JEE Basic Concepts in Chemistry Questions

The osmotic pressure $$ \pi $$ exerted by the solution equals the hydrostatic pressure of a liquid column of height $$ h $$.

Hydrostatic pressure formula: $$\pi = \rho g h$$

Given data (convert to SI units):
  • Density, $$ \rho = 1.00 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3} $$
  • Acceleration due to gravity, $$ g = 10 \text{ m s}^{-2} $$
  • Height, $$ h = 2.00 \text{ cm} = 0.02 \text{ m} $$

Hence
$$\pi = 1000 \times 10 \times 0.02 = 200 \text{ Pa}$$

For an ideal dilute solution, van’t Hoff equation applies:
$$\pi = c R T$$

Here $$ c $$ is the molar concentration (mol m$$^{-3}$$). The solution contains 2.00 g of solute per dm$$^{3}$$.
1 dm$$^{3} = 0.001 \text{ m}^{3}$$, so mass concentration is
$$\frac{2.00 \text{ g}}{0.001 \text{ m}^{3}} = 2000 \text{ g m}^{-3}$$

If the molar mass of the macromolecule is $$ M \text{ g mol}^{-1} $$, then
$$c = \frac{2000}{M} \text{ mol m}^{-3}$$

Substituting $$ c $$ into van’t Hoff equation:
$$\pi = \frac{2000}{M} R T$$

Solve for $$ M $$:
$$M = \frac{2000 R T}{\pi}$$

Insert the given constants $$ R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} $$ and $$ T = 300 \text{ K} $$:
$$M = \frac{2000 \times 8.3 \times 300}{200}$$

Calculate step by step:
$$8.3 \times 300 = 2490$$
$$2000 \times 2490 = 4\,980\,000$$
$$\frac{4\,980\,000}{200} = 24\,900$$

Thus
$$M \approx 24\,900 \text{ g mol}^{-1} = 2.49 \times 10^{4} \text{ g mol}^{-1}$$

Comparing with the required form $$ X \times 10^{4} \text{ g mol}^{-1} $$, we have $$ X \approx 2.49 $$.

Therefore, the value of $$ X $$ lies in the required range 2.4 - 2.55.

Mass of HNO₃ = 30 g. In 75% solution: mass of solution = 30/0.75 = 40 g. Volume = 40/1.25 = 32 mL.

The correct answer is Option 2: 32.

3 M NaCl solution with density 1.25 g/mL.

Determine the mass of solute in 1 L of solution

Molarity = 3 M means 3 moles of NaCl per litre of solution.

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

Mass of NaCl = 3 × 58.5 = 175.5 g.

Determine the mass of 1 L of solution

Mass of solution = Volume × Density = 1000 mL × 1.25 g/mL = 1250 g.

Calculate mass of solvent

Mass of solvent (water) = 1250 - 175.5 = 1074.5 g = 1.0745 kg.

Calculate molality

$$\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3}{1.0745} \approx 2.79 \text{ m}$$

The answer is Option 2: 2.79 m.

The number of atoms present in a given sample is obtained from the relation
$$\text{Number of atoms} = \frac{\text{mass of sample}}{\text{molar mass}} \times N_A$$
where $$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$$ is Avogadro’s constant.

In this question every element is taken in the same mass, $$m = 10^{-9}\, \text{g}$$. Since $$N_A$$ and $$m$$ are common to all, the factor that decides the number of atoms is only the molar mass $$M$$. Smaller $$M$$ ⇒ larger $$\dfrac{m}{M}$$ ⇒ larger number of atoms.

Approximate molar masses of the given elements are:
$$M_{\text{Pb}} \approx 207\ \text{g mol}^{-1}$$
$$M_{\text{Po}} \approx 209\ \text{g mol}^{-1}$$
$$M_{\text{Pr}} \approx 141\ \text{g mol}^{-1}$$
$$M_{\text{Pt}} \approx 195\ \text{g mol}^{-1}$$

Comparing their molar masses:
$$141\lt 195\lt 207\lt 209$$

Praseodymium (Pr) has the smallest molar mass, so for the same sample mass $$10^{-9}\, \text{g}$$ it provides the largest number of atoms.

Therefore, the element with the highest number of atoms is Praseodymium (Pr).
Answer: Option B (Pr)

The balanced chemical equation is $$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+CO_2(g)+H_2O(l)$$.

Step 1: Calculate moles of each reactant.

Molarity formula: $$M = \frac{\text{moles}}{\text{volume in L}} \; \Rightarrow \; \text{moles} = M \times V$$.

For $$HCl$$:
Volume $$V = 250\ \text{mL} = 0.250\ \text{L}$$,
Molarity $$M = 0.76\ \text{mol L}^{-1}$$.
Therefore, moles of $$HCl = 0.76 \times 0.250 = 0.19\ \text{mol}$$.

Molar mass of $$CaCO_3$$:
$$40 + 12 + 3(16) = 40 + 12 + 48 = 100\ \text{g mol}^{-1}$$.

Given mass = $$1000\ \text{g}$$, so moles of $$CaCO_3$$ are
$$\frac{1000}{100} = 10\ \text{mol}$$.

Step 2: Identify the limiting reagent.

From the equation, $$2$$ moles of $$HCl$$ react with $$1$$ mole of $$CaCO_3$$.
Required moles of $$HCl$$ for $$10$$ moles $$CaCO_3$$:
$$2 \times 10 = 20\ \text{mol}$$.

Available moles of $$HCl$$ = $$0.19\ \text{mol} \lt 20\ \text{mol}$$, so $$HCl$$ is the limiting reagent.

Step 3: Moles of $$CaCl_2$$ formed.

Stoichiometry: $$2\ \text{mol HCl} \rightarrow 1\ \text{mol CaCl}_2$$.
Thus, moles of $$CaCl_2 = \frac{0.19}{2} = 0.095$$.

Step 4: Mass of $$CaCl_2$$ produced.

Molar mass of $$CaCl_2$$:
$$40 + 2(35.5) = 40 + 71 = 111\ \text{g mol}^{-1}$$.

Mass = moles $$\times$$ molar mass = $$0.095 \times 111 = 10.545\ \text{g}$$.

Final Answer: $$10.545\ \text{g}$$ of $$CaCl_2$$ will be formed. This matches Option C.

Empirical formula: C: 54.2/12=4.52, H: 9.2/1=9.2, O: 36.6/16=2.29.

