrect. Lead sulphite is white, not black. The black precipitate formed is lead sulphide ($$PbS$$), not lead sulphite.

So Statement-II is false.

Therefore, Statement-I is true but Statement-II is false.

The correct answer is Option 3.

We need to identify which compound is white in color.

Analysis of each compound:

- Ammonium sulphide: $$(NH_4)_2S$$ is colorless in solution (may appear yellowish due to polysulfides). Not a white solid in typical form.

- Lead sulphate: $$PbSO_4$$ is a white crystalline solid. It is an insoluble white precipitate commonly encountered in qualitative analysis.

- Lead iodide: $$PbI_2$$ is yellow in color (bright yellow precipitate).

- Ammonium arsinomolybdate: This is yellow in color (canary yellow precipitate, used in testing for arsenic).

The correct answer is Option (2): lead sulphate.

In the Lassaigne's test for halogens, the sodium fusion extract may contain $$Na_2S$$ (if sulphur is present) and $$NaCN$$ (if nitrogen is present) along with $$NaX$$ (sodium halide).

When silver nitrate is added to test for halogens, both $$Na_2S$$ and $$NaCN$$ would also react with $$AgNO_3$$:

$$Na_2S + 2AgNO_3 \rightarrow Ag_2S \downarrow (\text{black}) + 2NaNO_3$$

$$NaCN + AgNO_3 \rightarrow AgCN \downarrow (\text{white}) + NaNO_3$$

These precipitates ($$Ag_2S$$ and $$AgCN$$) would interfere with the detection of silver halides.

To avoid this interference, the Lassaigne's extract is first boiled with dilute $$HNO_3$$. The dilute nitric acid decomposes $$Na_2S$$ and $$NaCN$$:

$$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$$

$$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$$

The gases $$H_2S$$ and $$HCN$$ escape on boiling, eliminating the interference. After this treatment, only the halide ions remain in solution and can be detected cleanly with $$AgNO_3$$.

The correct answer is Option 4: $$Na_2S$$ and $$NaCN$$ are decomposed by $$HNO_3$$.

Find the percentage of nitrogen using Kjeldahl's method.

Volume = 10 mL = 0.01 L, Molarity = 2 M.

$$ \text{Moles of } H_2SO_4 = 2 \times 0.01 = 0.02 \text{ mol} $$

The reaction is: $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$

1 mol $$H_2SO_4$$ neutralises 2 mol $$NH_3$$.

$$ \text{Moles of } NH_3 = 2 \times 0.02 = 0.04 \text{ mol} $$

Each mole of $$NH_3$$ contains 1 mole of N (atomic mass = 14 g/mol).

$$ \text{Mass of N} = 0.04 \times 14 = 0.56 \text{ g} $$

$$ \% N = \frac{\text{Mass of N}}{\text{Mass of compound}} \times 100 = \frac{0.56}{1} \times 100 = 56\% $$

The answer is 56%.

Step 1: Find the Distance Traveled by the Solvent Front

From the provided TLC diagram:

• Total height of the plate: $$10\text{ cm}$$
• Distance from bottom edge to baseline: $$1\text{ cm}$$
• Distance from top edge to solvent front: $$1\text{ cm}$$

The total distance traveled by the solvent from the baseline is:

$$\text{Distance of solvent front} = 10\text{ cm} - 1\text{ cm} - 1\text{ cm} = 8\text{ cm}$$

Step 2: Calculate the $$R_f$$ Values for Spots A and B

The $$R_f$$ value is calculated as the distance traveled by the compound from the baseline divided by the distance traveled by the solvent front from the baseline.

• For Spot A:

  Distance from bottom edge = $$5\text{ cm}$$, so distance from baseline = $$5\text{ cm} - 1\text{ cm} = 4\text{ cm}$$.

  $$R_f(A) = \frac{4\text{ cm}}{8\text{ cm}} = 0.5$$



• For Spot B:

  Distance from bottom edge = $$7\text{ cm}$$, so distance from baseline = $$7\text{ cm} - 1\text{ cm} = 6\text{ cm}$$.

  $$R_f(B) = \frac{6\text{ cm}}{8\text{ cm}} = 0.75$$

Step 3: Solve for $$x$$

The problem states that the $$R_f$$ value of B is a multiple of A expressed as:

$$R_f(B) = (x \times 10^{-1}) \times R_f(A)$$

Substitute the calculated $$R_f$$ values into the relation:

$$0.75 = (x \times 10^{-1}) \times 0.5$$

$$x \times 10^{-1} = \frac{0.75}{0.5} = 1.5$$

$$x = 1.5 \times 10 = 15$$

Conclusion:

Comparing the relative positions shows that the $$R_f$$ value of B is exactly $$1.5$$ times that of A, which gives a value of $$15$$ for $$x$$.

Answer: 15

$$R_f = \frac{\text{Distance traveled by the substance from the base line}}{\text{Distance traveled by the solvent from the base line}}$$

• For Sample A:
  $$R_f(A) = \frac{5.0}{12.5}$$
• For Sample C:
  $$R_f(C) = \frac{10.0}{12.5}$$

$$\text{Ratio} = \frac{R_f(A)}{R_f(C)} = \frac{\frac{5.0}{12.5}}{\frac{10.0}{12.5}} = \frac{5.0}{10.0} = 0.5$$

We are given that the ratio is equal to $$x \times 10^{-2}$$:

$$0.5 = x \times 10^{-2}$$ 

$$x = \frac{0.5}{10^{-2}} = 0.5 \times 10^2 = 50$$

The retardation factor ($$R_f$$) in thin layer chromatography is defined as:

$$ R_f = \frac{\text{Distance moved by compound}}{\text{Distance moved by solvent}} $$

Given:

Distance moved by compound = $$3.5$$ cm

Distance moved by solvent = $$5$$ cm

Substituting:

$$ R_f = \frac{3.5}{5} = 0.7 $$

Expressing in the required form:

$$ R_f = 7 \times 10^{-1} $$

Therefore, the answer is $$\boxed{7}$$.

Assertion (A): An aqueous solution of KOH used in volumetric analysis should be standardized or its concentration should be checked before use.

Reason (R): KOH solution absorbs atmospheric $$CO_2$$ on standing.

The reason statement is correct because potassium hydroxide reacts readily with carbon dioxide present in air.

$$2KOH+CO_2\rightarrow K_2CO_3+H_2O$$

As a result, part of the KOH is converted into potassium carbonate, causing the concentration of the KOH solution to change with time.

The assertion is also correct. KOH is not a primary standard because it absorbs both moisture and $$CO_2$$ from the atmosphere. Therefore, the concentration calculated during preparation does not remain constant for long periods.

Before using KOH solution in volumetric analysis, its exact concentration must be determined by standardization against a suitable primary standard.

Thus, the concentration of KOH solution is checked before use because it absorbs atmospheric $$CO_2$$ and undergoes the reaction

$$2KOH+CO_2\rightarrow K_2CO_3+H_2O$$

Therefore, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct explanation of Assertion (A).

A - (ii), B - (iii), C - (iv), D - (i)

In qualitative analysis of organic compounds (Lassaigne's test):

A. Nitrogen detection: The sodium fusion extract containing NaCN is treated with FeSO₄ and then with dilute H₂SO₄. The formation of Prussian blue colour confirms nitrogen. The product is Fe₄[Fe(CN)₆]₃ → Match with III.

B. Sulphur detection: The sodium fusion extract containing Na₂S is treated with sodium nitroprusside Na₂[Fe(CN)₅NO]. A violet colour confirms sulphur → Match with I.

C. Phosphorus detection: The compound is heated with an oxidizing agent, dissolved in HNO₃, and treated with ammonium molybdate (NH₄)₂MoO₄. A yellow precipitate of ammonium phosphomolybdate confirms phosphorus → Match with IV.

D. Halogen detection: The sodium fusion extract is acidified with HNO₃ and treated with AgNO₃. The formation of a precipitate (AgCl, AgBr, or AgI) confirms halogens → Match with II.

The correct matching is: A → III, B → I, C → IV, D → II.

In paper chromatography:

• The cellulose paper contains water molecules trapped in its pores
• This water acts as the stationary phase
• The mobile phase is an organic solvent or a solvent mixture that moves up the paper by capillary action

The water present in the pores of the paper forms the stationary phase.

This matches option 1.

Lassaigne's test detects nitrogen and halogens in organic compounds by fusing the compound with sodium metal. The key reactions are:

For nitrogen: $$\text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN}$$

For halogens: $$\text{Na} + \text{X} \rightarrow \text{NaX}$$

For the test to detect nitrogen, the compound must contain carbon (to form NaCN).

Let us examine each option:

Option A: $$\text{N}_2\text{H}_4 \cdot \text{HCl}$$ (Hydrazine hydrochloride) — Contains N and Cl, but no carbon. Without carbon, NaCN cannot form, so nitrogen detection fails.

Option B: $$\text{CH}_3\text{NH}_2 \cdot \text{HCl}$$ (Methylamine hydrochloride) — Contains C, N, and Cl. Carbon is present, so NaCN forms (positive nitrogen test). Cl is present, so NaCl forms (positive halogen test). Both tests are positive.

Option C: $$\text{NH}_4\text{Cl}$$ — Contains N and Cl, but no carbon. Nitrogen test fails.

Option D: $$\text{NH}_2\text{OH} \cdot \text{HCl}$$ (Hydroxylamine hydrochloride) — Contains N and Cl, but no carbon. Nitrogen test fails.

The answer is Option B: $$\text{CH}_3\text{NH}_2 \cdot \text{HCl}$$.

1.  TLC Plate (Retention Factor)

• On a standard silica gel TLC plate, the stationary phase (silica) is polar, and the solvent (mobile phase) is non-polar/less polar.
• More polar compounds stick strongly to the silica and travel a shorter distance (lower spot).
• Less polar compounds move easily with the solvent and travel a longer distance (higher spot).

Looking at the heights of the spots on your TLC plate:

• A travelled the highest hence Least polar
• C travelled to the middle hence Moderately polar
• B stayed the lowest hence Most polar

2. Column Chromatography Elution Order

Column chromatography works on the exact same principle, but the compounds drain out from the bottom of the column one by one (elute):

• The least polar compound interacts the least with the silica gel and rushes out first.
• The most polar compound binds tightly to the silica gel and comes out last.

Therefore, the order of elution from first to last is:

A,C ,B

Correct Answer:

Option C: A, C, B

In column chromatography, the component that elutes first has weaker adsorption to the stationary phase (adsorbent).

Since A eluted first, A has weaker adsorption. This means B has stronger adsorption (retained longer on the column).

In terms of R$$_f$$ value (from TLC): A would have a higher R$$_f$$ (moves further) and B would have a lower R$$_f$$ (moves less, stronger adsorption).

The correct answer is Option 1: low R$$_f$$, stronger adsorption.

We need to identify organic compound X that, when treated with sodium peroxide in a Carius tube, forms a mineral acid Y, which on addition of $$BaCl_2$$ gives a precipitate Z used for quantitative estimation.

In the Carius method for elemental analysis, the organic compound is heated with a strong oxidizing agent (sodium peroxide, $$Na_2O_2$$) in a sealed tube. Different elements are converted to different products:

- Sulfur (S) is oxidized to sulfate ($$SO_4^{2-}$$)

- Halogens are converted to halide ions

- Phosphorus is converted to phosphate

The mineral acid Y is formed from the oxidation product. When $$BaCl_2$$ is added to Y and forms precipitate Z, the precipitate must be $$BaSO_4$$ (barium sulfate), because:

$$BaSO_4$$ is a white precipitate that is insoluble in water and dilute acids. This is the standard test for sulfate ions, and it is used for quantitative estimation of sulfur in organic compounds. The mineral acid Y is therefore $$H_2SO_4$$ (sulfuric acid).

Identify compound X (must contain sulfur).

Among the options:

(A) Cytosine - $$C_4H_5N_3O$$ - contains C, H, N, O but NO sulfur.

(B) A nucleotide - nucleotides contain C, H, N, O, P but typically NO sulfur.

(C) Methionine - $$C_5H_{11}NO_2S$$ - this is an amino acid that CONTAINS SULFUR.

(D) Chloroxylenol - $$C_8H_9ClO$$ - contains C, H, Cl, O but NO sulfur.

Only methionine contains sulfur. When methionine is treated with $$Na_2O_2$$ in a Carius tube, the sulfur is oxidized to $$Na_2SO_4$$, which on acidification gives $$H_2SO_4$$ (mineral acid Y). Adding $$BaCl_2$$ gives $$BaSO_4$$ precipitate (Z), which is weighed for quantitative estimation of sulfur.

The correct answer is Option 3: Methionine.

A. Reverberatory furnace is used for copper smelting (IV)

B. Electrolytic cell (Hall-Heroult process) is used for aluminium extraction (II)

C. Blast furnace is used for pig iron production (I)

D. Zone refining is used for purification of silicon and germanium (III)

Matching: A-IV, B-II, C-I, D-III.

We need to identify the color change when SO$$_2$$ gas is passed through acidified K$$_2$$Cr$$_2$$O$$_7$$ solution.

Reaction:

SO$$_2$$ is a reducing agent. It reduces the orange dichromate ion (Cr$$_2$$O$$_7^{2-}$$, where Cr is in +6 state) to green chromic ion (Cr$$^{3+}$$):

$$\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + 3\text{SO}_2 \to \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$$

The Cr$$^{3+}$$ ions in solution give a green color.

This color change from orange to green is a confirmatory test for the presence of SO$$_3^{2-}$$ ions (which produce SO$$_2$$ on treatment with acid).

The correct answer is Option 3: Green.

Let us check each pair for mismatch:

Zinc - Liquation: Liquation is a method used for purifying metals with low melting points. Zinc is typically purified by electrolysis or distillation, not by liquation. Liquation is used for tin, bismuth, and lead. So this is a mismatch.

Copper - Electrolysis: Copper is commonly refined by electrolytic refining. This is correct.

Nickel - Mond process: The Mond process (using Ni(CO)₄) is used for purifying nickel. This is correct.

Titanium - van Arkel Method: The van Arkel method (iodide decomposition) is used for purifying titanium and zirconium. This is correct.

The mismatch is Option 1: Zinc - Liquation.

Match the cations with their group reactions in qualitative analysis.

Group II (H$$_2$$S in dilute HCl): Pb$$^{2+}$$, Cu$$^{2+}$$, Bi$$^{3+}$$, Cd$$^{2+}$$, As$$^{3+}$$, Sb$$^{3+}$$, Sn$$^{2+}$$

So P (Pb$$^{2+}$$, Cu$$^{2+}$$) → H$$_2$$S gas in presence of dilute HCl = i

Group III (NH$$_4$$OH in presence of NH$$_4$$Cl): Al$$^{3+}$$, Fe$$^{3+}$$, Cr$$^{3+}$$

So Q (Al$$^{3+}$$, Fe$$^{3+}$$) → NH$$_4$$OH in presence of NH$$_4$$Cl = iii

Group IV (H$$_2$$S in presence of NH$$_4$$OH): Co$$^{2+}$$, Ni$$^{2+}$$, Mn$$^{2+}$$, Zn$$^{2+}$$

So R (Co$$^{2+}$$, Ni$$^{2+}$$) → H$$_2$$S in presence of NH$$_4$$OH = iv

Group V ((NH$$_4$$)$$_2$$CO$$_3$$ in presence of NH$$_4$$OH): Ba$$^{2+}$$, Ca$$^{2+}$$, Sr$$^{2+}$$

So S (Ba$$^{2+}$$, Ca$$^{2+}$$) → (NH$$_4$$)$$_2$$CO$$_3$$ in presence of NH$$_4$$OH = ii

Matching: P→i, Q→iii, R→iv, S→ii

The correct answer is Option D: P→i, Q→iii, R→iv, S→ii.

In the wet test (qualitative analysis), Ba²⁺ cations belong to Group V (alkaline earth metals).

Ba²⁺ is detected by precipitation as barium carbonate (BaCO₃) using ammonium carbonate as the group reagent:

$$Ba^{2+} + (NH_4)_2CO_3 \to BaCO_3 \downarrow + 2NH_4^+$$

BaCO₃ is a white precipitate.

Note: While BaSO₄ is also an insoluble precipitate, in the systematic qualitative analysis scheme, Ba²⁺ is detected using carbonate precipitation in Group V analysis.

This matches option 2: BaCO₃.

We need to identify which transition element cation does NOT belong to Group IV in qualitative inorganic analysis.

Group IV cations are precipitated by H$$_2$$S in the presence of NH$$_4$$Cl and NH$$_4$$OH (basic medium). This group includes: Co$$^{2+}$$, Ni$$^{2+}$$, Mn$$^{2+}$$, and Zn$$^{2+}$$.

Fe$$^{3+}$$ belongs to Group III, where cations are precipitated by NH$$_4$$OH + NH$$_4$$Cl (as hydroxides). Fe$$^{3+}$$ precipitates as Fe(OH)$$_3$$ (brown precipitate) in Group III.

Among the given options:

- Fe$$^{3+}$$ → Group III (NOT Group IV) ✓

- Zn$$^{2+}$$ → Group IV

- Co$$^{2+}$$ → Group IV

- Ni$$^{2+}$$ → Group IV

The correct answer is Option 1: Fe$$^{3+}$$.

In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate (BaSO$$_4$$).

Molar mass of BaSO$$_4$$ = 137 + 32 + 64 = 233 g/mol.

$$\text{Mass of S} = \frac{32}{233} \times 1.4439 = \frac{46.2048}{233} = 0.19831 \text{ g}$$

$$\% \text{S} = \frac{0.19831}{0.471} \times 100 = 42.10\%$$

Rounding to the nearest integer: $$\approx 42\%$$.

The percentage of sulphur in the compound is $$\mathbf{42}$$.

1. Points of TLC

• Stationary Phase: Usually a polar substance like silica gel attached to the TLC plate.
• Mobile Phase: The solvent used to move the compounds. In this case, hexane is used, which is a highly non-polar solvent.
• Polarity vs. Distance: Since the stationary phase is polar and the mobile phase (hexane) is non-polar, a more polar compound will adhere more strongly to the plate and travel a shorter distance. A less polar (more non-polar) compound will dissolve better in hexane and travel further up the plate.

2. Identifying the Most Polar Compound

Looking at the distances traveled by each compound from the baseline:

• Distance traveled by Compound A = 6 cm
• Distance traveled by Compound B = 4 cm
• Distance traveled by Compound C = 2 cm

Since Compound C traveled the shortest distance (2 cm), it interacted the strongest with the polar stationary phase. Therefore, Compound C is the most polar compound.

3. Calculating the Rf Value

The Retardation Factor (Rf) is calculated using the formula:

From the given image:

• Distance traveled by Compound C = 2 cm
• Total distance traveled by the solvent front = 8 cm

Plugging these values into the formula:

4. Matching the Question Format

We can convert 0.25 into scientific notation matching this layout:

Correct Answer: 25

In the Carius method, the organic bromine is converted to AgBr.

Moles of AgBr formed:

$$n_{AgBr} = \frac{0.376}{188} = 0.002 \text{ mol}$$

Since 1 mol AgBr contains 1 mol Br:

Mass of Br = $$0.002 \times 80 = 0.16$$ g

Percentage of bromine:

$$\% Br = \frac{0.16}{0.400} \times 100 = 40\%$$

The percentage of bromine is $$\mathbf{40}$$%.

The retardation factor ((R_f)) in thin layer chromatography is defined as

$$R_f=\frac{\text{Distance travelled by the compound}}{\text{Distance travelled by the solvent front}}.$$

The compound travels a distance of

$$3.5\ \text{cm}.$$

The solvent front is (0.5\ \text{cm}) below the top edge of the plate, while the baseline is (0.5\ \text{cm}) above the bottom edge. Hence, the total distance travelled by the solvent front is

$$6-0.5-0.5=5.0\ \text{cm}.$$

Therefore,

$$R_f=\frac{3.5}{5.0}=0.7.$$

Expressing the value in the form (x\times10^{-1}),

$$0.7=7\times10^{-1}.$$

Hence,

$$\boxed{R_f=7\times10^{-1}},$$

so the required value is

$$\boxed{7}.$$

We need to identify which technique is not used to spot (visualize) components on a thin layer chromatographic (TLC) plate.

Common Techniques for Visualizing TLC Spots:

Option A: $$I_2$$ (Solid) — Used

Iodine vapour chamber is a widely used technique. The TLC plate is placed in a closed chamber with iodine crystals. The iodine vapour gets adsorbed by the organic compounds, forming brown spots.

Option B: U.V. Light — Used

Many TLC plates contain a fluorescent indicator. Under UV light (254 nm or 366 nm), the spots appear as dark patches against a fluorescent background.

Option C: Visualisation agent as a component of mobile phase — Not Used

The mobile phase (solvent) is used to separate the components by carrying them up the plate. The visualization agent is never added as part of the mobile phase because it would interfere with the separation process. Visualization is done after the separation is complete.

Option D: Spraying of an appropriate reagent — Used

Spraying reagents like ninhydrin (for amino acids), sulfuric acid, or other specific reagents onto the plate is a standard technique to make invisible spots visible.

Hence, the correct answer is Option C: Visualisation agent as a component of mobile phase.

We need to identify the formula of the purple-coloured compound formed in Lassaigne's test for sulphur using sodium nitroprusside. In this test the organic compound is fused with sodium metal to form sodium sulphide $$Na_2S$$ if sulphur is present.

$$2Na + S \rightarrow Na_2S$$

The sodium extract is then treated with sodium nitroprusside solution $$Na_2[Fe(CN)_5(NO)]$$ and the sulphide ions react to form a purple-coloured complex:

$$Na_2S + Na_2[Fe(CN)_5(NO)] \rightarrow Na_4[Fe(CN)_5(NOS)]$$

In this reaction, the $$S^{2-}$$ ion reacts with the $$NO^+$$ group in nitroprusside to form the thionitrosyl group $$NOS^-$$, resulting in the purple-coloured complex $$Na_4[Fe(CN)_5(NOS)]$$.

Option A: $$Na_4[Fe(CN)_5(NOS)]$$ — This is the correct formula of the purple complex formed in Lassaigne's test for sulphur. Option B: $$NaFe[Fe(CN)_6]$$ — This is Turnbull's blue, not related to the sulphur test. Option C: $$Na[Cr(NH_3)_2(NCS)_4]$$ — This is Reinecke's salt. Option D: $$Na_2[Fe(CN)_5(NO)]$$ — This is sodium nitroprusside itself (the reagent, not the product).

Hence, the correct answer is Option A: $$Na_4[Fe(CN)_5(NOS)]$$.

We need to evaluate two statements about Lassaigne's Test when both nitrogen and sulphur are present in an organic compound.

Evaluate Statement I.

Statement I: When both nitrogen and sulphur are present in an organic compound, sodium thiocyanate (NaSCN) is formed during sodium fusion.

This is correct. When an organic compound containing both N and S is fused with sodium metal:

$$Na + C + N + S \rightarrow NaSCN$$

Sodium thiocyanate is formed instead of NaCN and Na$$_2$$S separately.

Evaluate Statement II.

Statement II: Excess sodium decomposes sodium thiocyanate to give NaCN and Na$$_2$$S.

This is correct. When excess sodium is used in the fusion:

$$NaSCN + 2Na \rightarrow NaCN + Na_2S$$

The excess sodium breaks down the thiocyanate, allowing both nitrogen (as NaCN) and sulphur (as Na$$_2$$S) to be detected separately in subsequent tests. This is why excess sodium is recommended when both elements are suspected to be present.

Conclusion.

Both Statement I and Statement II are correct.

The correct answer is Option A: Both Statement I and Statement II are correct.

Analyzing the Assertion:

"A mixture contains benzoic acid and naphthalene. The pure benzoic acid can be separated out by the use of benzene."

Both benzoic acid and naphthalene are soluble in benzene. So benzene cannot be used to selectively dissolve one and leave the other behind. The standard method to separate benzoic acid from naphthalene would be to use a sodium bicarbonate solution (which dissolves benzoic acid as sodium benzoate, leaving naphthalene behind) or to use hot water (benzoic acid is soluble in hot water, naphthalene is not).

Therefore, the Assertion is false.

Analyzing the Reason:

"Benzoic acid is soluble in hot water."

This is a well-known fact. Benzoic acid has limited solubility in cold water but dissolves readily in hot water. This is the basis for its purification by recrystallization.

Therefore, the Reason is true.

Hence, the correct answer is Option D: Assertion is false but Reason is true.

We need to match each mixture with its appropriate purification process.

A. Chloroform & Aniline → III. Distillation

Chloroform (boiling point ≈ 61°C) and aniline (boiling point ≈ 184°C) are both liquids that are miscible with each other. Since there is a large difference in their boiling points (about 123°C), they can be easily separated by simple distillation.

B. Benzoic acid & Naphthalene → IV. Crystallisation

Both benzoic acid and naphthalene are solids. Benzoic acid is soluble in hot water and crystallises out upon cooling, while naphthalene is insoluble in water. So benzoic acid can be separated by dissolving the mixture in hot water, filtering to remove naphthalene, and then crystallising benzoic acid from the filtrate.

C. Water & Aniline → I. Steam distillation

Aniline is sparingly soluble in water and is steam volatile. Steam distillation is used to separate aniline from water, as aniline distils over with steam at a temperature below its normal boiling point.

D. Naphthalene & Sodium chloride → II. Sublimation

Naphthalene sublimes (converts directly from solid to vapor) upon heating, while sodium chloride does not. So heating the mixture causes naphthalene to sublime, leaving behind sodium chloride.

The correct matching is: A-III, B-IV, C-I, D-II

The correct answer is Option D.

We need to find the most appropriate technique for separating a mixture of $$100 \text{ mg}$$ of $$p$$-nitrophenol and picric acid.

The total quantity of the mixture is $$100 \text{ mg}$$, which is a very small amount (milligram scale).

Steam distillation is used to separate volatile compounds that are immiscible with water. While $$p$$-nitrophenol is steam volatile, this technique is generally used for larger quantities and is not ideal for milligram-scale separations.

A 2-5 ft long column of silica gel is used for preparative column chromatography. This technique is suitable for separating gram-scale quantities (typically 1-100 g). For just $$100 \text{ mg}$$, using such a long column would result in significant loss of sample and poor separation efficiency.

Sublimation is used when one component sublimes and the other does not. Neither $$p$$-nitrophenol nor picric acid readily sublimes under normal conditions, so this technique is not appropriate here.

Preparative TLC is specifically designed for separating small quantities of compounds, typically in the range of $$1 \text{ mg}$$ to $$100 \text{ mg}$$. The mixture is spotted on a thick TLC plate, developed in a suitable solvent, and the separated bands are scraped off and extracted. This is the most appropriate technique for $$100 \text{ mg}$$ of a mixture.

Since the quantity is only $$100 \text{ mg}$$, preparative TLC is the most suitable technique for this separation.

Hence, the correct answer is Option D: Preparative TLC (Thin Layer Chromatography).

We are given an assertion-reason question about Thin Layer Chromatography (TLC).

Assertion A: Thin layer chromatography is an adsorption chromatography.

This is true. In TLC, the separation of components occurs based on the principle of adsorption. The components of the mixture are adsorbed to different extents on the stationary phase (the thin layer of adsorbent), and they move at different rates when a solvent (mobile phase) moves up the plate by capillary action.

Reason R: A thin layer of silica gel is spread over a glass plate of suitable size in TLC which acts as an adsorbent.

This is also true. In TLC, a thin layer (about 0.2 mm thick) of an adsorbent such as silica gel ($$SiO_2 \cdot xH_2O$$) or alumina ($$Al_2O_3$$) is coated uniformly on a glass plate. This layer acts as the stationary phase and serves as the adsorbent.

Now, the reason directly explains why TLC is classified as adsorption chromatography — it is because the stationary phase (silica gel layer) functions as an adsorbent, and separation occurs through differential adsorption of components on this layer. So R is the correct explanation of A.

Hence, the correct answer is Option A.

In the Carius method for estimation of bromine, the organic compound is heated with fuming $$HNO_3$$ in the presence of $$AgNO_3$$. The bromine in the compound is converted to $$AgBr$$.

Given:

Mass of organic compound = $$0.45 \text{ g}$$

Mass of $$AgBr$$ formed = $$0.36 \text{ g}$$

Molar mass of $$AgBr = 188 \text{ g mol}^{-1}$$

Atomic mass of $$Br = 80 \text{ g mol}^{-1}$$

$$ \text{Moles of } AgBr = \frac{0.36}{188} $$

Since 1 mole of $$AgBr$$ contains 1 mole of $$Br$$:

$$ \text{Moles of } Br = \frac{0.36}{188} $$

$$ \text{Mass of } Br = \frac{0.36}{188} \times 80 = \frac{28.8}{188} = 0.15319 \text{ g} $$

$$ \% \text{ of } Br = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 $$

$$ \% \text{ of } Br = \frac{0.15319}{0.45} \times 100 = 34.04\% $$

The correct answer is Option A: $$34.04\%$$.

1. Identify the Major Product R

When 4-bromobenzyl alcohol is treated with red phosphorus and bromine (red P / Br₂), it acts as a brominating agent (PBr₃ formed in situ). This reagent selectively replaces the aliphatic hydroxyl group (-OH) with a bromine atom (-Br). The aromatic ring-bound bromine remains unaffected.

• Starting material: 4-bromobenzyl alcohol (C₇H₇OBr)
• Major product R: 4-bromobenzyl bromide
• Molecular formula of R: C₇H₆Br₂

2. Calculate the Molar Mass of Compound R (C₇H₆Br₂)

Using the given atomic masses:

• Carbon (C): 7 atoms × 12 = 84
• Hydrogen (H): 6 atoms × 1 = 6
• Bromine (Br): 2 atoms × 80 = 160

Molar mass of R = 84 + 6 + 160 = 250 g/mol

3. Analysis of Carius Method for Estimation of Bromine

In the Carius method, all bromine atoms present in the organic compound are converted quantitatively into Silver Bromide (AgBr) precipitates.

1 mole of C₇H₆Br₂ contains 2 moles of Br atoms, which will yield 2 moles of AgBr.

• Molar mass of AgBr = Mass of Ag + Mass of Br = 108 + 80 = 188 g/mol

4. Stoichiometric Calculation

From the relationship established above:

• 1 mole of R (250 g) gives 2 moles of AgBr (2 × 188 g = 376 g)

Now, we calculate the amount of AgBr formed from 1.00 g of R:

• Mass of AgBr formed = (376 g / 250 g) × 1.00 g
• Mass of AgBr formed = 1.504 g

Rounding off to two decimal places as shown in standard numerical formats:

• Mass of AgBr formed = 1.50 g

In the Lassaigne's test (sodium fusion extract test) for halogens, the sodium fusion extract is boiled with concentrated $$HNO_3$$ before adding $$AgNO_3$$.