Ratio: 4.52/2.29 : 9.2/2.29 : 2.29/2.29 = 1.97 : 4.02 : 1 ≈ 2:4:1

Empirical formula: C₂H₄O. Empirical mass = 44. Molar mass = 132 = 3×44.

Molecular formula: C₆H₁₂O₃.

The correct answer is Option 4: C₆H₁₂O₃.

Ksp for Cr(OH)₃ = 1.6×10⁻³⁰. Find molar solubility.

$$Cr(OH)_3 \rightleftharpoons Cr^{3+} + 3OH^-$$

If solubility = s: $$K_{sp} = s(3s)^3 = 27s^4$$

$$s^4 = \frac{1.6 \times 10^{-30}}{27}$$

$$s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$$

The correct answer is Option 3.

The balanced reaction of magnesium with dilute hydrochloric acid is
$$Mg + 2\,HCl \rightarrow MgCl_2 + H_2$$

From the equation, $$1$$ mole of $$Mg$$ produces $$1$$ mole of $$H_2$$.

Step 1 - Convert volume of $$H_2$$ to moles
At STP, $$1$$ mole of any ideal gas occupies $$22.4$$ L $$= 22\,400$$ mL.
Given volume of $$H_2$$ $$= 220$$ mL.

Number of moles of $$H_2$$ produced
$$n_{H_2}= \frac{220 \text{ mL}}{22\,400 \text{ mL mol}^{-1}}$$

$$n_{H_2}= 9.821\times10^{-3} \text{ mol}$$ (keep four significant figures).

Step 2 - Relate moles of $$H_2$$ to moles of $$Mg$$
The stoichiometry is $$1:1$$.
Therefore,
$$n_{Mg}= n_{H_2}= 9.821\times10^{-3} \text{ mol}$$.

Step 3 - Calculate mass of magnesium
Molar mass of $$Mg = 24 \text{ g mol}^{-1}$$.

Mass of $$Mg$$ required
$$m_{Mg}= n_{Mg}\times M_{Mg}= (9.821\times10^{-3})\times24$$

$$m_{Mg}= 2.357\times10^{-1} \text{ g}= 0.2357 \text{ g}$$.

Convert grams to milligrams:
$$0.2357 \text{ g}= 235.7 \text{ mg}\approx 236 \text{ mg}$$.

Hence, the mass of magnesium needed is approximately $$236$$ mg.
Correct option: Option C.

The combustion analysis gives the following data:

Mass of organic compound burnt $$= 0.210 \text{ g}$$
Mass of $$CO_2$$ formed $$= 0.307 \text{ g}$$
Mass of $$H_2O$$ formed $$= 0.127 \text{ g}$$

Step 1: Calculate mass of carbon in the sample.

Molar mass of $$CO_2 = 44 \text{ g mol}^{-1}$$ and each mole contains $$12 \text{ g}$$ of C.
Therefore,

$$\text{Mass of C} = 0.307 \times \frac{12}{44} \text{ g}$$

$$\text{Mass of C} = 0.08372 \text{ g}$$

Step 2: Calculate mass of hydrogen in the sample.

Molar mass of $$H_2O = 18 \text{ g mol}^{-1}$$ and each mole contains $$2 \text{ g}$$ of H.
Therefore,

$$\text{Mass of H} = 0.127 \times \frac{2}{18} \text{ g}$$

$$\text{Mass of H} = 0.01411 \text{ g}$$

Step 3: Calculate mass of oxygen in the sample.

Total mass of C and H in the sample:

$$0.08372 + 0.01411 = 0.09783 \text{ g}$$

Hence,

$$\text{Mass of O} = 0.210 - 0.09783 = 0.11217 \text{ g}$$

Step 4: Convert masses to percentages.

Percentage of hydrogen:

$$\%H = \frac{0.01411}{0.210} \times 100 = 6.72\%$$

Percentage of oxygen:

$$\%O = \frac{0.11217}{0.210} \times 100 = 53.41\%$$

Step 5: State the result.

The percentages of hydrogen and oxygen in the organic compound are $$6.72\%$$ and $$53.41\%$$ respectively.

Thus, the correct choice is Option B.

We need to find the initial mass of $$CO_2$$.

After removing $$10^{21}$$ molecules from the sample, $$2.8 \times 10^{-3}$$ mol of $$CO_2$$ remains.

The moles of $$CO_2$$ removed can be calculated as $$ n_{removed} = \frac{10^{21}}{6.02 \times 10^{23}} = \frac{1}{602} = 1.66 \times 10^{-3} \text{ mol} $$.

Adding this to the remaining moles gives the total initial moles: $$ n_{initial} = 2.8 \times 10^{-3} + 1.66 \times 10^{-3} = 4.46 \times 10^{-3} \text{ mol} $$.

The molar mass of $$CO_2$$ is $$12 + 2(16) = 44$$ g/mol.

Therefore, the initial mass is $$ x = 4.46 \times 10^{-3} \times 44 = 0.19624 \text{ g} = 196.2 \text{ mg} $$.

The correct answer is Option 3: 196.2 mg.

We need to identify which statements are correct among (A) through (E).

(A) "Weight of a substance is the amount of matter present in it."

This is INCORRECT. The amount of matter present in a substance is its mass, not weight. Weight is the force exerted by gravity on an object ($$W = mg$$).

(B) "Mass is the force exerted by gravity on an object."

This is INCORRECT. The force exerted by gravity is weight, not mass. Mass is an intrinsic property measuring the amount of matter.

(C) "Volume is the amount of space occupied by a substance."

This is CORRECT. This is the standard definition of volume.

(D) "Temperatures below 0°C are possible in Celsius scale, but in Kelvin scale negative temperature is not possible."

This is CORRECT. The Kelvin scale starts at absolute zero (0 K = -273.15°C), so negative Kelvin temperatures are not physically meaningful in classical thermodynamics.

(E) "Precision refers to the closeness of various measurements for the same quantity."

This is CORRECT. Precision measures the reproducibility of measurements (how close repeated measurements are to each other), as opposed to accuracy (closeness to the true value).

Correct statements: (C), (D), and (E).

The correct answer is Option B: (C), (D) and (E) Only.

During combustion, all carbon in the compound converts to $$CO_2$$ and all hydrogen converts to $$H_2O$$. We first calculate the moles of each element obtained from the given products.

Step 1: Moles of carbon
Mass of $$CO_2 = 1.46 \text{ g}$$. Molar mass of $$CO_2 = 44 \text{ g mol}^{-1}$$.
$$\text{Moles of }CO_2 = \frac{1.46}{44} = 0.03318 \text{ mol}$$.
Each mole of $$CO_2$$ contains one mole of C, so $$\text{Moles of C} = 0.03318 \text{ mol}$$.