Purpose:

During sodium fusion, if nitrogen or sulphur is present in the organic compound, sodium cyanide ($$NaCN$$) or sodium sulphide ($$Na_2S$$) may form. These interfere with the silver nitrate test for halogens because:

$$NaCN + AgNO_3 \to AgCN \downarrow$$ (white precipitate)

$$Na_2S + 2AgNO_3 \to Ag_2S \downarrow$$ (black precipitate)

These precipitates would give false positive results for halogens. Boiling the extract with concentrated $$HNO_3$$ decomposes these interfering compounds:

$$NaCN + HNO_3 \to NaNO_3 + HCN \uparrow$$

$$Na_2S + 2HNO_3 \to 2NaNO_3 + H_2S \uparrow$$

The volatile $$HCN$$ and $$H_2S$$ escape, leaving only the halide ions in solution for accurate detection with $$AgNO_3$$.

The correct answer is Option B: to decompose cyanide or sulphide of sodium.

We need to identify the anion in an inorganic salt based on the described test results. When BaCl₂ is added to the water extract of the salt, a white precipitate forms, indicating the presence of an anion that forms an insoluble barium salt. Possible candidates are SO₄²⁻ (BaSO₄ — white), SO₃²⁻ (BaSO₃ — white) and CO₃²⁻ (BaCO₃ — white). The white precipitate dissolves in dilute HCl with the release of a gas X having a characteristic odour. BaSO₄ does not dissolve in dilute HCl, so SO₄²⁻ is ruled out. BaSO₃ dissolves in dilute HCl releasing SO₂, which has a suffocating, pungent odour: $$\text{BaSO}_3 + 2\text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} + \text{SO}_2 \uparrow$$ BaCO₃ also dissolves in HCl but releases CO₂, which is odourless.

The evolution of a gas with a characteristic pungent odour thus confirms the presence of the SO₃²⁻ anion. The answer is Option B: SO₃²⁻.

We need to match each anion with the gas evolved when it reacts with dilute $$H_2SO_4$$.

(A) $$CO_3^{2-}$$ reacts with $$H_2SO_4$$ to give carbon dioxide: $$CO_3^{2-} + H_2SO_4 \rightarrow CO_2 + H_2O + SO_4^{2-}$$. Carbon dioxide is a colourless gas that is evolved with brisk effervescence and turns lime water milky: $$CO_2 + Ca(OH)_2 \rightarrow CaCO_3\downarrow + H_2O$$. Therefore it matches with (IV).

(B) $$S^{2-}$$ reacts with $$H_2SO_4$$ to give hydrogen sulfide: $$S^{2-} + H_2SO_4 \rightarrow H_2S + SO_4^{2-}$$. Hydrogen sulfide is a colourless gas with a rotten egg smell and turns lead acetate paper black: $$Pb(CH_3COO)_2 + H_2S \rightarrow PbS\downarrow(\text{black}) + 2CH_3COOH$$. Therefore it matches with (I).

(C) $$SO_3^{2-}$$ reacts with $$H_2SO_4$$ to give sulfur dioxide: $$SO_3^{2-} + H_2SO_4 \rightarrow SO_2 + H_2O + SO_4^{2-}$$. Sulfur dioxide is a colourless gas with a pungent smell; it acts as a reducing agent and turns acidified potassium dichromate solution green: $$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3(\text{green}) + H_2O$$. Therefore it matches with (II).

(D) $$NO_2^-$$ reacts with $$H_2SO_4$$ to form nitrous acid which decomposes to nitrogen dioxide: $$NO_2^- + H_2SO_4 \rightarrow HNO_2 \rightarrow NO_2(\text{brown fumes})$$. The brown fumes oxidize iodide ions to iodine, turning starch solution blue: $$2NO_2 + 2KI \rightarrow 2KNO_2 + I_2$$ and $$I_2 + \text{starch} \rightarrow \text{blue colour}$$. Therefore it matches with (III).

The correct answer is Option D: A - IV, B - I, C - II, D - III.

A separating funnel is used to separate two immiscible liquids having different densities.

When the mixture is allowed to stand undisturbed, the heavier liquid settles at the bottom while the lighter liquid forms the upper layer.

Option $$\mathrm{A}$$ represents the correct structure of a separating funnel.

It contains a stopper at the top, allowing the funnel to be closed and shaken safely during liquid-liquid extraction.

It also contains a stopcock at the bottom, which allows controlled removal of the denser lower layer.

The tapered shape helps produce a clear interface between the two liquid layers, enabling precise separation. Thus, the right option is A.

1. Given Data:

• Mass of the organic compound (W) = 0.166 g
• Initial volume of H2​SO4​ (Vinitial​) = 50.0 mL Filo
• Normality of H2​SO4​ (N1​) = 0.5 N Competishun
• Volume of NaOH solution (V2​) = 30.0 mL Filo
• Normality of NaOH solution (N2​) = 0.25 N Filo

2. Concept and Formula:

In Kjeldahl's method, the mass percentage of nitrogen (%N) in an organic compound is given by the formula:

%N=Mass of the organic compound (in g)1.4×Milliequivalents of acid consumed by NH3​​

Where:

Milliequivalents (meq)=Normality (N)×Volume (in mL)

3. Calculation of Milliequivalents:

• Total initial milliequivalents of H2​SO4​:meqinitial​=50.0 mL×0.5 N=25.0 meq
• Milliequivalents of NaOH used in titration:meqNaOH​=30.0 mL×0.25 N=7.5 meq

Note on Interpretation:

If we consider the 7.5 meq to be the leftover unreacted acid, then the acid consumed by ammonia would be:

25.0 meq−7.5 meq=17.5 meq

Substituting this into the percentage formula yields:

%N=0.1661.4×17.5​≈147.6%

Since a mass percentage greater than 100% is physically impossible, the terminology "the used acid required" indicates that the milliequivalents of acid effectively consumed by the ammonia gas corresponds directly to the titration value:

Milliequivalents of acid consumed by NH3​=meqNaOH​=7.5 meq

4. Calculating the Mass Percentage of Nitrogen:

Substituting the values into the Kjeldahl's formula:

%N=0.1661.4×7.5​%N=0.16610.5​≈63.25%

Rounding to the nearest integer, we get:

63%

Final Answer:

The mass percentage of nitrogen in the organic compound is 63% (or 63.25%).

We need to find the percentage of nitrogen in the compound using Kjeldahl's method. The mass of the organic compound is 0.55 g, and 12.5 mL (0.0125 L) of H$$_2$$SO$$_4$$ solution of molarity 1 M was used to neutralise the ammonia. Thus, the moles of H$$_2$$SO$$_4$$ used are 1 × 0.0125 = 0.0125 mol.

The reaction of ammonia with sulfuric acid is $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$, so 1 mole of H$$_2$$SO$$_4$$ reacts with 2 moles of NH$$_3$$ and the moles of NH$$_3$$ that reacted are 2 × 0.0125 = 0.025 mol. Each mole of NH$$_3$$ contains 1 mole of nitrogen, giving 0.025 mol of nitrogen whose mass is 0.025 × 14 = 0.35 g. Therefore, the percentage of nitrogen is $$\% N = \frac{0.35}{0.55} \times 100 = 63.6\% \approx 64\%$$. The percentage of nitrogen in the compound is 64%.

We are given that 0.2 g of an organic compound produces $$N_2$$ gas with volume 22.400 mL at STP using the Dumas method.

Since the volume of $$N_2$$ must be expressed in litres, we have:

$$ V_{N_2} = 22.400 \text{ mL} = 0.02240 \text{ L} $$

Substituting this into the molar volume relation at STP (22.4 L) gives:

$$ n_{N_2} = \frac{V}{V_m} = \frac{0.02240}{22.4} = 0.001 \text{ mol} $$

This yields the mass of nitrogen as:

$$ m_{N_2} = n \times M = 0.001 \times 28 = 0.028 \text{ g} $$

From the above, the percentage of nitrogen in the compound is:

$$ \% N = \frac{m_{N_2}}{m_{\text{compound}}} \times 100 = \frac{0.028}{0.2} \times 100 = 14\% $$

Therefore, the percentage of nitrogen in the compound is 14.

We need to find the percentage of nitrogen in the organic compound using Kjeldahl's method.

Mass of organic compound = 0.25 g

Volume of $$H_2SO_4$$ neutralised = 2.5 mL = 0.0025 L

Molarity of $$H_2SO_4$$ = 2 M

Moles of $$H_2SO_4$$ = Molarity $$\times$$ Volume (in L) = $$2 \times 0.0025 = 0.005 \text{ mol}$$

The neutralisation reaction is: $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$

So 1 mole of $$H_2SO_4$$ reacts with 2 moles of $$NH_3$$.

Moles of $$NH_3$$ = $$2 \times 0.005 = 0.01 \text{ mol}$$

Each mole of $$NH_3$$ contains 1 mole of nitrogen.

Moles of nitrogen = 0.01 mol

Mass of nitrogen = $$0.01 \times 14 = 0.14 \text{ g}$$

Percentage of nitrogen = $$\dfrac{0.14}{0.25} \times 100 = 56\%$$

Hence, the answer is 56.

We need to find the percentage of bromine in an organic compound given that 0.5 g of the compound produced 0.40 g of silver bromide.

The molar mass of AgBr is $$M_{AgBr} = M_{Ag} + M_{Br} = 108 + 80 = 188 \text{ g/mol}$$.

The number of moles of AgBr formed is $$\text{Moles of AgBr} = \frac{0.40}{188} \text{ mol}$$. In the Carius method, all bromine is converted to AgBr, and since each molecule of AgBr contains one atom of Br, the number of moles of Br is the same: $$\text{Moles of Br} = \frac{0.40}{188} \text{ mol}$$.

The mass of bromine is calculated as $$\text{Mass of Br} = \text{Moles of Br} \times M_{Br} = \frac{0.40}{188} \times 80$$ which gives $$\text{Mass of Br} = \frac{32}{188} = 0.1702 \text{ g}$$.

The percentage of bromine in the compound is then $$\% Br = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 = \frac{0.1702}{0.5} \times 100 = 34.04\%$$, which rounds to 34 %.

Therefore, the correct answer is 34 %.

We need to find the percentage of chlorine in an organic compound using Carius estimation. The mass of the organic compound is 0.25 g, the mass of AgCl obtained is 0.40 g, and the molar masses are Ag = 108 g/mol and Cl = 35.5 g/mol.

The molar mass of AgCl is calculated as $$M_{AgCl} = M_{Ag} + M_{Cl} = 108 + 35.5 = 143.5 \text{ g/mol}$$, and the moles of AgCl are given by $$\text{Moles of AgCl} = \dfrac{0.40}{143.5} = 2.787 \times 10^{-3} \text{ mol}$$.

Since 1 mole of AgCl contains 1 mole of Cl, the moles of Cl are $$\text{Moles of Cl} = \text{Moles of AgCl} = 2.787 \times 10^{-3} \text{ mol}$$ and its mass is $$\text{Mass of Cl} = 2.787 \times 10^{-3} \times 35.5 = 0.09894 \text{ g}$$.

The percentage of chlorine is $$\% \text{ of Cl} = \dfrac{\text{Mass of Cl}}{\text{Mass of compound}} \times 100 = \dfrac{0.09894}{0.25} \times 100 = 39.58\%$$.

Alternatively, using the direct formula $$\% \text{ of Cl} = \dfrac{\text{Mass of AgCl} \times M_{Cl}}{\text{Mass of compound} \times M_{AgCl}} \times 100$$ we obtain $$\% \text{ of Cl} = \dfrac{0.40 \times 35.5}{0.25 \times 143.5} \times 100 = \dfrac{14.2}{35.875} \times 100 = 39.58\%$$.

Rounding to the nearest integer gives $$\approx 40\%$$.

The percentage of chlorine present in the compound is 40%.

We need to find the ratio of Rf values of substances A and B.

Recall the formula for Rf value

$$R_f = \dfrac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent front}}$$

Calculate Rf for substance A

$$R_f(A) = \dfrac{2.08}{3.25} = 0.64$$

Calculate Rf for substance B

$$R_f(B) = \dfrac{1.05}{3.25} = 0.3231$$

Find the ratio

$$\dfrac{R_f(A)}{R_f(B)} = \dfrac{0.64}{0.3231} = \dfrac{2.08}{1.05} = 1.981$$

Round to the nearest integer

$$\dfrac{R_f(A)}{R_f(B)} \approx 2$$

The answer is 2.

We begin by recalling what “de-ionized water” means. De-ionized (or demineralised) water is water from which all the dissolved ions, both cations like $$\text{Na}^+, \text{K}^+, \text{Ca}^{2+}$$ and anions like $$\text{Cl}^-, \text{SO}_4^{2-}, \text{HCO}_3^-$$, have been removed. So, to obtain de-ionized water we need a process that can simultaneously take out positive as well as negative ions.

Now we analyse each method named in the options.

Option A tells us about the Synthetic resin method. In this method we employ two kinds of ion-exchange resins:

• A cation-exchange resin, often represented as $$\text{R-H}$$. The functional group releases $$\text{H}^+$$ ions and picks up any metal cation present in water according to the general exchange reaction

$$\text{R-H} + \text{M}^{n+} \; \longrightarrow \; \text{R-M}^{(n-)} + n\,\text{H}^+$$

• An anion-exchange resin, commonly written as $$\text{R'-OH}$$. This resin releases an $$\text{OH}^-$$ ion and absorbs any anion present:

$$\text{R'-OH} + \text{A}^- \; \longrightarrow \; \text{R'-A} + \text{OH}^-$$

After the water has passed successively through both resins, the liberated $$\text{H}^+$$ and $$\text{OH}^-$$ ions combine:

$$\text{H}^+ + \text{OH}^- \; \longrightarrow \; \text{H}_2\text{O}$$

Since every cation and every anion has been exchanged out of the water, the effluent is practically free from all ions; that is exactly the definition of de-ionized water. Thus, the synthetic resin method fulfils the requirement completely.

Option B refers to Calgon’s method. Here we add sodium hexametaphosphate $$\text{(NaPO}_3\text{)}_6$$ (popularly called Calgon) to water. The compound forms soluble complexes specifically with hardness-producing calcium and magnesium ions, preventing them from forming scale. Only $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ are affected; other ions remain. Therefore, this method softens water but does not totally de-ionize it.

Option C is Clark’s method. We add calculated amounts of lime $$\text{Ca(OH)}_2$$ to precipitate carbonate hardness as $$\text{CaCO}_3$$ and $$\text{Mg(OH)}_2$$. Again, only temporary hardness is removed; many ions stay dissolved. So Clark’s method cannot yield de-ionized water.

Option D mentions the Permutit (zeolite) method. A natural or synthetic zeolite $$\text{Na}_2\text{Z}$$ exchanges its $$\text{Na}^+$$ ions with $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ in hard water as per

$$\text{Na}_2\text{Z} + \text{Ca}^{2+} \; \longrightarrow \; \text{CaZ} + 2\,\text{Na}^+$$

This removes only the calcium and magnesium ions responsible for hardness; all other cations and every anion remain. Therefore, the permutit method merely softens water; it does not create de-ionized water.

Comparing the four methods, only the synthetic resin (ion-exchange) method eliminates both cations and anions completely, giving pure de-ionized water. Thus it is the most suitable technique for the desired purpose.

Hence, the correct answer is Option A.

In the Lassaigne's test (sodium fusion test) for halogens, the sodium extract may contain sodium cyanide ($$NaCN$$) and sodium sulphide ($$Na_2S$$) along with sodium halides. Both $$CN^-$$ and $$S^{2-}$$ ions can interfere with the silver nitrate test because they form insoluble precipitates with $$Ag^+$$: $$AgCN$$ (white) and $$Ag_2S$$ (black).