Step 2: Mass of carbon
$$\text{Mass of C} = 0.03318 \times 12 = 0.398 \text{ g}$$.

Step 3: Moles of hydrogen
Mass of $$H_2O = 0.567 \text{ g}$$. Molar mass of $$H_2O = 18 \text{ g mol}^{-1}$$.
$$\text{Moles of }H_2O = \frac{0.567}{18} = 0.0315 \text{ mol}$$.
Each mole of $$H_2O$$ contains two moles of H, so $$\text{Moles of H} = 2 \times 0.0315 = 0.0630 \text{ mol}$$.

Step 4: Mass of hydrogen
$$\text{Mass of H} = 0.0630 \times 1 = 0.063 \text{ g}$$.

Step 5: Mass and moles of oxygen in the compound
Total mass of sample = 1.00 g.
Mass of C + H already determined = $$0.398 + 0.063 = 0.461 \text{ g}$$.
Remaining mass is oxygen: $$\text{Mass of O} = 1.00 - 0.461 = 0.539 \text{ g}$$.
Moles of O: $$\text{Moles of O} = \frac{0.539}{16} = 0.0337 \text{ mol}$$.

Step 6: Find simplest mole ratio
Divide each mole value by the smallest, $$0.03318$$:

$$\frac{\text{C}}{0.03318} = 1.00$$
$$\frac{\text{H}}{0.03318} = \frac{0.0630}{0.03318} \approx 1.90 \approx 2$$
$$\frac{\text{O}}{0.03318} = \frac{0.0337}{0.03318} \approx 1.02 \approx 1$$

Hence the empirical formula is $$CH_2O$$.

Step 7: Empirical formula mass
$$\text{E.F. mass} = 12 + 2(1) + 16 = 30 \text{ g mol}^{-1}$$.

Therefore, the empirical formula mass of compound (X) is $$30 \text{ g}$$.

Option A is correct.

The reaction: $$5I^- + IO_3^- + 6H^+ \to 3I_2 + 3H_2O$$

0.1 M KI solution: moles of $$I^-$$ in 200 mL = $$0.1 \times 0.2 = 0.02$$ mol.

(A) 200 mL of KI has 0.02 mol $$I^-$$. From stoichiometry, 5 mol $$I^-$$ reacts with 1 mol $$IO_3^-$$.

Moles of $$KIO_3 = 0.02/5 = 0.004$$ mol. ✓

(B) 200 mL has 0.02 mol $$I^-$$. From stoichiometry, 5 mol $$I^-$$ reacts with 6 mol $$H^+$$.

Moles of $$H^+ = 0.02 \times 6/5 = 0.024$$ mol. Moles of $$H_2SO_4 = 0.024/2 = 0.012$$ mol ≠ 0.006. ✗

(C) 0.5 L of KI has $$0.1 \times 0.5 = 0.05$$ mol $$I^-$$. From stoichiometry, 5 mol $$I^-$$ produces 3 mol $$I_2$$.

Moles of $$I_2 = 0.05 \times 3/5 = 0.03$$ mol ≠ 0.005. ✗

(D) In this reaction, $$IO_3^-$$ goes from I(+5) to I₂(0), gaining 5 electrons per I atom.

Equivalent weight = Molecular weight / 5. ✓

The correct answer is Option 1: (A) and (D) only.

0.01 mole of compound X containing 10% hydrogen produces 0.9 g H₂O on combustion. Find molar mass.

Find mass of hydrogen in the water produced

Mass of H in H₂O = $$\frac{2}{18} \times 0.9 = 0.1$$ g

Find mass of compound X

Since X contains 10% hydrogen:

Mass of H = 10% of mass of X

$$0.1 = 0.10 \times m_X$$

$$m_X = 1$$ g

Find molar mass

0.01 mol of X has mass 1 g:

$$M = \frac{1}{0.01} = 100$$ g/mol

The answer is 100.

We need to find the mass of benzoic acid that produces $$CO_2$$ occupying 11.2 L at STP.

Benzoic acid reacts with aqueous sodium bicarbonate: $$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$$ The stoichiometry shows that 1 mole of benzoic acid produces 1 mole of $$CO_2$$.

At STP (Standard Temperature and Pressure: 0 degC, 1 atm), 1 mole of any ideal gas occupies 22.4 L. $$\text{Moles of } CO_2 = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ mol}$$

From the 1:1 stoichiometry: $$\text{Moles of benzoic acid} = \text{Moles of } CO_2 = 0.5 \text{ mol}$$

The molar mass of benzoic acid ($$C_6H_5COOH = C_7H_6O_2$$) is $$(7 \times 12) + (6 \times 1) + (2 \times 16) = 84 + 6 + 32 = 122 \text{ g/mol}$$

$$X = 0.5 \times 122 = 61 \text{ g}$$

The answer is 61 g.

We need to find the final concentration when 20 mL of 2 M NaOH is added to 400 mL of 0.5 M NaOH.

When two solutions of the same solute are mixed, the total moles of solute are conserved:

$$C_{final} = \frac{n_1 + n_2}{V_{total}}$$

where $$n = C \times V$$ (moles = concentration times volume).

Solution 1: $$n_1 = C_1 \times V_1 = 2 \times 20 = 40$$ mmol (using mL for volume)

Solution 2: $$n_2 = C_2 \times V_2 = 0.5 \times 400 = 200$$ mmol

$$V_{total} = 20 + 400 = 420 \text{ mL}$$

$$C_{final} = \frac{n_1 + n_2}{V_{total}} = \frac{40 + 200}{420} = \frac{240}{420} = \frac{4}{7} \text{ M}$$

$$C_{final} = \frac{4}{7} \approx 0.5714 \text{ M} = 57.14 \times 10^{-2} \text{ M}$$

Rounding to the nearest integer: $$57 \times 10^{-2}$$ M.

The answer is 57.

The reaction: $$Fe_3O_4(s) + 4CO(g) \to 3Fe(l) + 4CO_2(g)$$

Given: $$2.32 \times 10^3$$ kg $$Fe_3O_4$$ and $$2.8 \times 10^2$$ kg CO.

Moles of $$Fe_3O_4 = \frac{2.32 \times 10^6 \text{ g}}{232 \text{ g/mol}} = 10^4$$ mol.

Moles of CO $$= \frac{2.8 \times 10^5 \text{ g}}{28 \text{ g/mol}} = 10^4$$ mol.

From stoichiometry: 1 mol $$Fe_3O_4$$ needs 4 mol CO.

$$10^4$$ mol $$Fe_3O_4$$ needs $$4 \times 10^4$$ mol CO, but only $$10^4$$ mol CO is available.

CO is the limiting reagent.

From 4 mol CO, we get 3 mol Fe.
From $$10^4$$ mol CO: Fe produced $$= \frac{3}{4} \times 10^4 = 7500$$ mol.

Mass of Fe $$= 7500 \times 56 = 420000$$ g $$= 420$$ kg.

The answer is 420.

Mass of organic compound taken: $$m_{\text{compound}} = 292 \text{ mg} = 0.292 \text{ g}$$.

Volume of $$N_2$$ collected: $$V = 50 \text{ mL} = 0.050 \text{ L}$$.
Temperature: $$T = 300 \text{ K}$$.
Total pressure: $$P_{\text{total}} = 715 \text{ mm Hg}$$.
Aqueous tension (vapour pressure of water): $$P_{\text{H}_2O} = 15 \text{ mm Hg}$$.

First remove the vapour-pressure contribution to get the pressure of dry nitrogen.
$$P_{N_2} = P_{\text{total}} - P_{\text{H}_2O} = 715 - 15 = 700 \text{ mm Hg}$$.

Convert this pressure into atmospheres (1 atm = 760 mm Hg):
$$P_{N_2} = \frac{700}{760} \text{ atm} = 0.9211 \text{ atm}$$.

Use the ideal-gas equation $$PV = nRT$$. Take $$R = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}$$.

$$n = \frac{P V}{R T} = \frac{0.9211 \times 0.050}{0.0821 \times 300} \text{ mol}$$.

$$n = 0.00187 \text{ mol of } N_2$$.

Molar mass of $$N_2$$ is $$28 \text{ g mol}^{-1}$$, so the mass of nitrogen obtained is
$$m_N = n \times 28 = 0.00187 \times 28 = 0.0524 \text{ g} = 52.4 \text{ mg}$$.

Percentage of nitrogen in the organic compound:
$$\%N = \frac{m_N}{m_{\text{compound}}} \times 100 = \frac{0.0524}{0.292} \times 100 = 17.94 \% \approx 18 \%$$.

Hence, the percentage composition of nitrogen in the organic compound is 18 %.

The mineral limestone is almost entirely $$CaCO_3$$. When limestone is heated it undergoes thermal decomposition (calcination):

$$CaCO_3 \rightarrow CaO + CO_2$$ $$-(1)$$

Step 1 : Calculate the mass of pure $$CaCO_3$$ present in the given limestone.

Limestone given = $$150 \text{ kg}$$, but it is only $$75\%$$ pure.
Mass of pure $$CaCO_3 = 150 \times \frac{75}{100} = 112.5 \text{ kg}$$ $$-(2)$$

Step 2 : Use stoichiometry of reaction $$(1)$$.

Molar mass data:
$$CaCO_3 : 40 + 12 + 3 \times 16 = 100 \text{ g mol}^{-1}$$
$$CaO : 40 + 16 = 56 \text{ g mol}^{-1}$$

From the balanced equation $$(1)$$, $$100 \text{ g}$$ of $$CaCO_3$$ produce $$56 \text{ g}$$ of $$CaO$$.

Therefore, the mass ratio is
$$\frac{\text{mass of }CaO}{\text{mass of }CaCO_3} = \frac{56}{100} = 0.56$$ $$-(3)$$

Step 3 : Find the mass of $$CaO$$ obtained from the pure $$CaCO_3$$ in $$2$$.

Mass of $$CaO = 112.5 \times 0.56 = 63.0 \text{ kg}$$ $$-(4)$$

Hence, the amount of calcium oxide produced (nearest integer) = 63 kg.

Find the molarity of a 70% (mass/mass) aqueous solution of a monobasic acid with density 1.25 g/mL and molar mass 70 g/mol.

Assuming 100 g of solution, the mass of acid is 70% of 100 g = 70 g, and the mass of water is 100 − 70 = 30 g.

Using the density, the volume of the solution is given by:

$$ V = \frac{\text{mass}}{\text{density}} = \frac{100 \text{ g}}{1.25 \text{ g/mL}} = 80 \text{ mL} = 0.080 \text{ L} $$

The number of moles of acid is

$$ n = \frac{70}{70} = 1 \text{ mol} $$

Thus the molarity is

$$ M = \frac{n}{V} = \frac{1}{0.080} = 12.5 \text{ M} $$

Expressing this in the required form gives $$12.5 = 125 \times 10^{-1}$$ M.

The answer is 125.

MX₂ → M²⁺ + 2X⁻. i = 1+2α where α is degree of ionization.

i = normal mass/observed mass = 164/65.6 = 2.5.

2.5 = 1+2α. α = 0.75 = 75%.

The answer is 75.

Molar masses (in $$\mathrm{g\;mol^{-1}}$$): $$C_4H_{10}=58$$, $$O_2 = 32$$, $$H_2O = 18$$.

Moles of butane mixed
$$n_{C_4H_{10}}=\frac{174.0\times10^{3}}{58}=3000\;\text{mol}$$

Moles of oxygen mixed
$$n_{O_2}=\frac{320.0\times10^{3}}{32}=10000\;\text{mol}$$

Balanced equation (fractional coefficients as given) :
$$C_4H_{10}+ \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$$

Stoichiometric ratios
$$1\;\text{mol}\;C_4H_{10} \;:\; \frac{13}{2}=6.5\;\text{mol}\;O_2$$
$$1\;\text{mol}\;C_4H_{10} \;:\; 5\;\text{mol}\;H_2O$$

Oxygen needed for the available butane
$$3000 \times 6.5 = 19500\;\text{mol}\;O_2$$

Only $$10000\;\text{mol}\;O_2$$ are present, so $$O_2$$ is the limiting reagent.

From the equation, $$6.5\;\text{mol}\;O_2 \rightarrow 5\;\text{mol}\;H_2O$$, or
$$n_{H_2O} = n_{O_2}\times\frac{5}{6.5} = 10000\times\frac{5}{6.5} = 10000\times\frac{10}{13} = 7692.3077\;\text{mol}$$

Mass of water formed
$$m_{H_2O} = 7692.3077 \times 18 = 1.3846 \times 10^{5}\;\text{g} = 138.46\;\text{kg}$$

Density of liquid water $$\approx 1\;\text{kg L}^{-1}$$, therefore
$$V_{H_2O} = 138.46\;\text{kg}\times\frac{1\;\text{L}}{1\;\text{kg}} \approx 138\;\text{L}$$

Hence, the volume of water produced is about $$\mathbf{138\;L}$$.