To remove these interfering ions, the sodium extract is first boiled with dilute nitric acid ($$HNO_3$$) before adding silver nitrate. The nitric acid decomposes $$NaCN$$ and $$Na_2S$$: $$NaCN + HNO_3 \rightarrow NaNO_3 + HCN\uparrow$$ and $$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S\uparrow$$. The volatile $$HCN$$ and $$H_2S$$ escape, leaving only the halide ions to react with silver nitrate.

Therefore, the compound added to the sodium extract before addition of silver nitrate is nitric acid, which is option (1).

First, we recall the working principle of simple distillation. Simple distillation is suitable for separating two miscible liquids if their boiling points differ appreciably. The empirical guideline taught in textbooks is that the boiling-point difference should be at least $$20^{\circ}\text{C}$$. If this condition is met, the vapour of the lower-boiling liquid becomes sufficiently richer in that component, condenses first, and a good separation is obtained without the need for a fractionating column.

Now we look at the two liquids mentioned in the Assertion.

We have propanone (commonly called acetone) with a boiling point of approximately $$56^{\circ}\text{C}$$, and we have propan-1-ol (commonly called 1-propanol) with a boiling point of approximately $$97^{\circ}\text{C}$$.

Let us calculate the difference in their boiling points:

$$\Delta T_{\text{b}} \;=\; 97^{\circ}\text{C} \;-\; 56^{\circ}\text{C} \;=\; 41^{\circ}\text{C}$$

We clearly see that $$\Delta T_{\text{b}} = 41^{\circ}\text{C} > 20^{\circ}\text{C}.$$

Because the boiling-point gap is more than the threshold of $$20^{\circ}\text{C},$$ the rule just recalled tells us that simple distillation will indeed be effective for this pair. Thus the statement “A simple distillation can be used to separate a mixture of propanol and propanone” is correct. Hence, Assertion (A) is true.

Next, we examine the Reason. The Reason states: “Two liquids with a difference of more than $$20^{\circ}\text{C}$$ in their boiling points can be separated by simple distillations.” This is exactly the same guideline we used above, so the Reason is also a correct scientific statement.

Furthermore, the Reason directly supplies the explanation for why simple distillation works in the specific case cited in the Assertion. We used that very rule—$$\Delta T_{\text{b}} > 20^{\circ}\text{C}$$—to justify the truth of (A). Therefore, (R) is not only true but also the correct explanation of (A).

Among the options, this correspondence matches Option B.

Hence, the correct answer is Option B.

In the scheme of qualitative inorganic analysis, metal cations are separated group-wise by taking advantage of differences in the solubility of their salts. The sequence of groups and the reagents employed are fixed, so each radical always appears in the same group. Before matching, let us recall the standard group distribution:

We have

Group I: insoluble chlorides with dilute $$\mathrm{HCl},$$ which contains $$\mathrm{Ag^+,\; Pb^{2+},\; Hg_2^{2+}}.$$

Next,

Group II: sulphides precipitated by passing } \mathrm{H_2S} \text{ in acidic medium. This group is split into two sub-groups:

Group IIA (Cu group): $$\mathrm{Cu^{2+},\; Cd^{2+},\; Pb^{2+},\; Bi^{3+},\; Hg^{2+}},$$

Group IIB (As group): $$\mathrm{As^{3+/5+},\; Sb^{3+/5+},\; Sn^{2+/4+}}.$$

Then, on adding $$\mathrm{NH_4Cl}$$ and excess $$\mathrm{NH_4OH},$$ the cations whose hydroxides are insoluble are brought down as follows:

Group III: $$\mathrm{Fe^{3+},\; Al^{3+},\; Cr^{3+}}.$$

If any cations still remain in solution, bubbling $$\mathrm{H_2S}$$ through the alkaline filtrate (containing $$\mathrm{NH_4OH}$$ and $$\mathrm{NH_4Cl}$$) precipitates sulphides of

Group IV: $$\mathrm{Zn^{2+},\; Mn^{2+},\; Co^{2+},\; Ni^{2+}}.$$

Using this standard information, let us place each ion from List - I into its correct qualitative group from List - II.

(a) $$\mathrm{Mn^{2+}}$$
From the above summary, $$\mathrm{Mn^{2+}}$$ is precipitated as $$\mathrm{MnS}$$ in the alkaline medium of Group IV. So, $$\mathrm{Mn^{2+}} \longrightarrow \text{Group IV} \; (iii).$$

(b) $$\mathrm{As^{3+}}$$
Arsenic belongs to the As sub-group; its sulphide $$\mathrm{As_2S_3}$$ is precipitated by $$\mathrm{H_2S}$$ in acid medium and is soluble in yellow ammonium sulphide, a characteristic of Group IIB. Thus, $$\mathrm{As^{3+}} \longrightarrow \text{Group IIB} \; (iv).$$

(c) $$\mathrm{Cu^{2+}}$$
Copper falls in the Cu sub-group; its black sulphide $$\mathrm{CuS}$$ precipitates directly with $$\mathrm{H_2S}$$ in acid medium and does not dissolve in yellow ammonium sulphide, placing it firmly in Group IIA. Hence, $$\mathrm{Cu^{2+}} \longrightarrow \text{Group IIA} \; (ii).$$

(d) $$\mathrm{Al^{3+}}$$
Aluminium hydroxide $$\mathrm{Al(OH)_3}$$ is gelatinous and precipitates on adding $$\mathrm{NH_4OH}$$ in the presence of $$\mathrm{NH_4Cl},$$ i.e. during Group III testing. Therefore, $$\mathrm{Al^{3+}} \longrightarrow \text{Group III} \; (i).$$

Collecting all the pairings, we obtain

$$(a)-(iii)$$, $$\; (b)-(iv)$$, $$\; (c)-(ii)$$, $$\; (d)-(i).$$

This sequence corresponds exactly to Option A in the given choices.

Hence, the correct answer is Option A.

The qualitative analysis of organic compounds often employs the well-known lassaigne’s test, more popularly called the sodium fusion test. In this procedure the organic compound is first fused with metallic sodium. The highly reactive sodium converts any hetero-elements present in the molecule into simple, water-soluble inorganic salts, which are then identified by conventional ionic group tests.

We must recall which hetero-elements are actually converted into such ionic forms:

• $$\text{Nitrogen (N)}$$ is converted to $$\text{NaCN}$$.
• $$\text{Sulfur (S)}$$ is converted to $$\text{Na}_2\text{S}$$.
• $$\text{Halogens (X = Cl, Br, I)}$$ are converted to $$\text{NaX}$$.
• $$\text{Phosphorus (P)}$$, if present, is oxidised to $$\text{Na}_2\text{HPO}_4$$ by adding an oxidising agent (generally $$\text{NaNO}_3$$) during the fusion.

These inorganic ions are then detected by their specific confirmatory tests:

$$\text{CN}^- \; \overset{\text{FeSO}_4/\text{FeCl}_3}{\rightarrow}\; \text{Prussian blue}$$,

$$\text{S}^{2-} \; \overset{\text{Acetate‐buffered Pb}^{2+}}{\rightarrow}\; \text{Black PbS}$$,

$$\text{X}^- \; \overset{\text{AgNO}_3}{\rightarrow}\; \text{AgX precipitate}$$,

$$\text{HPO}_4^{2-} \; \overset{\text{(NH}_4)_2\text{MoO}_4 + \text{HNO}_3}{\rightarrow}\; \text{Canary yellow } (\text{NH}_4)_3\text{PO}_4\cdot12\text{MoO}_3$$.

On the other hand, $$\text{Oxygen}$$ and $$\text{Carbon}$$ are always present in organic compounds and do not need special detection; moreover they do not form such simple ionic derivatives with sodium that can be identified by ordinary wet tests during a sodium fusion.

Hence, the elements that are detected by sodium fusion extract are sulfur, nitrogen, phosphorus and halogens.

Looking at the given choices, Option A lists exactly these four elements.

Hence, the correct answer is Option A.

A high-boiling organic liquid that decomposes near its boiling point cannot be purified by simple distillation or fractional distillation, because heating it to its boiling point at normal pressure would cause decomposition. Steam distillation is useful for steam-volatile compounds but is not a general method for all high-boiling liquids.

Reduced pressure (vacuum) distillation lowers the boiling point of the liquid by reducing the external pressure. This allows the compound to boil and be collected at a temperature well below its decomposition temperature. Therefore, the correct purification technique for a high-boiling organic compound that decomposes near its boiling point is reduced pressure distillation.

The answer is option (4).

In chromatography, the separation and purification of compounds depend on several factors: the differential adsorption or partition of the compound between the stationary and mobile phases (which relates to solubility), the flow rate of the solvent system (mobility), and the length of the column or TLC plate (which affects resolution).

However, the physical state of the pure compound (whether it is a solid, liquid, or gas in its pure form) does not directly affect the chromatographic separation process. What matters is how the compound interacts with the stationary and mobile phases in the dissolved state during chromatography, not what physical state the isolated pure compound exists in.

Therefore, the purification in chromatography is independent of the physical state of the pure compound. The correct answer is Option (4).

Let us analyze both statements about the retardation factor ($$R_f$$) in chromatography.

Statement-I claims that $$R_f$$ can be measured in meter/centimetre. The retardation factor is defined as $$R_f = \frac{\text{Distance moved by solute}}{\text{Distance moved by solvent front}}$$. Since it is a ratio of two distances, it is a dimensionless quantity with no units. Therefore, Statement-I is false.

Statement-II claims that the $$R_f$$ value of a compound remains constant in all solvents. This is also false because the $$R_f$$ value depends on the nature of the solvent (mobile phase), the adsorbent (stationary phase), the thickness of the layer, and the temperature. Changing the solvent will change the $$R_f$$ value of a compound.

Since both statements are false, the correct answer is option (C): Both Statement-I and Statement-II are false.

In the Carius method for the estimation of halogens in an organic compound, the compound is heated with fuming nitric acid in a sealed tube (Carius tube). The organic matter is oxidised and the halogen is converted to the corresponding halide ion (X$$^-$$).

Silver nitrate (AgNO$$_3$$) is added to the resulting solution. The halide ions react with $$Ag^+$$ to form an insoluble silver halide precipitate (AgCl, AgBr, or AgI), which is then filtered, washed, dried and weighed to determine the amount of halogen present.

Therefore, the Carius method uses fuming nitric acid in the presence of $$AgNO_3$$.

We first recall that a refining process is selected by comparing the physical or chemical properties of the required metal with those of its common impurities. One such important physical property is the melting point.

The method called liquation is based on the principle that, if a metal melts at a much lower temperature than its accompanying solid impurities, then gentle heating can melt the metal while leaving the impurities unmelted. The molten metal can then be separated from the solid residue by simple flow or by skimming.

Mathematically we say that the refining will work well when

$$T_{\text{m,\,metal}} \;<\; T_{\text{m,\,impurities}}$$

where $$T_{\text{m}}$$ denotes the melting point. Because the inequality $$T_{\text{m,\,metal}}\;<\;T_{\text{m,\,impurities}}$$ is satisfied for low-melting metals like Sn, Pb and Bi, these metals are purified economically by liquation.

On the other hand,

• The chromatographic method exploits differences in adsorption, not melting points.
• Electrolysis relies on differences in electrode potentials.
• Zone refining uses repeated melting and solidification in a moving zone; it is best suited for very high-purity, high-melting semiconductors such as Si or Ge.

Hence, for metals whose melting points are low, the generally used refining process is liquation, which corresponds to Option D.

Hence, the correct answer is Option D.

Fractional distillation is a purification method applicable to metals with relatively low boiling points, where the impurities have significantly different boiling points from the metal being purified. For this method to be economically viable, the metal must have a sufficiently low boiling point so that the energy cost of vaporising and redistilling the metal is not prohibitive.

Among the given options, zinc (Zn) has a boiling point of about $$907^\circ\text{C}$$, which is the lowest among the metals listed. Iron (Fe) boils at about $$2862^\circ\text{C}$$, copper (Cu) at about $$2562^\circ\text{C}$$, and nickel (Ni) at about $$2913^\circ\text{C}$$. The very high boiling points of Fe, Cu, and Ni make fractional distillation extremely energy-intensive and therefore not economically practical for these metals.

Zinc's relatively low boiling point makes it the metal that can be economically purified by fractional distillation. Zinc and its impurities (such as cadmium, which boils at $$767^\circ\text{C}$$) have boiling points different enough to allow effective separation. Zinc is indeed industrially purified by this method.

The correct answer is option 2 (Zn).

Indium is a semiconductor-grade metal that requires very high purity. The method used for its purification is zone refining.

Zone refining is based on the principle that impurities are more soluble in the molten state (melt) than in the solid state. A narrow zone of the metal is melted using a circular heater, and this molten zone is slowly moved along the length of the rod. As the molten zone advances, impurities preferentially remain in the melt and are swept to one end of the rod. The process is repeated several times to achieve ultra-high purity.

This method is particularly suited for elements like indium, germanium, silicon, and gallium that are used in semiconductor applications where extremely high purity (99.9999%) is required.

Therefore, the correct answer is option (3): zone refining.

In metallurgy, two important thermal processes are calcination and roasting, which differ based on the ore type and conditions used.

Reaction (A): $$\text{ZnCO}_{3(s)} \xrightarrow{\Delta} \text{ZnO}_{(s)} + \text{CO}_{2(g)}$$

This is calcination — the process of heating a carbonate ore in the absence of air (or limited air) to decompose it into the metal oxide and carbon dioxide. Carbonate ores are typically subjected to calcination.

Reaction (B): $$2\text{ZnS}_{(s)} + 3\text{O}_{2(g)} \xrightarrow{\Delta} 2\text{ZnO}_{(s)} + 2\text{SO}_{2(g)}$$

This is roasting — the process of heating a sulphide ore in excess of air (in the presence of oxygen) to convert it into the metal oxide, with the release of sulphur dioxide gas. Sulphide ores are treated by roasting.

The key distinction is that calcination involves carbonate (or hydrated) ores heated in limited/no air, while roasting involves sulphide ores heated in excess air. Therefore, (A) is calcination and (B) is roasting.

Mercury is a liquid metal with a low boiling point, so it is purified by distillation refining, giving $$a \to (ii)$$.

Copper is refined industrially by electrolytic refining, where impure copper is the anode and pure copper deposits on the cathode in a $$\text{CuSO}_4$$ solution, giving $$b \to (iii)$$.

Silicon is a semiconductor that requires extremely high purity. It is purified by zone refining, where a narrow molten zone is moved along a rod so that impurities concentrate at one end, giving $$c \to (iv)$$.

Nickel is purified by the Mond process, a vapour phase refining method. Impure nickel reacts with $$\text{CO}$$ at about 330 K to form volatile $$\text{Ni(CO)}_4$$, which is then decomposed at higher temperature (450 K) to deposit pure nickel, giving $$d \to (i)$$.

The correct matching is $$a$$-(ii), $$b$$-(iii), $$c$$-(iv), $$d$$-(i), which corresponds to option (3).

We match each test/reagent with the species it detects.

(a) Lassaigne's Test: In this test, the organic compound is fused with sodium metal to convert the elements present (nitrogen, sulphur, phosphorus, and halogens) into ionic compounds that can be detected in aqueous solution. Therefore, (a) matches with (iii) — N, S, P, and halogen.

(b) Cu(II) oxide: When an organic compound is heated with copper(II) oxide, the carbon in the compound is oxidised to $$CO_2$$, which can be detected by passing it through lime water (turning it milky). This test is used to detect the presence of carbon. Therefore, (b) matches with (i) — Carbon.

(c) Silver nitrate: The sodium fusion extract, when treated with silver nitrate solution, gives precipitates of silver halides ($$AgCl$$, $$AgBr$$, or $$AgI$$), which are used to specifically identify halogens. Therefore, (c) matches with (iv) — Halogen specifically.