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

The formula for percentage of sulphur is $$\% S = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of compound}} \times 100$$.

Here the molar mass of $$BaSO_4$$ is $$137 + 32 + 4(16) = 233$$ g/mol.

For a compound mass of 160 mg and a $$BaSO_4$$ precipitate of 466 mg, substitution into the formula gives $$\% S = \frac{32}{233} \times \frac{466}{160} \times 100$$.

This can be rewritten as $$= \frac{32 \times 466}{233 \times 160} \times 100$$ and then as $$= \frac{14912}{37280} \times 100$$.

Since $$\frac{466}{233} = 2$$, the calculation simplifies to $$\% S = \frac{32 \times 2}{160} \times 100 = \frac{64}{160} \times 100 = 40\%$$.

The answer is 40%.

For fortification, the required concentration is $$12\,\text{ppm}$$ of iron.

Step 1 : Relate ppm to mass.
By definition $$1\,\text{ppm} = 1\,\text{mg}$$ of solute per $$1\,\text{kg}$$ of sample. Therefore $$12\,\text{ppm} = 12\,\text{mg}$$ Fe per $$1\,\text{kg}$$ wheat.

Step 2 : Calculate grams of iron needed for $$150\,\text{kg}$$ wheat.
Required Fe = $$12\,\text{mg kg}^{-1} \times 150\,\text{kg}$$ $$= 1800\,\text{mg} = 1.8\,\text{g}$$

Step 3 : Find molar mass of $$FeSO_4\cdot7H_2O$$.

Molar mass = $$M_{Fe}+M_S+4M_O+7(2M_H+M_O)$$
$$=56 + 32 + 4(16) + 7(2\cdot1 + 16)$$ $$=56 + 32 + 64 + 7(18)$$ $$=56 + 32 + 64 + 126$$ $$=278\,\text{g mol}^{-1}$$

Step 4 : Determine mass fraction of iron in the salt.
Mass fraction of Fe $$= \frac{56}{278}$$ $$\approx 0.2014$$

Step 5 : Compute mass of $$FeSO_4\cdot7H_2O$$ required.
Let $$m$$ be the grams of hydrate needed.
Iron supplied by this mass = $$0.2014\,m$$ (g).
Set equal to required iron: $$0.2014\,m = 1.8$$
$$\Rightarrow m = \frac{1.8}{0.2014}$$ $$\Rightarrow m \approx 8.94\,\text{g}$$

Step 6 : Round to the nearest integer (as asked).
$$m \approx 9\,\text{g}$$

Hence, about 9 grams of $$FeSO_4\cdot7H_2O$$ are needed to fortify 150 kg of wheat to a level of 12 ppm iron.

81 g Al reacts with 128 g O₂. Find mass of Al₂O₃ produced.

Write the balanced equation

$$4Al + 3O_2 \rightarrow 2Al_2O_3$$

Find moles

Moles of Al = 81/27 = 3 mol

Moles of O₂ = 128/32 = 4 mol

Find limiting reagent

For 3 mol Al, we need $$\frac{3}{4} \times 3 = 2.25$$ mol O₂. We have 4 mol O₂.

So Al is the limiting reagent.

Calculate product

3 mol Al produces $$\frac{2}{4} \times 3 = 1.5$$ mol Al₂O₃

Mass = 1.5 × 102 = 153 g

The answer is 153.

Find the empirical formula mass of compound X with C: 14.5 %, Cl: 64.46 %, H: 1.8 %.

The total given percentages of C, Cl, and H add to 14.5 + 64.46 + 1.8 = 80.76 %, leaving 100 - 80.76 = 19.24 % for oxygen.

This corresponds to moles of each element per 100 g of compound: C: $$\frac{14.5}{12} = 1.208$$, H: $$\frac{1.8}{1} = 1.8$$, O: $$\frac{19.24}{16} = 1.2025$$, Cl: $$\frac{64.46}{35.5} = 1.816$$.

Dividing by the smallest mole value (1.2025 for O) gives the ratios C: $$1.208/1.2025 \approx 1.005 \approx 1$$, H: $$1.8/1.2025 \approx 1.497 \approx 1.5$$, O: $$1.2025/1.2025 = 1$$, Cl: $$1.816/1.2025 \approx 1.510 \approx 1.5$$.

Multiplying these ratios by 2 yields C = 2, H = 3, O = 2, Cl = 3, so the empirical formula is $$C_2H_3O_2Cl_3$$.

The empirical formula mass is calculated as $$2(12) + 3(1) + 2(16) + 3(35.5) = 24 + 3 + 32 + 106.5 = 165.5$$, which in units of $$\times 10^{-1}$$ becomes $$165.5 = 1655 \times 10^{-1}$$.

The answer is 1655.

We need to find the composition of a mixture of CaCO$$_3$$ and MgCO$$_3$$ that weighs 2.21 g and leaves a residue of 1.152 g after ignition.

When carbonates are heated (ignited), they decompose to form the corresponding metal oxides and carbon dioxide:

$$\text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2$$

$$\text{MgCO}_3 \xrightarrow{\Delta} \text{MgO} + \text{CO}_2$$

We set up variables by letting $$x$$ represent the mass of CaCO$$_3$$ and $$y$$ represent the mass of MgCO$$_3$$, and since their total mass is 2.21 g, we have

$$x + y = 2.21 \quad \text{...(1)}$$

Next, we calculate the mass of oxide produced from each carbonate.

Since the molar mass of CaCO$$_3$$ is 100, $$x$$ g of CaCO$$_3$$ yields $$\frac{56}{100}\times x$$ g of CaO, given that the molar mass of CaO is 56.

Similarly, because the molar mass of MgCO$$_3$$ is 84, $$y$$ g of MgCO$$_3$$ produces $$\frac{40}{84}\times y$$ g of MgO, with MgO having a molar mass of 40.