(d) Black precipitate with acetic acid and lead acetate: When the sodium fusion extract containing sulphide ions is treated with acetic acid and lead acetate, a black precipitate of lead sulphide ($$PbS$$) is formed: $$Na_2S + Pb(CH_3COO)_2 \rightarrow PbS \downarrow + 2CH_3COONa$$. Therefore, (d) matches with (ii) — Sulphur.

The correct match is (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).

First we note the classical qualitative analysis scheme for metal cations. The cations are divided into analytical groups according to the reagent that first causes their precipitation. We recall the two steps involved here:

Step 1 - Addition of concentrated HCl. The reagent supplies a very high concentration of $$\text{Cl}^-$$ ions. According to the scheme, only the “Group I” cations—Ag}^+,\,$$\text{Pb}^{2+}$$,\,$$\text{Hg}_2^{2+—form$$ practically insoluble chlorides and precipitate at this stage:

Ag $$^+ + \text{Cl}^- \longrightarrow \text{AgCl}\downarrow$$

Pb $$^{2+} + 2\text{Cl}^- \longrightarrow \text{PbCl}_2\downarrow$$

None of the ions present in the given mixture—Al}^{3+},\,$$\text{Zn}^{2+}$$,\,$$\text{Ca}^{2+}$$,\,$$\text{Fe}^{3+}$$,\,$$\text{Ni}^{2+}$$,\,$$\text{Ba}^{2+}$$,\,$$\text{Cu}^{2+—belong$$ to this chloride group. Hence after adding conc. HCl no precipitate forms so far; every ion remains in solution.

Step 2 - Passing H$$_2$$S through the strongly acidic solution. We first state the relevant equilibrium that controls the availability of sulfide ions:

H $$_2\text{S}\rightleftharpoons\text{H}^+ + \text{HS}^- \quad\text{and}\quad \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-}$$

Because concentrated HCl supplies a large $$[\text{H}^+]$$, the above equilibria lie far to the left, so the actual concentration of $$\text{S}^{2-}$$ ions in the solution is extremely low. Therefore, only those metal sulfides whose solubility product $$K_{sp}$$ is exceedingly small (Group II cations) can satisfy

$$[$$ M $$^{n+}]\,[$$ S $$^{2-}] \gt K_{sp}$$

and precipitate in this acidic medium. The recognised Group II cations are Cu}^{2+},\,$$\text{Cd}^{2+}$$,\,$$\text{Bi}^{3+}$$,\,$$\text{Pb}^{2+}$$,\,$$\text{Hg}^{2+}$$,\,$$\text{As}^{3+/5+}$$,\,$$\text{Sb}^{3+/5+}$$,\,$$\text{Sn}^{2+/4+$$. Among our list, only

Cu $$^{2+} + \text{S}^{2-} \longrightarrow \text{CuS}\downarrow$$

belongs to this category; CuS has such a tiny $$K_{sp}$$ that it precipitates even with the minute sulfide ion concentration present.

The other given ions behave differently:

• $$\text{Fe}^{3+}$$ and $$\text{Al}^{3+}$$ form hydroxide precipitates in basic medium (Group III), not sulfides in acid.
• $$\text{Ni}^{2+}$$ and $$\text{Zn}^{2+}$$ form sulfides only when the medium is basic (Group IV); in acidic solution their sulfides remain far too soluble.
• $$\text{Ca}^{2+}$$ and $$\text{Ba}^{2+}$$ precipitate later as carbonates (Group V); their sulfides are readily soluble in acid.

Hence, out of the entire set, only one cation—Cu$$^{2+}$$—precipitates as CuS at this stage.

Hence, the correct answer is Option A (1).

We have an organic compound whose mass is given as $$0.471\ \text{g}$$. During sulphur estimation by the classical Carius method, all the sulphur present in the compound is converted into barium sulfate, $$\text{BaSO}_{4}$$. The experiment yields $$1.44\ \text{g}$$ of $$\text{BaSO}_{4}$$.

First, we need to know how much sulphur is present in this $$\text{BaSO}_{4}$$. For that we recall the molar mass (also called molecular weight) of barium sulfate:

$$\text{BaSO}_{4}: \quad M = M_{\text{Ba}} + M_{\text{S}} + 4\,M_{\text{O}}$$

Substituting the atomic masses, we get

$$M = 137\ \text{u} + 32\ \text{u} + 4 \times 16\ \text{u} = 137 + 32 + 64 = 233\ \text{u}$$

This means that in every $$233\ \text{g}$$ of $$\text{BaSO}_{4}$$, there are exactly $$32\ \text{g}$$ of sulphur. Hence the mass ratio of sulphur to barium sulfate is

$$\frac{m_{\text{S}}}{m_{\text{BaSO}_{4}}} = \frac{32}{233}$$

Using this ratio, the actual mass of sulphur in our sample of $$\text{BaSO}_{4}$$ is

$$m_{\text{S}} = 1.44\ \text{g} \times \frac{32}{233}$$

Carrying out the multiplication step by step,

$$1.44 \times 32 = 46.08$$

and now dividing by $$233$$,

$$m_{\text{S}} = \frac{46.08}{233}\ \text{g} \approx 0.198\ \text{g}$$

So, $$0.198\ \text{g}$$ of sulphur is present in the original $$0.471\ \text{g}$$ sample of the organic compound.

The percentage of sulphur is obtained from the general formula

$$\%\text{S} = \frac{\text{mass of S in the sample}}{\text{mass of the organic compound}} \times 100$$

Substituting the values just found,

$$\%\text{S} = \frac{0.198}{0.471} \times 100$$

Performing the division first,

$$\frac{0.198}{0.471} \approx 0.420$$

and then multiplying by $$100$$,

$$\%\text{S} \approx 42.0\%$$

Rounding to the nearest integer as required, we obtain $$42\%$$.

So, the answer is $$42\%$$.

We start with the un-balanced skeletal equation given for Duma’s method:

$$C_2H_7N +\left(2x+\dfrac{y}{2}\right)CuO \;\longrightarrow\; x\,CO_2 +\dfrac{y}{2}H_2O +\dfrac{z}{2}N_2 +\left(2x+\dfrac{y}{2}\right)Cu$$

To obtain the numerical values of the unknown coefficients, we now equate the number of atoms of each element on the two sides of the arrow.

Carbon balance
There are $$2$$ carbon atoms in one molecule of $$C_2H_7N$$ on the left. On the right, every $$CO_2$$ molecule contains one carbon atom, and the coefficient in front of $$CO_2$$ is $$x$$. So we write

$$x = 2$$

Hydrogen balance
On the left, the compound $$C_2H_7N$$ provides $$7$$ hydrogen atoms. On the right, each molecule of $$H_2O$$ has two hydrogens and its coefficient is $$\dfrac{y}{2}$$, giving a total of

$$\left(\dfrac{y}{2}\right)\times 2 = y$$ hydrogen atoms.

Equating the two sides, we have

$$y = 7$$

Nitrogen balance
The left side supplies one nitrogen atom. On the right, a molecule of $$N_2$$ contains two nitrogen atoms and its coefficient is $$\dfrac{z}{2}$$, giving

$$\left(\dfrac{z}{2}\right)\times 2 = z$$ nitrogen atoms. Hence

$$z = 1$$

Oxygen balance (consistency check)
The coefficient in front of $$CuO$$ is $$2x + \dfrac{y}{2}$$, so the number of oxygen atoms on the left is also $$2x + \dfrac{y}{2}$$. On the right, oxygen appears in $$CO_2$$ and $$H_2O$$. Counting them gives

$$\underbrace{x\times 2}_{\text{from } CO_2} + \underbrace{\left(\dfrac{y}{2}\right)\times 1}_{\text{from } H_2O} = 2x + \dfrac{y}{2}$$

Substituting $$x = 2$$ and $$y = 7$$, both sides give $$2(2) + \dfrac{7}{2} = 4 + \dfrac{7}{2}$$, confirming that oxygen is already balanced.

Thus, the integer value required for $$y$$ is

$$7$$

Hence, the correct answer is Option 7.

In the Carius method for estimation of bromine, the organic compound is burnt in excess silver nitrate, and all the bromine is converted to silver bromide (AgBr), which is weighed.

The molar mass of AgBr = 108 + 80 = 188 g mol$$^{-1}$$.

Moles of AgBr formed: $$n_{\text{AgBr}} = \frac{0.2397}{188} = 1.275 \times 10^{-3}$$ mol

Since each mole of AgBr contains one mole of Br, the mass of bromine in the sample is:

$$m_{\text{Br}} = 1.275 \times 10^{-3} \times 80 = 0.1020 \text{ g}$$

The percentage of bromine in the organic compound is:

$$\% \text{Br} = \frac{m_{\text{Br}}}{m_{\text{compound}}} \times 100 = \frac{0.1020}{0.15} \times 100 = 68.0\%$$

The percentage of bromine in the organic compound is $$\mathbf{68}$$%.

In Kjeldahl’s method, the nitrogen present in the organic compound is finally converted into ammonia. That ammonia is absorbed by a known, measured volume of standard acid. By knowing how much of the acid is neutralised we can back-calculate the amount of nitrogen. Here we will move in the reverse direction: we already know the mass of nitrogen, and we have to find the volume of 1 M $$\mathrm{H_2SO_4}$$ that would be neutralised.

We are given that $$0.8\ \text{g}$$ of the organic compound contains $$42\%$$ nitrogen by mass. First we calculate the actual mass of nitrogen present:

$$\text{Mass of nitrogen}=0.8\ \text{g}\times \dfrac{42}{100}=0.336\ \text{g}$$

Next we convert this mass of nitrogen to moles. The atomic mass of nitrogen is $$14\ \text{g mol}^{-1}$$, so

$$\text{Moles of }N=\dfrac{0.336\ \text{g}}{14\ \text{g mol}^{-1}}=0.024\ \text{mol}$$

In the Kjeldahl procedure each atom of nitrogen becomes one molecule of ammonia, $$\mathrm{NH_3}$$. Therefore,

$$\text{Moles of }NH_3=\text{Moles of }N=0.024\ \text{mol}$$

Now we write the neutralisation reaction between ammonia and sulphuric acid:

$$2\,\mathrm{NH_3}+ \mathrm{H_2SO_4}\;\longrightarrow\; (\mathrm{NH_4})_2\mathrm{SO_4}$$

From the balanced equation, $$2$$ moles of $$\mathrm{NH_3}$$ require $$1$$ mole of $$\mathrm{H_2SO_4}$$. Hence the moles of $$\mathrm{H_2SO_4}$$ needed are

$$\text{Moles of }H_2SO_4=\dfrac{\text{Moles of }NH_3}{2}=\dfrac{0.024}{2}=0.012\ \text{mol}$$

The concentration (molarity) of the acid is given as $$1\ \text{M}$$, meaning

$$\text{Molarity}=\dfrac{\text{Moles of solute}}{\text{Volume of solution in litres}}$$

Re-arranging for volume, we get

$$\text{Volume}=\dfrac{\text{Moles}}{\text{Molarity}}=\dfrac{0.012\ \text{mol}}{1\ \text{mol L}^{-1}}=0.012\ \text{L}$$

Converting litres to millilitres ($$1\ \text{L}=1000\ \text{mL}$$):

$$0.012\ \text{L}\times1000=12\ \text{mL}$$

So, the answer is $$12$$.

We are given that $$0.5\ \text{g}$$ of compound A produces $$0.3849\ \text{g}$$ of silver chloride ($$\text{AgCl}$$) when treated with $$\text{AgNO}_3$$ in the Carius method. Our task is to calculate the percentage of chlorine present in compound A.

First, we note the atomic masses supplied in the question:

$$\text{Atomic mass of Ag}=107.87,\qquad \text{Atomic mass of Cl}=35.5$$

So, the molar mass of silver chloride is obtained by simple addition:

$$M_{\text{AgCl}} = M_{\text{Ag}} + M_{\text{Cl}} = 107.87 + 35.5 = 143.37\ \text{g mol}^{-1}$$

Next, we determine the number of moles of $$\text{AgCl}$$ actually formed. The basic formula for moles is

$$\text{moles} = \dfrac{\text{given mass}}{\text{molar mass}}$$

Substituting the values, we get

$$n_{\text{AgCl}} = \dfrac{0.3849}{143.37} = 0.002685\ \text{mol}\ (\text{approximately}).$$

In every mole of $$\text{AgCl}$$ there is exactly one mole of chlorine. Hence, the moles of chlorine present are the same as the moles of $$\text{AgCl}$$ formed:

$$n_{\text{Cl}} = n_{\text{AgCl}} = 0.002685\ \text{mol}.$$

We now convert these moles of chlorine to mass using the relation

$$\text{mass} = \text{moles} \times \text{atomic mass}.$$

Thus,

$$m_{\text{Cl}} = 0.002685 \times 35.5 = 0.0953\ \text{g}\ (\text{to four significant figures}).$$

Finally, the percentage of chlorine in compound A is calculated by the standard definition

$$\%\text{Cl} = \dfrac{\text{mass of Cl in the sample}}{\text{mass of the sample}} \times 100.$$

Substituting our values,

$$\%\text{Cl} = \dfrac{0.0953}{0.5} \times 100 = 19.06\%.$$

Rounding off to the nearest integer as instructed, we obtain $$19\%.$

Hence, the correct answer is Option 19.

We are asked to calculate how many moles of copper(II) oxide, CuO, take part in the Dumas combustion of a given mass of the amine “N, N-dimethylaminopentane”.

First we write the molecular formula of the compound. A pentyl group contributes $$\mathrm{C_5H_{11}}$$ and two additional methyl groups are attached to the nitrogen, giving two more carbon atoms and six more hydrogens. Hence we have

$$\mathrm{C_{5+2}H_{11+6}N}=\mathrm{C_7H_{17}N}.$$

Its molar mass is obtained term by term:

$$M=\underbrace{7\times 12}_{\text{carbon}}+\underbrace{17\times 1}_{\text{hydrogen}}+\underbrace{1\times 14}_{\text{nitrogen}} =84+17+14 =115\ \text{g mol}^{-1}.$$

The sample mass is $$57.5\ \text{g}$$, so the number of moles of the amine present is

$$n_{\text{amine}}=\frac{57.5\ \text{g}}{115\ \text{g mol}^{-1}}=0.500\ \text{mol}.$$

Now we recall the stoichiometry of the Dumas combustion. For a general molecule $$\mathrm{C_xH_yN_z},$$ complete oxidation with CuO satisfies the balanced relation

$$\mathrm{C_xH_yN_z} \;+\;(2x+\tfrac{y}{2})\,\mathrm{CuO} \;\longrightarrow\; x\,\mathrm{CO_2}+ \tfrac{y}{2}\,\mathrm{H_2O}+ \tfrac{z}{2}\,\mathrm{N_2} + (2x+\tfrac{y}{2})\,\mathrm{Cu}.$$

This equality is obtained as follows. Each carbon atom requires two oxygen atoms to become $$\mathrm{CO_2}$$, and each pair of hydrogen atoms requires one oxygen atom to become $$\mathrm{H_2O}$$. Therefore, per mole of the organic compound, the total number of oxygen atoms demanded is $$2x+\dfrac{y}{2}$$. Since every mole of CuO supplies exactly one oxygen atom, the same numerical amount, $$2x+\dfrac{y}{2},$$ gives the moles of CuO used.

Substituting $$x=7$$ and $$y=17$$ for our specific amine, we get

$$n_{\text{CuO per mol amine}}=2(7)+\frac{17}{2}=14+\frac{17}{2}=14+8.5=22.5\ \text{mol}.$$

We actually have only half a mole of the amine, so the total moles of CuO consumed become

$$n_{\text{CuO total}} =0.500\ \text{mol amine}\;\times\;22.5\ \frac{\text{mol CuO}}{\text{mol amine}} =11.25\ \text{mol}.$$

It is convenient to express $$11.25$$ as a product with $$10^{-2}$$:

$$11.25\ \text{mol}=1125\times10^{-2}\ \text{mol}.$$

The problem asks for the nearest integer appearing as the coefficient in “$$\times10^{-2}$$”, which is clearly $$1125$$.

So, the answer is $$1125$$.

$$R_f = \frac{\text{Distance travelled by compound}}{\text{Distance travelled by solvent}}$$
$$R_f = \ \frac{\ 2cm}{5cm}\ =\ 4\ \times\ 10^{-1}$$
X = 4

In Carius method, the halogen present in the organic compound is converted completely into its silver halide. Here the silver halide obtained is silver bromide, $$\mathrm{AgBr}$$.

The given data are:
    Mass of organic compound, $$w = 0.2 \text{ g}$$
    Mass of silver bromide, $$m_{\mathrm{AgBr}} = 0.188 \text{ g}$$
    Atomic mass of silver, $$M_{\mathrm{Ag}} = 108$$
    Atomic mass of bromine, $$M_{\mathrm{Br}} = 80$$

First, we need the molar mass of silver bromide. By definition,

$$M_{\mathrm{AgBr}} = M_{\mathrm{Ag}} + M_{\mathrm{Br}} = 108 + 80 = 188 \text{ g mol}^{-1}.$$

Now, to obtain the mass of bromine actually present in the precipitate, we use the proportion coming from the formula weight. The fraction of bromine in one mole of $$\mathrm{AgBr}$$ is

$$\frac{M_{\mathrm{Br}}}{M_{\mathrm{AgBr}}} = \frac{80}{188}.$$

Therefore, the mass of bromine in the given precipitate is

$$m_{\mathrm{Br}} = \frac{80}{188}\times m_{\mathrm{AgBr}}.$$

Substituting $$m_{\mathrm{AgBr}} = 0.188 \text{ g}$$, we get

$$m_{\mathrm{Br}} = \frac{80}{188}\times 0.188 = 0.188 \times \frac{80}{188} = 0.188 \times 0.4255319 = 0.08 \text{ g}.$$

We now calculate the percentage of bromine in the original compound sample. By definition,

$$\%\ \mathrm{Br} = \frac{\text{mass of Br in sample}}{\text{mass of sample}}\times 100 = \frac{0.08}{0.2}\times 100 = 0.4 \times 100 = 40\%.$$

After rounding to the nearest integer, the percentage remains $$40$$.

Hence, the correct answer is Option C.

We recall a basic physical-chemistry fact: the normal boiling point of a liquid is the temperature at which its vapour pressure equals the external pressure. If we lower the external pressure, the temperature needed to reach that vapour pressure also lowers. This principle is used in “distillation under reduced pressure”, sometimes called “vacuum distillation”. The relation is expressed by the Clausius-Clapeyron equation, stated first:

$$\dfrac{d\ln P}{dT}= \dfrac{\Delta H_{\text{vap}}}{RT^{2}}$$

Here $$P$$ is the vapour pressure, $$T$$ is the absolute temperature, $$R$$ is the gas constant and $$\Delta H_{\text{vap}}$$ is the molar enthalpy of vaporisation. Although we are not required to integrate it fully for this problem, the equation tells us qualitatively that if we reduce $$P$$, then for a fixed $$\Delta H_{\text{vap}}$$, the corresponding equilibrium temperature $$T$$ must also decrease.

Now, let us apply this idea to glycerol, $$\mathrm{C_{3}H_{8}O_{3}}$$. Glycerol is a highly viscous, trihydric alcohol formed as a by-product in the saponification (soap-making) process:

$$\text{Fat or oil (ester)} + 3\,\text{NaOH} \;\longrightarrow\; \text{Soap (sodium salt of fatty acid)} + \text{Glycerol}$$

Important physical data: the normal (1 atm) boiling point of glycerol is about $$290^\circ\text{C}$$. At this high temperature glycerol tends to undergo partial decomposition, charring and colouration, which is undesirable for industrial recovery. Therefore we must choose a separation technique that enables evaporation at a lower temperature so that thermal decomposition does not occur.

We examine the four given options in turn.

Option A — Fractional distillation: This is effective when two liquids have different but reasonably low boiling points and do not decompose. Glycerol itself has a single high boiling point and decomposition risk, so mere fractional distillation at atmospheric pressure is not suitable.

Option B — Differential distillation: This is essentially batch distillation without a fractionating column, again performed at atmospheric pressure. The same problem of high temperature persists, hence it is also unsuitable.

Option C — Steam distillation: Steam distillation works nicely for volatile, water-immiscible substances whose boiling points are normally >100 °C but which form a heterogeneous azeotrope with water. Glycerol, however, is completely miscible with water and therefore does not distil over with steam; it stays in the aqueous layer. Consequently steam distillation cannot separate glycerol.

Option D — Distillation under reduced pressure: By purposefully lowering the external pressure, we lower the boiling temperature according to the relation we wrote earlier. For glycerol, reducing the pressure to about $$10^{2}$$-$$10^{3}$$ Pa (a few mm Hg) brings its boiling point down to roughly $$150^\circ\text{C}$$ or even lower, which is well below the temperature at which significant decomposition occurs. Industrial soap plants therefore connect the glycerol-containing liquor to a vacuum still, remove water first, and then distil pure glycerol safely under reduced pressure.

Thus, among the four techniques listed, only “distillation under reduced pressure” specifically addresses the high-boiling, decomposition-prone nature of glycerol.

Hence, the correct answer is Option D.

In column chromatography that uses silica gel as the stationary phase, the surface of the silica bears many $$\text{Si-OH}$$ (silanol) groups. These groups are polar and weakly acidic. Compounds reaching the column therefore compete with the solvent for hydrogen-bonding or acid-base interaction sites on the silica.

The governing principle is straightforward: the stronger a compound binds to the polar silica surface, the more slowly it travels down the column. Conversely, compounds that bind only weakly are swept out first by the mobile phase. Hence, we must compare the relative strength of interaction (and therefore the effective polarity on silica) of the three given compounds.

We list each structure and its key interaction sites:

• Compound (C) acetophenone, $$C_{6}H_{5}COCH_{3}$$, contains a single carbonyl oxygen. The carbonyl oxygen is only a hydrogen-bond acceptor and cannot donate a hydrogen bond. Moreover, there are no basic lone pairs that become protonated by the acidic silica surface. Thus, interaction with silica is limited to weak dipole attractions; overall, acetophenone is the least strongly adsorbed.

• Compound (A) benzanilide, $$C_{6}H_{5}CONHC_{6}H_{5}$$, is an amide. The carbonyl oxygen can accept a hydrogen bond and the $$N-H$$ can donate one. However, the lone pair on the nitrogen is delocalised into the carbonyl group, so basicity is strongly reduced; the molecule is also sterically bulky because of two phenyl rings, which lessens surface contact. Hence, benzanilide interacts with silica more than acetophenone but less than a free amine.

• Compound (B) aniline, $$C_{6}H_{5}NH_{2}$$, possesses a free amino group. On the acidic silica surface this amino group can become partially protonated 

$$RNH_2+Si-OH\ \longrightarrow\ RNH_3^+\ +\ Si-O^-$$

generating a strong ionic (acid-base) attraction in addition to hydrogen bonding. Therefore aniline is most strongly retained on silica.

Combining these observations, the adsorption (polarity) order on silica is

$$\text{aniline (B)} \;>\; \text{benzanilide (A)} \;>\; \text{acetophenone (C)}.$$

Because the stationary phase is polar and the mobile phase is the relatively polar mixture hexane : ethyl acetate (20 : 80), the compounds with weaker adsorption (lower effective polarity on silica) travel the fastest and elute first. Reversing the above order therefore gives the elution sequence:

$$\text{acetophenone (C)} \;\longrightarrow\; \text{benzanilide (A)} \;\longrightarrow\; \text{aniline (B)}.$$

Thus the column first furnishes (C), then (A), and finally (B).

Hence, the correct answer is Option C.

We have a mixture that contains three compounds dissolved in the organic solvent ethyl acetate:

$$$\text{m-chloroaniline (basic amine)}$$$ $$$\text{m-chlorophenol (weak acid, }pK_a\approx 10\text{)}$$$ $$$\text{m-chlorobenzoic acid (stronger acid, }pK_a\approx 4\text{)}$$$

The separation is carried out by successive extractions, using the difference in acid-base strength of the functional groups. A compound converts to its water-soluble ionic form only if the reagent present is strong enough to protonate or deprotonate it. We trace each step.

First extraction: a saturated solution of $$\text{NaHCO}_3$$ (a weak base) is shaken with the organic layer.

The reaction possible is:

$$$\text{m-chlorobenzoic acid (HA)} + \text{HCO}_3^- \;\longrightarrow\; \text{m-chlorobenzoate ion (A}^-)\;+\;\text{H}_2\text{CO}_3$$$

Because $$pK_a(\text{benzoic acid})\approx 4$$ is sufficiently low, $$\text{HCO}_3^-$$ can remove its proton. Phenol, with $$pK_a\approx 10$$, is not acidic enough to be deprotonated by $$\text{HCO}_3^-$$, and the basic amine is certainly not affected by a base. Thus only the salt of m-chlorobenzoic acid transfers to the aqueous layer. We label this aqueous layer as Fraction A.

So, after the first extraction:

$$$\text{Fraction A (aqueous)}:\; \text{m-chlorobenzoate ion }(\;\longrightarrow\; \text{recovers m-chlorobenzoic acid on acidification)}$$$ $$$\text{Organic phase remaining}:\; \text{m-chlorophenol + m-chloroaniline}$$$

Second extraction: the residual organic layer is now shaken with dilute $$\text{NaOH}$$ (a stronger base than $$\text{NaHCO}_3$$).

The reaction that occurs is:

$$$\text{m-chlorophenol (ArOH)} + \text{OH}^- \;\longrightarrow\; \text{m-chlorophenoxide ion (ArO}^-)\;+\;\text{H}_2\text{O}$$$

This time the base is strong enough to deprotonate the phenol. The resulting phenoxide ion is water-soluble and goes into the new aqueous layer, which we label as Fraction B. The amine does not react with base, so it stays in the organic phase.

Thus after the second extraction:

$$$\text{Fraction B (aqueous)}:\; \text{m-chlorophenoxide ion }(\;\longrightarrow\; \text{recovers m-chlorophenol on acidification)}$$$ $$$\text{Organic phase left}:\; \text{m-chloroaniline}$$$

The final organic layer is labelled Fraction C, and it contains only the unreacted base.

Summarising the contents in their neutral, isolated forms obtained after the usual acidification of the aqueous fractions:

$$$\text{Fraction A} \;:\; \text{m-chlorobenzoic acid}$$$ $$$\text{Fraction B} \;:\; \text{m-chlorophenol}$$$ $$$\text{Fraction C} \;:\; \text{m-chloroaniline}$$$

Matching this order with the given options, we see that Option B lists the compounds in exactly this sequence.

Hence, the correct answer is Option B.

We have two hydrocarbons in the same flask: isohexane and 3-methylpentane. Their normal boiling points are given as $$60^{\circ}\text C$$ and $$63^{\circ}\text C$$ respectively.

First, let us examine the temperature data. Writing the temperatures in an inequality form, we obtain

$$60^{\circ}\text C \; < \; 63^{\circ}\text C.$$

Now we calculate the difference in their boiling points:

$$\Delta T \;=\; 63^{\circ}\text C \;-\; 60^{\circ}\text C \;=\; 3^{\circ}\text C.$$

This small temperature gap is crucial. In laboratory practice, the rule of thumb is that:

If } \Delta T \;\lt\; 25^{\circ}\text C,\; \text{use fractional distillation.

Fractional distillation provides a fractionating column that offers many theoretical plates, giving repeated condensation-evaporation cycles. These cycles allow liquids whose boiling points are very close to be separated effectively. Simple distillation, on the other hand, is suitable only when the boiling-point difference is fairly large (usually $$\ge 25^{\circ}\text C$$), so it would not work efficiently here.

Therefore, the best separation technique is fractional distillation.

Next, we identify which component will come out first. The general statement is:

The liquid with the lower boiling point vaporizes (and hence distills) first.

From the inequality $$60^{\circ}\text C < 63^{\circ}\text C,$$ we see that isohexane (boiling at $$60^{\circ}\text C$$) has the lower boiling point, whereas 3-methylpentane (boiling at $$63^{\circ}\text C$$) has the higher boiling point.

Consequently, isohexane will rise through the fractionating column first and will be collected in the distillate before 3-methylpentane.

Putting both conclusions together, we state: the mixture should be separated by fractional distillation, and isohexane will be the first liquid to distill.

Hence, the correct answer is Option A.

In the Carius method we directly obtain the mass of halogen present. Here,

$$\text{Mass of compound taken}=0.172\ \text{g}$$

$$\text{Mass of bromine found}=0.080\ \text{g}$$

We first calculate the percentage of bromine in the unknown sample. The general relation is

$$\%\,\text{Br}=\frac{\text{mass of Br obtained}}{\text{mass of sample}}\times 100$$

Substituting the given numbers,

$$\%\,\text{Br}=\frac{0.080}{0.172}\times 100$$

$$\%\,\text{Br}=0.465116\ldots\times 100$$

$$\%\,\text{Br}=46.5116\% \approx 46.5\%$$

Now we compare this experimental value with the theoretical bromine percentages of each option. For every structure we first write its molecular formula, then its molar mass, and finally the percentage of bromine.

Option A : $$CH_3CHBr_2$$

Molecular formula: $$C_2H_4Br_2$$

Molar mass:

$$M=2(12)+4(1)+2(79.9)=24+4+159.8=187.8\ \text{g mol}^{-1}$$

Theoretical bromine percentage:

$$\%\,\text{Br}=\frac{2(79.9)}{187.8}\times 100=\frac{159.8}{187.8}\times 100=85.06\%$$

This is far higher than the required $$46.5\%$$, so Option A is rejected.

Option B : p-bromoaniline

Molecular formula: an aniline ring $$C_6H_5NH_2$$ with one Br substituent gives $$C_6H_6BrN$$.

Molar mass:

$$M=6(12)+6(1)+79.9+14=72+6+79.9+14=171.9\ \text{g mol}^{-1}$$

Theoretical bromine percentage:

$$\%\,\text{Br}=\frac{79.9}{171.9}\times 100=46.509\%\approx 46.5\%$$

This matches the experimental value almost exactly.

Option C : 3,5-dibromoaniline

Molecular formula: $$C_6H_5Br_2N$$

Molar mass:

$$M=6(12)+5(1)+2(79.9)+14=72+5+159.8+14=250.8\ \text{g mol}^{-1}$$

Theoretical bromine percentage:

$$\%\,\text{Br}=\frac{159.8}{250.8}\times 100=63.74\%$$

This value is much higher than $$46.5\%$$, so Option C is also ruled out.

Option D : $$H_3CCH_2Br$$

Molecular formula: $$C_2H_5Br$$

Molar mass:

$$M=2(12)+5(1)+79.9=24+5+79.9=108.9\ \text{g mol}^{-1}$$

Theoretical bromine percentage:

$$\%\,\text{Br}=\frac{79.9}{108.9}\times 100=73.40\%$$

Again this is far above $$46.5\%$$, so Option D is discarded.