The total residue mass therefore leads to the second equation:

$$\frac{56}{100}x + \frac{40}{84}y = 1.152$$

which simplifies to $$0.56x + 0.4762y = 1.152 \quad \text{...(2)}$$

Substituting $$x = 2.21 - y$$ from equation (1) into equation (2) yields:

$$0.56(2.21 - y) + 0.4762y = 1.152$$

which expands to $$1.2376 - 0.56y + 0.4762y = 1.152$$

or $$1.2376 - 0.0838y = 1.152$$

Rearranging gives $$-0.0838y = 1.152 - 1.2376 = -0.0856$$,

and hence $$y = \frac{0.0856}{0.0838} = 1.0215 \approx 1.023 \, \text{g}$$

Then $$x = 2.21 - 1.023 = 1.187 \, \text{g}$$

Verification shows $$0.56 \times 1.187 + 0.4762 \times 1.023 = 0.6647 + 0.4872 = 1.1519 \approx 1.152$$, confirming consistency.

The mixture contains 1.187 g of CaCO$$_3$$ and 1.023 g of MgCO$$_3$$.

The correct answer is Option (1): 1.187 g CaCO$$_3$$ + 1.023 g MgCO$$_3$$.

Organic compound has 42.1% carbon, 6.4% hydrogen, and the remainder is oxygen, with a molecular weight of 342.

First, we calculate the percentage of oxygen: % O = 100 - 42.1 - 6.4 = 51.5%.

Next, we determine the moles of each element in one mole of the compound:

C: $$\frac{42.1}{100} \times 342 / 12 = \frac{143.98}{12} = 12.0$$ → 12 atoms

H: $$\frac{6.4}{100} \times 342 / 1 = 21.89$$ → 22 atoms

O: $$\frac{51.5}{100} \times 342 / 16 = \frac{176.13}{16} = 11.0$$ → 11 atoms

Therefore, the molecular formula is $$C_{12}H_{22}O_{11}$$ (sucrose). We can verify this by checking the molar mass: $$12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342$$, which matches the given molecular weight.

The correct answer is Option C: $$C_{12}H_{22}O_{11}$$.

We need to identify the incorrect statement about Dalton's Atomic Theory.

Dalton's Atomic Theory states:

(i) Matter consists of indivisible atoms.

(ii) All atoms of a given element have identical properties including identical mass.

(iii) Chemical reactions involve reorganization of atoms (atoms are neither created nor destroyed).

(iv) Compounds are formed when atoms of different elements combine in fixed, simple whole number ratios.

Option 1: "Chemical reactions involve reorganization of atoms" - This is correct per Dalton's theory.

Option 2: "Matter consists of indivisible atoms" - This is correct per Dalton's theory.

Option 3: "Compounds are formed when atoms of different elements combine in any ratio" - This is INCORRECT. Dalton stated atoms combine in fixed simple ratios, not "any ratio."

Option 4: Combines "any ratio" with "identical mass" - while the "any ratio" part is wrong, the "identical mass" part is actually a correct postulate of Dalton.

Option 3 directly contradicts Dalton's law of definite proportions by stating "any ratio" instead of "fixed ratio."

The correct answer is Option 3.

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. 900g glucose = 5 mol. O₂ needed = 30 mol = 960 g.

Option (3): 960.

We need to find the number of moles of methane required to produce 11 g of $$CO_2$$ after complete combustion.

Write the balanced equation for the complete combustion of methane

$$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$

From the balanced equation, 1 mole of $$CH_4$$ produces 1 mole of $$CO_2$$. This is a 1:1 molar ratio.

Calculate the moles of $$CO_2$$ produced

The molar mass of $$CO_2 = 12 + 2(16) = 44$$ g/mol.

$$\text{Moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{11}{44} = 0.25 \text{ mol}$$

Use the stoichiometric ratio to find moles of $$CH_4$$

Since the molar ratio of $$CH_4$$ to $$CO_2$$ is 1:1:

$$\text{Moles of } CH_4 = \text{Moles of } CO_2 = 0.25 \text{ mol}$$

Verification: 0.25 mol of $$CH_4 = 0.25 \times 16 = 4$$ g of methane, which upon complete combustion produces 0.25 mol of $$CO_2 = 0.25 \times 44 = 11$$ g. This matches the given data.

The correct answer is Option (4): 0.25.

We need to identify the incorrect postulates of Dalton's atomic theory.

(A) Atoms of different elements differ in mass. This is a correct postulate. ✓

(B) Matter consists of divisible atoms. Dalton stated that atoms are indivisible. So this is an incorrect statement of Dalton's theory. ✗

(C) Compounds are formed when atoms of different elements combine in a fixed ratio. This is a correct postulate. ✓

(D) All the atoms of a given element have different properties including mass. Dalton stated that all atoms of a given element are identical in all properties including mass. So this is an incorrect statement. ✗

(E) Chemical reactions involve reorganisation of atoms. This is a correct postulate. ✓

The incorrect postulates are (B) and (D).

The correct answer is Option (2): (B), (D) only.

Let the empirical formula of the oxide be $$M_xY_2O_4$$, where $$x$$ moles of metal $$M$$ are present per formula unit.

M exists in two oxidation states: $$+2$$ and $$+3$$.
Given fraction of $$M^{2+}$$ ions = $$\dfrac{1}{3}$$.
Therefore, fraction of $$M^{3+}$$ ions = $$1 - \dfrac{1}{3} = \dfrac{2}{3}$$.

Total number of $$M$$ ions = $$x$$.
Number of $$M^{2+}$$ ions = $$\dfrac{1}{3}x$$.
Number of $$M^{3+}$$ ions = $$\dfrac{2}{3}x$$.

Calculate the positive charge contributed by each metal:

Charge from $$M^{2+}$$ ions:
$$\left(\dfrac{1}{3}x\right)\times(+2)=\dfrac{2x}{3}$$

Charge from $$M^{3+}$$ ions:
$$\left(\dfrac{2}{3}x\right)\times(+3)=\dfrac{6x}{3}=2x$$

Total positive charge from $$M$$ ions:
$$\dfrac{2x}{3}+2x=\dfrac{8x}{3}$$

Metal $$Y$$ is only in the $$+3$$ state and its subscript is 2, so
Charge from $$Y$$ ions = $$2\times(+3)=+6$$.

Oxygen is in the $$-2$$ state.
Charge from oxygen = $$4\times(-2)=-8$$.

For overall electrical neutrality, the sum of all charges must be zero:

$$\dfrac{8x}{3}+6-8=0$$

Simplify:

$$\dfrac{8x}{3}-2=0 \; \Longrightarrow \; \dfrac{8x}{3}=2 \; \Longrightarrow \; x=\dfrac{2\times3}{8}=0.75$$

Hence, $$X = 0.75$$.