Only p-bromoaniline shows a bromine percentage that coincides with the experimental value. Therefore the unknown compound must be p-bromoaniline.

Hence, the correct answer is Option B.

We have to find the mass percentage of bromine present in the organic compound whose mass is given as $$1.6\;\text{g}$$. During the Carius estimation the bromine present in the compound is completely converted into silver bromide, $$\text{AgBr}$$, whose mass is reported to be $$1.88\;\text{g}$$. Hence the whole problem boils down to: how much bromine is contained in $$1.88\;\text{g}$$ of $$\text{AgBr}$$, and what fraction of the original $$1.6\;\text{g}$$ sample does that bromine represent?

First we need the molar mass of silver bromide. Using the given atomic masses

$$M_{\text{Ag}} = 108\;\text{g mol}^{-1}, \qquad M_{\text{Br}} = 80\;\text{g mol}^{-1}$$

we write

$$M_{\text{AgBr}} = M_{\text{Ag}} + M_{\text{Br}} = 108 + 80 = 188\;\text{g mol}^{-1}.$$

Now, the number of moles of $$\text{AgBr}$$ actually formed is obtained from the basic relation

$$n = \dfrac{m}{M},$$

where $$m$$ is mass and $$M$$ is molar mass. Substituting the known quantities,

$$n_{\text{AgBr}} = \dfrac{1.88\;\text{g}}{188\;\text{g mol}^{-1}} = 0.01\;\text{mol}.$$

In the silver bromide crystal lattice each formula unit contains exactly one bromine atom, so the amount (in moles) of bromine captured is numerically equal to the number of moles of $$\text{AgBr}$$ formed. Thus

$$n_{\text{Br}} = n_{\text{AgBr}} = 0.01\;\text{mol}.$$

To get the actual mass of bromine present we again employ

$$m = n \times M.$$

Hence, substituting,

$$m_{\text{Br}} = 0.01\;\text{mol} \times 80\;\text{g mol}^{-1} = 0.8\;\text{g}.$$

We now know that $$0.8\;\text{g}$$ of bromine came from the original $$1.6\;\text{g}$$ sample. The percentage composition by mass is therefore computed from the standard formula

$$\%\text{Br} = \dfrac{m_{\text{Br}}}{m_{\text{sample}}}\times 100.$$

Substituting the values,

$$\%\text{Br} = \dfrac{0.8\;\text{g}}{1.6\;\text{g}}\times 100 = 50.$$

Hence, the correct answer is Option 50.

$$H_2O$$ : Sugar $$\rightarrow$$ Recrystallization (Q)

Sugar is a solid that is highly soluble in water at elevated temperatures but less soluble at lower temperatures. When a saturated solution of sugar in water is cooled down, sugar separates out as pure crystals, making Recrystallization the ideal purification technique.

$$H_2O$$ : Aniline $$\rightarrow$$ Steam distillation (R) Aniline is an organic liquid that is virtually immiscible with water but volatile in steam. Steam distillation is used to separate and purify substances that are steam-volatile and insoluble in water, allowing them to distill at a temperature lower than their actual boiling point (which prevents thermal decomposition).

$$H_2O$$ : Toluene $$\rightarrow$$ Differential extraction (S) Toluene and water are two completely immiscible liquid layers. When an organic compound is present in an aqueous medium, it can be separated using a separating funnel by adding an organic solvent (like toluene) in which the compound is more soluble. This separation technique is called Differential extraction.

Hence,

(A) → (Q); (B) → (R); (C) → (S)

First, let us recall what happens inside a column chromatographic set-up. We pack a finely divided solid, such as silica gel or alumina, into a long vertical glass tube. This packed solid is called the stationary phase. Next, a liquid or a mixture of liquids, called the mobile phase or eluent, is allowed to flow down the column under gravity. The sample that we wish to separate is introduced at the top of this column and is carried downward by the flowing solvent.

Now, the key physical phenomenon that actually brings about the separation is the tendency of different components of the mixture to cling to, or leave, the surface of the solid. In surface science, the word used for this surface attachment is “adsorption,” not “absorption.”

Adsorption means the molecules accumulate only on the surface of the solid. Absorption, on the other hand, would imply that the molecules pass into the bulk of another phase, like water soaking into a sponge; that is not what happens here. So we can immediately eliminate any option that speaks of “absorption.”

Let us examine all the options one by one:

Option A speaks of the differential adsorption of substances on the solid phase. This is precisely what we observe: each solute molecule spends part of its time adsorbed (stuck) on the stationary phase and part of its time dissolved (moving) in the mobile phase. Because different solutes have different affinities for the solid surface, they travel through the column at different average speeds and emerge separated.

Option B uses the term “differential absorption,” which, as explained, is incorrect terminology for the surface phenomenon actually taking place.

Option C mentions gravitational force. While gravity certainly makes the liquid flow downward, it is not the principle responsible for the separation of components.

Option D mentions capillary action. In a packed column the liquid does not move primarily by capillary rise; instead it flows downward under gravity, so capillary forces are not central to the chromatographic principle either.

Therefore, the only scientifically accurate description of the principle is the one stated in Option A.

Hence, the correct answer is Option A.

From the data of the Dumas combustion we know the exact numbers of moles of the three products that are formed:

$$n_{\,CO_2}=6,\qquad n_{\,H_2O}=4,\qquad n_{\,N_2}=1$$

First we convert these moles of the products into the moles of the original elements present in the unknown compound.

Each mole of $$CO_2$$ contains exactly one mole of carbon atoms because the formula is $$CO_2$$. Therefore

$$n_{\,C}=n_{\,CO_2}=6\; \text{mol}$$

Each mole of $$H_2O$$ contains two moles of hydrogen atoms since the subscripts of hydrogen in water are 2. Thus

$$n_{\,H}=2\,n_{\,H_2O}=2\times4=8\; \text{mol}$$

Each mole of $$N_2$$ contains two moles of nitrogen atoms because nitrogen gas is di-atomic ($$N_2$$). Hence

$$n_{\,N}=2\,n_{\,N_2}=2\times1=2\; \text{mol}$$

So the mole (atom) ratio of the elements in the burnt sample is

$$C:H:N = 6:8:2$$

We obtain the simplest whole-number ratio (the empirical ratio) by dividing every number by the greatest common divisor, which is 2:

$$\frac{6}{2} : \frac{8}{2} : \frac{2}{2} = 3 : 4 : 1$$

Therefore the empirical formula is

$$C_3H_4N$$

The options offered all have even numbers of carbon and hydrogen atoms, so we scale the empirical formula by the smallest integer that converts it to one of the listed formulas. Multiplying by 2 gives

$$2\,(C_3H_4N)=C_6H_8N_2$$

This molecular formula exactly matches Option C.

Hence, the correct answer is Option C.

We begin by recalling the principle of differential extraction. In this technique two immiscible liquids are shaken together so that a solute can distribute itself preferentially into one of the layers. For the liquids to separate properly inside the separating funnel, they must satisfy two conditions: they must be immiscible, and they must have different densities, because density decides which layer will settle at the bottom.

Now we examine the pair given in the question: dichloromethane (abbreviated as DCM, chemical formula $$CH_2Cl_2$$) and water $$H_2O$$. The first thing to note is that DCM is an organochlorine solvent, whereas water is a polar protic solvent. Because of their very different intermolecular forces, they do not mix appreciably; thus they are indeed immiscible. So options that claim miscibility or formation of a turbid/colloidal mixture can already be ruled out.

Next we must decide which of the two liquids will occupy the lower position in the separating funnel. This is governed by the density relation

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

The literature values at room temperature are

$$\rho_{\text{DCM}} \approx 1.33\ \text{g\,cm}^{-3}$$

$$\rho_{H_2O} \approx 1.00\ \text{g\,cm}^{-3}$$

Since $$1.33\ \text{g\,cm}^{-3} > 1.00\ \text{g\,cm}^{-3}$$, dichloromethane is denser than water. According to the basic rule, the denser liquid settles at the bottom, while the less dense liquid forms the upper layer. Therefore DCM will form the lower layer and water will form the upper layer inside the separating funnel.

This description matches the statement in Option C: “DCM and $$H_2O$$ would stay as lower and upper layer respectively in the S.F.”

Hence, the correct answer is Option C.

First, let us recall the definition of the retention factor, commonly written as $$R_f,$$ which is used for comparing how far a substance travels on a chromatographic plate.

The formula is stated as

$$ R_f=\frac{\text{Distance travelled by the centre of the spot of the solute}}{\text{Distance travelled by the solvent front}} $$

Since the solute spot can never move farther than the solvent front, the numerator is always smaller than or at most equal to the denominator. Therefore, from the above formula we immediately obtain

$$ 0 < R_f \le 1 . $$

Now we analyse each option one by one in the light of the definition.

Option A: “Higher $$R_f$$ value means higher adsorption.” We know that stronger adsorption to the stationary phase causes a compound to move more slowly; hence the distance in the numerator becomes smaller. This gives a smaller $$R_f,$$ not a higher one. Therefore, a higher $$R_f$$ actually corresponds to lower adsorption. So the statement in Option A is wrong.

Option B: “The value of $$R_f$$ cannot be more than one.” From the inequality $$0 < R_f \le 1$$ derived above, this statement is perfectly true. Hence Option B is correct and therefore not the answer we seek.

Option C: “$$R_f$$ value depends on the type of chromatography.” Different chromatographic techniques (paper, thin-layer, column, etc.) employ different stationary and mobile phases, thus altering adsorption-desorption equilibria. Consequently, $$R_f$$ indeed varies with the type of chromatography. So Option C is a correct statement.

Option D: “$$R_f$$ value of dependent on the mobile phase.” Changing the polarity or composition of the mobile phase directly affects how far a solute moves, hence changing $$R_f.$$ Therefore this statement is also correct.

Among the four given choices, only Option A makes an incorrect claim about $$R_f.$$

Hence, the correct answer is Option A.

First, let us recall some basic terminology used in chromatographic techniques. The substance that forms the stationary surface on which separation occurs is called the adsorbent. The solvent or mixture of solvents that moves through this stationary phase is known as the mobile phase. Any compound that is being separated or analysed and therefore gets held, at least temporarily, on the stationary phase is referred to as the adsorbate.

With these definitions in mind, we identify which substance of Item I plays which role mentioned in Item II.

We have benzaldehyde, $$\mathrm{C_6H_5CHO}$$. This is an organic compound that one would normally inject or spot on the chromatographic plate or load into the column to be separated from other components. Thus benzaldehyde itself is the species that gets adsorbed on the surface of the stationary phase. Because the substance being separated is, by definition, the adsorbate, we can say

$$\text{Benzaldehyde} \; \longrightarrow \; \text{Adsorbate}.$$

Hence, benzaldehyde of Item I matches with $$\text{Item II : R (Adsorbate)}$$.

Now we look at alumina, chemically written as $$\mathrm{Al_2O_3}$$. In chromatographic practice, finely powdered alumina is commonly packed in columns or coated on plates to provide the surface to which different molecules can adhere with different strengths. The solid that provides this surface is precisely what is known as the adsorbent. Therefore,

$$\text{Alumina} \; \longrightarrow \; \text{Adsorbent}.$$

So alumina of Item I associates with $$\text{Item II : Q (Adsorbent)}$$.

Finally, we examine acetonitrile, $$\mathrm{CH_3CN}$$. It is a polar aprotic solvent widely employed as the flowing liquid in high-performance liquid chromatography (HPLC) and other chromatographic systems. Because it is the liquid that moves through the stationary phase carrying the sample along, acetonitrile functions as the mobile phase. Therefore,

$$\text{Acetonitrile} \; \longrightarrow \; \text{Mobile phase}.$$

This gives the match of acetonitrile with $$\text{Item II : P (Mobile phase)}$$.

Collecting all three pairings obtained above, we have

$$\begin{aligned} (A)&\; \text{Benzaldehyde} &\rightarrow&\; (R)\\ (B)&\; \text{Alumina} &\rightarrow&\; (Q)\\ (C)&\; \text{Acetonitrile} &\rightarrow&\; (P) \end{aligned}$$

Comparing this ordered list with the options provided, we see that it coincides exactly with Option B.

Hence, the correct answer is Option B.

We start by recalling a well-known qualitative test that analysts employ for detecting the presence of the element phosphorus in an organic compound. The sequence of reagents is distinctive: first the compound is fused or oxidised with an oxidising agent such as sodium peroxide, $$\text{Na}_2\text{O}_2$$, in order to convert any phosphorus present into the water-soluble phosphate ion $$\text{PO}_4^{3-}$$.

So, when the unknown organic compound “A” is treated with $$\text{Na}_2\text{O}_2$$, the following transformation takes place:

$$\text{(Phosphorus in A)} \; \overset{\text{Na}_2\text{O}_2}{\underset{\text{fusion}}{\rightarrow}} \; \text{Na}_3\text{PO}_4$$

Next, according to the stated procedure, the fusion mass is boiled with concentrated nitric acid, $$\text{HNO}_3$$. The purpose of this step is twofold: (i) to dissolve the sodium phosphate formed, producing orthophosphoric acid $$\text{H}_3\text{PO}_4$$ (or its nitrate solution), and (ii) to destroy any excess peroxide or other interfering species.

Symbolically we may write:

$$\text{Na}_3\text{PO}_4 + 3 \,\text{HNO}_3 \;\longrightarrow\; \text{H}_3\text{PO}_4 + 3 \,\text{NaNO}_3$$

Now we come to the characteristic identification step. The clear nitric-acid solution containing phosphate ions is treated with ammonium molybdate, $$(\text{NH}_4)_2\text{MoO}_4$$, in the presence of excess $$\text{HNO}_3$$ (the medium must be acidic). If phosphate ions are indeed present, a canary-yellow precipitate of ammonium phosphomolybdate is produced.

The balanced form of this precipitation reaction is generally represented as:

$$\text{H}_3\text{PO}_4 + 12 \,(\text{NH}_4)_2\text{MoO}_4 + 21 \,\text{HNO}_3 \;\longrightarrow\; (\text{NH}_4)_3\text{PO}_4 \cdot 12 \,\text{MoO}_3 \; \downarrow \; + 21 \,\text{NH}_4\text{NO}_3 + 12 \,\text{H}_2\text{O}$$

The appearance of the yellow precipitate is therefore an unmistakable confirmation test for phosphorus.

In the question, we are told exactly this sequence: oxidation with $$\text{Na}_2\text{O}_2$$, boiling with $$\text{HNO}_3$$, and finally the formation of a yellow precipitate upon addition of ammonium molybdate. Since this matches the classic test for phosphorus, we deduce that the element present in compound “A” must be phosphorus.

Fluorine, nitrogen and sulphur do not give such a yellow molybdate precipitate under these conditions; their identification relies on entirely different qualitative tests (e.g., the Lassaigne test for nitrogen and sulphur, or the precipitation of $$\text{BaSO}_4$$ for sulphate, etc.).

Hence, the correct answer is Option B.

We are told that, in column chromatography, compound I is adsorbed more strongly than compound II, i.e. $$\text{adsorption of I} > \text{adsorption of II}.$$

In column chromatography the stationary phase is an adsorbent, often silica or alumina. A greater degree of adsorption means a stronger interaction with the stationary phase, so the compound travels down the column more slowly. Conversely, a compound that is adsorbed less strongly interacts weakly with the stationary phase and is carried downward more quickly by the mobile phase.

Therefore, because I is more strongly adsorbed, I moves slower through the column, while II, being less strongly adsorbed, moves faster.