Option D which is: 0.75

We analyze both statements about the standardisation of NaOH using potassium hydrogen phthalate (KHP).

Statement I: Potassium hydrogen phthalate (KHP) is a primary standard for standardisation of sodium hydroxide solution.

KHP is indeed a well-known primary standard. It is a stable, pure, non-hygroscopic solid with a high molecular weight, making it ideal for accurately standardizing NaOH solutions. Statement I is CORRECT.

Statement II: In this titration, phenolphthalein can be used as indicator.

KHP is a weak acid (monoprotic, from phthalic acid). The titration of a weak acid (KHP) with a strong base (NaOH) has an equivalence point in the basic pH range (around pH 8-10). Phenolphthalein changes colour in the pH range 8.2-10.0, which coincides with the equivalence point. Statement II is CORRECT.

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

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Moles of NaCl = 5.85/58.5 = 0.1 mol

Volume = 500 mL = 0.5 L

Molarity = 0.1/0.5 = 0.2 M

The correct answer is Option 4: 0.2.

We need to find the number of silver atoms deposited on a plate.

The thickness of the silver coating is $$0.05$$ cm, the area of the plate is $$0.05$$ m² which equals $$0.05 \times 10^4$$ cm² = $$500$$ cm², the atomic mass of Ag is $$108$$ g/mol, and the density of silver is $$d = 7.9$$ g/cm³.

The volume of the silver deposited is given by the product of the area and thickness: $$V = \text{Area} \times \text{Thickness} = 500 \times 0.05 = 25 \text{ cm}^3$$.

The mass of silver deposited follows from mass = density × volume: $$\text{mass} = \text{density} \times \text{volume}$$, thus $$m = 7.9 \times 25 = 197.5 \text{ g}$$.

The number of moles of silver is $$n = \frac{m}{M} = \frac{197.5}{108} = 1.8287 \text{ mol}$$.

The number of silver atoms is given by $$N = n \times N_A$$ where $$N_A = 6.022 \times 10^{23}$$, hence $$N = 1.8287 \times 6.022 \times 10^{23} = 11.013 \times 10^{23}$$.

Rounding to the nearest integer gives $$N \approx 11 \times 10^{23}$$.

Answer: 11

10 mL of gaseous hydrocarbon on combustion gives 40 mL of $$CO_2$$ and 50 mL of water vapour (all at same T and P). We need the total number of carbon and hydrogen atoms in the hydrocarbon.

For a hydrocarbon $$C_xH_y$$:

$$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules. So the volume ratios equal the mole ratios.

From 10 mL of $$C_xH_y$$:

- $$CO_2$$ produced = 40 mL, so $$10x = 40$$, giving $$x = 4$$

- $$H_2O$$ produced = 50 mL, so $$10 \times \frac{y}{2} = 50$$, giving $$y = 10$$

The hydrocarbon is $$C_4H_{10}$$ (butane).

Total number of carbon and hydrogen atoms = $$x + y = 4 + 10 = 14$$.

The answer is 14.

We are asked to find the number of millimoles of $$Pb_3(PO_4)_2$$ formed when 72 mmol of $$PbCl_2$$ reacts with 50 mmol of $$(NH_4)_3PO_4$$.

The reaction is represented by the balanced chemical equation $$ 3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl $$. From this equation, the stoichiometric ratio is 3 mol of $$PbCl_2$$ to 2 mol of $$(NH_4)_3PO_4$$ to 1 mol of $$Pb_3(PO_4)_2$$.

Comparing the available amounts, 72 mmol of $$PbCl_2$$ would require $$\frac{2}{3} \times 72 = 48$$ mmol of $$(NH_4)_3PO_4$$, and since 50 mmol of $$(NH_4)_3PO_4$$ is available, there is more than enough of the phosphate. Conversely, 50 mmol of $$(NH_4)_3PO_4$$ would require $$\frac{3}{2} \times 50 = 75$$ mmol of $$PbCl_2$$, but only 72 mmol of $$PbCl_2$$ are present. Therefore, $$PbCl_2$$ is the limiting reagent.

According to the stoichiometry, 3 mmol of $$PbCl_2$$ yields 1 mmol of $$Pb_3(PO_4)_2$$, so the amount of product formed is $$ \text{mmol of } Pb_3(PO_4)_2 = \frac{72}{3} = 24 \text{ mmol} $$.

The answer is 24 mmol.

$$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$. 1 mole CH₄ gives 1 mole CO₂ (44 g).

22 g CO₂ = 22/44 = 0.5 mol. So 0.5 mol CH₄ needed = 50 × 10⁻² mol. x = 50.

The answer is $$\boxed{50}$$.

We need to find the molarity of 1 L of orthophosphoric acid $$H_3PO_4$$ with 70% purity by weight and specific gravity $$1.54 \, \text{g/cm}^3$$.

First, we recall that molarity is defined as the number of moles of solute per litre of solution: $$M = \frac{\text{Mass of solute in 1 L solution}}{\text{Molar mass of solute}}$$.

Next, the mass of 1 L of solution follows from its density and volume: $$\text{Mass of 1 L solution} = \text{density} \times \text{volume} = 1.54 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1540 \, \text{g}$$.

Since the acid is 70% by weight, the mass of pure $$H_3PO_4$$ in that litre is $$\text{Mass of } H_3PO_4 = \frac{70}{100} \times 1540 = 1078 \, \text{g}$$.

Dividing by its molar mass gives the number of moles: $$\text{Moles} = \frac{1078}{98} = 11.0 \, \text{mol}$$.

Because this amount is present in 1 L of solution, the molarity is $$M = 11.0 \, \text{M}$$.

The correct answer is 11 M.

Given: 84 g of NaOH, molar mass = 40 g/mol.

Moles of NaOH = $$\frac{84}{40} = 2.1$$ mol.

For a 3M solution: $$M = \frac{n}{V}$$, so $$V = \frac{n}{M} = \frac{2.1}{3} = 0.7$$ dm$$^3 = 7 \times 10^{-1}$$ dm$$^3$$.

The answer is $$\boxed{7}$$.

We need to find the amount of NaOH in 50 mL of NaOH solution, given that 50 mL of 0.5 M oxalic acid neutralises 25 mL of the NaOH solution.