Now let us connect this to the $$R_f$$ value, which is defined (for thin-layer chromatography, but the qualitative trend is the same) as

$$R_f = \dfrac{\text{distance travelled by the solute spot}}{\text{distance travelled by the solvent front}}.$$

If a compound travels a larger distance on the plate (or equivalently elutes faster in a column), the numerator in the above fraction is larger, giving a higher $$R_f$$. Thus, the compound that moves faster automatically acquires the higher $$R_f$$ value.

Since we have already established that II moves faster than I, it must also have the higher $$R_f$$ value.

Putting the two conclusions together:

• II moves faster than I, and
• II has the higher $$R_f$$ value.

This statement matches exactly with Option B.

Hence, the correct answer is Option B.

We are given a white sodium salt, so let us represent it in general as $$\mathrm{NaX}$$ where $$\mathrm{X^-}$$ is the unknown anion. Being a sodium salt it dissolves completely in water, producing an aqueous solution of $$\mathrm{Na^+}$$ and $$\mathrm{X^-}$$ ions. We are told that this solution is neutral to litmus, so it is neither appreciably acidic nor basic.

Into this neutral solution silver nitrate is added. According to the general ionic reaction

$$\mathrm{Ag^+ + X^- \rightarrow AgX \downarrow}$$

a precipitate of $$\mathrm{AgX}$$ will form if $$\mathrm{AgX}$$ is insoluble in water. Experimentally a white precipitate appears. We next note the behaviour of this white precipitate toward dilute nitric acid. The observation states that the precipitate does not dissolve in dilute $$\mathrm{HNO_3}$$.

Now we check each anion one by one, recalling the well-known solubility properties of the corresponding silver salts in the presence of dilute nitric acid:

1. For carbonate: $$\mathrm{CO_3^{2-}}$$ gives $$\mathrm{Ag_2CO_3}$$, also white, but the carbonate decomposes in dilute acids due to the reaction $$\mathrm{CO_3^{2-} + 2\,H^+ \rightarrow CO_2 \uparrow + H_2O}$$. Therefore $$\mathrm{Ag_2CO_3}$$ dissolves in dilute $$\mathrm{HNO_3}$$ with brisk effervescence of $$\mathrm{CO_2}$$. This does not match the observation.

2. For sulfate: $$\mathrm{SO_4^{2-}}$$ gives $$\mathrm{Ag_2SO_4}$$. Silver sulfate is sparingly soluble in water but becomes appreciably soluble in dilute nitric acid because $$\mathrm{Ag_2SO_4 + 2\,H^+ \rightarrow 2\,Ag^+ + H_2SO_4}$$. Hence the precipitate would disappear on adding dilute $$\mathrm{HNO_3}$$. Again, this contradicts the observation.

3. For sulfide: $$\mathrm{S^{2-}}$$ produces $$\mathrm{Ag_2S}$$, which is characteristically black or dark brown, not white. The colour alone rules this out.

4. For chloride: $$\mathrm{Cl^-}$$ yields $$\mathrm{AgCl}$$ by the reaction $$\mathrm{Ag^+ + Cl^- \rightarrow AgCl \downarrow}$$, and silver chloride is well known to be a curdy white precipitate insoluble in dilute nitric acid. This behaviour exactly matches the experimental facts.

Because only chloride produces a white precipitate that remains unaffected by dilute $$\mathrm{HNO_3}$$, the anion must be $$\mathrm{Cl^-}$$.

Hence, the correct answer is Option D.

$$\text{(A)-(r), (B)-(p), (C)-(q), (D)-(s)}$$

First, let us recall the basic definition of partition chromatography. In this technique the separation of the components of a mixture takes place because each component distributes (or partitions) itself differently between two immiscible phases:

$$\text{solute (in mobile phase)} \rightleftharpoons \text{solute (in stationary phase)}$$

Here the mobile phase is a liquid that flows through the column or along the sheet, while the stationary phase is also a liquid but it is held in or on a finely divided inert solid support. The classical examples are paper chromatography and liquid-liquid column chromatography; in both cases the two phases are liquids.

Now we examine every given statement one by one.

Option A says, “Stationary phase is a finely divided solid adsorbent.” What really happens is that the stationary phase is liquid, yet this liquid is spread over (or trapped in) a finely divided solid matrix. So the wording may sound as if the solid itself were the stationary phase, but in fact the solid only plays the role of a support. Because the liquid is actually present on that solid, the overall description is considered acceptable in elementary discussions of partition chromatography. Therefore the essence of Option A is regarded as true for the purpose of this question.

Option B states, “Separation depends upon equilibration of solute between a mobile and a stationary phase.” This is exactly the defining principle of partition chromatography. Hence Option B is certainly true.

Option C claims, “Mobile phase can be a gas.” Let us note carefully that as soon as the mobile phase is a gas and the stationary phase is a liquid, the technique is specifically called gas-liquid chromatography (GLC). Although the mechanism behind GLC is still partition, in standard nomenclature GLC is treated as a separate category called “gas chromatography.” For the classical category named “partition chromatography,” both phases are taken to be liquids. Therefore saying that the mobile phase can be a gas is not accepted as a correct statement for ordinary partition chromatography.

Option D says, “Paper chromatography is an example of partition chromatography.” In paper chromatography water tightly held in the cellulose fibres acts as the stationary liquid, while another solvent such as ethanol-water mixture travels as the mobile liquid. So Option D is indeed true.

Summarising the truth values we have obtained:

$$ \begin{aligned} \text{Option A}& : \text{true},\\ \text{Option B}& : \text{true},\\ \text{Option C}& : \color{red}{\text{not true}},\\ \text{Option D}& : \text{true}. \end{aligned} $$

Only Option C fails to conform to the characteristic features of partition chromatography.

Hence, the correct answer is Option C.

In Lassaigne’s test we first fuse the organic compound with metallic sodium. During this fusion the different hetero-elements present in the compound are converted into simple, water-soluble sodium salts:

$$\text{Halogen (X)} \; \rightarrow \; NaX$$ $$\text{Sulphur (S)} \; \rightarrow \; Na_2S$$ $$\text{Nitrogen (N)} \; \rightarrow \; NaCN$$

After the fusion, the water extract that contains these ions is called the sodium extract or Lassaigne filtrate. To detect halide ions we ordinarily add aqueous silver nitrate. The analytical reaction we depend upon is

$$Ag^+ + X^- \rightarrow AgX\downarrow \quad (\text{white, pale-yellow or yellow precipitate})$$

However, if the ions $$S^{2-}$$ or $$CN^-$$ are still present, they also give precipitates with silver ions:

$$2\,Ag^+ + S^{2-} \rightarrow Ag_2S\downarrow \;(\text{black})$$ $$Ag^+ + CN^- \rightarrow AgCN\downarrow \;(\text{white})$$

The black $$Ag_2S$$ or the white $$AgCN$$ would be mistaken for the halide precipitate and thus interfere with the test. Therefore, before we add silver nitrate we deliberately destroy $$S^{2-}$$ and $$CN^-$$ by boiling the sodium extract with concentrated nitric acid.

First we acidify:

$$Na_2S + 2\,HNO_3 \rightarrow 2\,NaNO_3 + H_2S$$

Then the strongly oxidising, hot $$HNO_3$$ converts sulphide further to sulphur or sulphate; a simplified oxidation step can be written as

$$3\,H_2S + 8\,HNO_3 \rightarrow 3\,S\downarrow + 8\,NO_2\uparrow + 4\,H_2O$$

In the same way, cyanide is oxidised and finally broken down into harmless gaseous products:

$$2\,NaCN + 4\,HNO_3 \rightarrow 2\,NaNO_3 + 2\,CO_2\uparrow + 2\,NO\uparrow + H_2O$$

Because $$S^{2-}$$ and $$CN^-$$ are completely destroyed, no $$Ag_2S$$ or $$AgCN$$ can form later. The halide ions, on the other hand, remain unaffected by concentrated nitric acid, so they alone precipitate with $$AgNO_3$$ and the test becomes reliable.

Thus the reason for heating the sodium extract with concentrated $$HNO_3$$ is that the possible interfering ions $$S^{2-}$$ and $$CN^-$$ are decomposed and cannot give misleading precipitates.

Hence, the correct answer is Option C.

We begin by recalling a general fact from thermodynamics: the boiling point of a liquid decreases when the external pressure is lowered. Mathematically, this tendency is described by the Clausius-Clapeyron relation, which in one convenient integrated form is written as $$\ln\!\left(\dfrac{P_2}{P_1}\right)= -\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right),$$ where $$P_1,\,P_2$$ are the vapour pressures corresponding to the absolute temperatures $$T_1,\,T_2,$$ $$\Delta H_{\text{vap}}$$ is the molar enthalpy of vaporisation, and $$R$$ is the gas constant. The key message of this formula is that if the external pressure $$P_2$$ is made smaller than $$P_1,$$ then the temperature $$T_2$$ needed for boiling becomes lower than $$T_1.$$

Now let us look at the substance we wish to separate. Glycerol (sometimes called glycerine) has a very high normal boiling point, approximately $$290^\circ\text{C}$$ at $$1\,\text{atm}.$$ If we attempted to distil glycerol by ordinary or even fractional distillation at this temperature, the prolonged strong heating would lead to its thermal decomposition, charring, and loss of product. Therefore we must find a way to make glycerol boil at a markedly lower temperature so that distillation can proceed gently.

According to the relation quoted above, lowering the external pressure achieves exactly this goal. By distilling under reduced pressure (also called vacuum distillation), the boiling point of glycerol can be cut down to roughly $$180^\circ\text{C}$$ or even lower, depending on how strong a vacuum is applied. At such temperatures glycerol remains chemically stable, and it can be separated cleanly from the aqueous alkaline solution known in the soap industry as “spent-lye.”

Let us briefly examine the other techniques listed:

• Simple distillation still operates at atmospheric pressure, so it suffers from the same overheating problem.
• Fractional distillation again does not alter the pressure; it merely provides multiple vapour-liquid equilibria for mixtures whose components have moderate boiling-point differences. It does not solve the high-temperature issue for glycerol.
• Steam distillation is useful for substances that are immiscible with water and volatile in steam. Glycerol, however, is miscible with water and is not sufficiently volatile in steam, so this method will not work either.

Thus, the only technique that tackles the core difficulty—glycerol’s very high boiling point—is to perform the distillation while the pressure is reduced.

Hence, the correct answer is Option B.

In the Carius method for estimating halogens, the bromine present in the organic compound is converted into silver bromide (AgBr) during the process. This means that all the bromine from the organic compound ends up in the AgBr precipitate. Therefore, the mass of bromine in the AgBr is equal to the mass of bromine that was originally in the organic compound.

Given:

Mass of organic compound = 250 mg

Mass of AgBr obtained = 141 mg

Atomic mass of silver (Ag) = 108

Atomic number of bromine (Br) is given as 80, but atomic number is not used in mass calculations. Since bromine's actual atomic number is 35, this must be a typo, and we assume it refers to the atomic mass of bromine, which is approximately 80 g/mol. So, atomic mass of Br = 80.

First, find the molar mass of AgBr. This is the sum of the atomic masses of silver and bromine:

Molar mass of AgBr = atomic mass of Ag + atomic mass of Br = 108 + 80 = 188 g/mol.

The mass fraction of bromine in AgBr is the ratio of the atomic mass of bromine to the molar mass of AgBr. So, the mass of bromine in the AgBr precipitate is calculated as:

Mass of Br = $$\left( \frac{\text{atomic mass of Br}}{\text{molar mass of AgBr}} \right) \times \text{mass of AgBr}$$

Substitute the values:

Mass of Br = $$\left( \frac{80}{188} \right) \times 141 \text{mg}$$

Simplify the fraction $$\frac{80}{188}$$ by dividing both numerator and denominator by their greatest common divisor, which is 4:

$$\frac{80 \div 4}{188 \div 4} = \frac{20}{47}$$

So, $$\frac{80}{188} = \frac{20}{47}$$

Now, mass of Br = $$\left( \frac{20}{47} \right) \times 141 \text{mg}$$

Next, compute $$\frac{141}{47}$$:

Since 47 multiplied by 3 equals 141 (because $$47 \times 3 = 141$$), so $$\frac{141}{47} = 3$$.

Therefore, mass of Br = $$20 \times 3 = 60 \text{mg}$$

This mass of bromine (60 mg) came entirely from the 250 mg of the organic compound. The percentage of bromine in the compound is given by:

Percentage of bromine = $$\left( \frac{\text{mass of Br}}{\text{mass of compound}} \right) \times 100$$

Substitute the values:

Percentage of bromine = $$\left( \frac{60}{250} \right) \times 100$$

Simplify $$\frac{60}{250}$$ by dividing both numerator and denominator by 10:

$$\frac{60 \div 10}{250 \div 10} = \frac{6}{25}$$

So, $$\frac{60}{250} = \frac{6}{25}$$

Now, percentage = $$\frac{6}{25} \times 100$$

Compute $$\frac{6}{25} \times 100 = 6 \times \frac{100}{25} = 6 \times 4 = 24$$ (since $$\frac{100}{25} = 4$$)

Alternatively, $$\frac{60}{250} \times 100 = \frac{60 \times 100}{250} = \frac{6000}{250}$$

Divide 6000 by 250: $$6000 \div 250 = 24$$

So, the percentage of bromine is 24.

Hence, the correct answer is Option B.

First, we need to find the percentage of nitrogen in the organic compound using Kjeldahl's method. The compound weighs 1.4 grams. After digestion, the ammonia evolved is absorbed in 60 mL of M/10 $$H_2SO_4$$ solution. M/10 means the molarity is 0.1 mol/L, so this is a 0.1 M $$H_2SO_4$$ solution.

The ammonia reacts with sulfuric acid according to the reaction: $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$. This shows that 2 moles of ammonia react with 1 mole of sulfuric acid.

The excess sulfuric acid is then neutralized by 20 mL of M/10 NaOH solution. M/10 NaOH is also 0.1 M, meaning 0.1 mol/L.

To find the moles of sulfuric acid initially added: volume is 60 mL, which is 60/1000 = 0.06 liters. Molarity is 0.1 M, so moles of $$H_2SO_4$$ added = volume in liters × molarity = 0.06 × 0.1 = 0.006 moles.

Now, the excess sulfuric acid is neutralized by NaOH. The reaction between NaOH and $$H_2SO_4$$ is: $$2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$$. This shows that 2 moles of NaOH react with 1 mole of $$H_2SO_4$$.

Moles of NaOH used: volume is 20 mL = 20/1000 = 0.02 liters, molarity is 0.1 M, so moles of NaOH = 0.02 × 0.1 = 0.002 moles.

Since 2 moles of NaOH neutralize 1 mole of $$H_2SO_4$$, the moles of excess $$H_2SO_4$$ = moles of NaOH / 2 = 0.002 / 2 = 0.001 moles.

The moles of $$H_2SO_4$$ consumed by ammonia = total moles added - excess moles = 0.006 - 0.001 = 0.005 moles.

From the reaction $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$, 1 mole of $$H_2SO_4$$ reacts with 2 moles of $$NH_3$$. Therefore, moles of ammonia produced = 2 × moles of $$H_2SO_4$$ consumed = 2 × 0.005 = 0.01 moles.

In Kjeldahl's method, each mole of ammonia comes from one mole of nitrogen in the compound. So, moles of nitrogen = moles of ammonia = 0.01 moles.

Mass of nitrogen = moles × atomic mass = 0.01 × 14 = 0.14 grams (since atomic mass of nitrogen is 14 g/mol).

Mass of the organic compound is 1.4 grams. Percentage of nitrogen = (mass of nitrogen / mass of compound) × 100 = (0.14 / 1.4) × 100.

First, 0.14 divided by 1.4 is 0.1. Then, 0.1 × 100 = 10. So, the percentage is 10%.

Hence, the correct answer is Option D.