The balanced neutralisation reaction is:

$$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$$

Oxalic acid ($$H_2C_2O_4$$) is diprotic, so 1 mole of oxalic acid reacts with 2 moles of NaOH.

$$\text{Moles of } H_2C_2O_4 = M \times V = 0.5 \times \frac{50}{1000} = 0.025 \text{ mol}$$

From the stoichiometry, moles of NaOH = 2 × moles of oxalic acid:

$$\text{Moles of NaOH in 25 mL} = 2 \times 0.025 = 0.05 \text{ mol}$$

Since the concentration is uniform, moles in 50 mL = 2 × moles in 25 mL:

$$\text{Moles of NaOH in 50 mL} = 2 \times 0.05 = 0.1 \text{ mol}$$

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

$$\text{Mass} = 0.1 \times 40 = 4 \text{ g}$$

Therefore, the amount of NaOH in 50 mL of the solution is 4 g.

We need to find the number of oxygen atoms in the chemical formula of fuming sulphuric acid.

Fuming sulphuric acid, also known as oleum, is a solution of sulphur trioxide ($$SO_3$$) in concentrated sulphuric acid ($$H_2SO_4$$). The chemical formula of oleum, also called pyrosulphuric acid or disulphuric acid, is:

$$H_2S_2O_7$$

It can be thought of as the combination of $$H_2SO_4$$ and $$SO_3$$:

$$H_2SO_4 + SO_3 \rightarrow H_2S_2O_7$$

Counting the atoms in $$H_2S_2O_7$$ gives 2 hydrogen atoms, 2 sulphur atoms, and 7 oxygen atoms. Therefore, the number of oxygen atoms in the chemical formula of fuming sulphuric acid is 7.

We need to find the mass of sodium acetate required to prepare a given solution.

The formula for mass from molarity is:

$$ m = M \times V \times M_w $$

where $$M$$ is molarity, $$V$$ is volume in litres, and $$M_w$$ is molar mass.

Given:

$$M = 0.35$$ M, $$V = 250$$ mL $$= 0.250$$ L, $$M_w = 82.02$$ g/mol

Substituting:

$$ m = 0.35 \times 0.250 \times 82.02 = 0.0875 \times 82.02 = 7.177 \text{ g} $$

Rounding to the nearest integer: $$m \approx 7$$ g.

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

The question asks for the number of white coloured salts among the given compounds. We will evaluate each compound based on standard colour properties.

Compound (A) SrSO₄ (Strontium sulfate):
Strontium sulfate is a white crystalline solid. It is insoluble in water and retains its white colour.
Colour: White

Compound (B) MgNH₄PO₄ (Magnesium ammonium phosphate):
Magnesium ammonium phosphate is a white crystalline precipitate, commonly formed in qualitative analysis for magnesium ions.
Colour: White

Compound (C) BaCrO₄ (Barium chromate):
Barium chromate is a yellow solid, often used as a yellow pigment. It is not white.
Colour: Yellow

Compound (D) Mn(OH)₂ (Manganese(II) hydroxide):
Manganese(II) hydroxide is initially white when freshly precipitated, though it oxidizes to brown on exposure to air. The pure compound is white.
Colour: White

Compound (E) PbSO₄ (Lead sulfate):
Lead sulfate is a white crystalline solid and insoluble in water.
Colour: White

Compound (F) PbCrO₄ (Lead chromate):
Lead chromate is a bright yellow solid, known as chrome yellow. It is not white.
Colour: Yellow

Compound (G) AgBr (Silver bromide):
Silver bromide is a pale yellow solid, sensitive to light. It is not white.
Colour: Pale yellow

Compound (H) PbI₂ (Lead iodide):
Lead iodide is a bright yellow solid at room temperature.
Colour: Yellow

Compound (I) CaC₂O₄ (Calcium oxalate):
Calcium oxalate is a white crystalline solid and insoluble in water.
Colour: White

Compound (J) Fe(OH)₂(CH₃COO) (Iron(II) acetate hydroxide):
This compound is a basic salt of iron(II). Iron(II) hydroxide is white when pure but oxidizes rapidly to green or brown. The acetate form is typically greenish or brown due to oxidation and is not considered white.
Colour: Not white (green/brown)

Summary of white salts:
- (A) SrSO₄: White
- (B) MgNH₄PO₄: White
- (D) Mn(OH)₂: White
- (E) PbSO₄: White
- (I) CaC₂O₄: White
Total: 5 white salts.

The compounds (C), (F), (G), (H), and (J) are not white.

Final Answer: 5

A metal chloride has 55% chlorine by weight, and 100 mL of its vapour at STP weighs 0.57 g. The molecular formula can be determined by finding the molar mass from the STP data.

At STP (Standard Temperature and Pressure), one mole of an ideal gas occupies 22400 mL, so the molar mass $$M$$ is calculated as $$M = \frac{\text{mass} \times 22400}{\text{volume}} = \frac{0.57 \times 22400}{100} = \frac{12768}{100} = 127.68\;\text{g/mol}$$.

Since 55% of the compound’s mass is chlorine, the mass of Cl in one mole is $$55\% \times 127.68 = 70.22$$ g. The number of Cl atoms per molecule is then $$\frac{70.22}{35.5} \approx 1.98 \approx 2$$.

The remaining mass corresponds to the metal: $$127.68 - 2(35.5) = 127.68 - 71 = 56.68$$ g/mol. This value is close to the atomic mass of Mn (54.94) or Fe (55.85), indicating that the metal is M and the formula is $$MCl_2$$.

Therefore, the molecular formula is $$MCl_2$$, corresponding to Option 3.

The Assertion states that hydrogen is an environment friendly fuel. This is true because when hydrogen burns, it produces only water as a byproduct:

$$2H_2 + O_2 \rightarrow 2H_2O$$

No harmful gases like CO$$_2$$, SO$$_2$$, or NO$$_x$$ are produced, making it environmentally friendly.

Now, the Reason states that the atomic number of hydrogen is 1 and it is a very light element. This is also true — hydrogen has atomic number 1 and is indeed the lightest element.

However, the reason is not the correct explanation of the assertion. Hydrogen being environment friendly is because its combustion product is water (non-polluting), not because of its atomic number or mass.

Hence, the correct answer is Option B: Both A and R are true but R is NOT the correct explanation of A.

Assertion A: 3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.

Reason R: Molar mass of hydrated oxalic acid is 126 g mol$$^{-1}$$.

Hydrated oxalic acid is $$H_2C_2O_4 \cdot 2H_2O$$.

Molar mass = $$2(1) + 2(12) + 4(16) + 2(18) = 2 + 24 + 64 + 36 = 126$$ g/mol.

Therefore, Reason R is TRUE.

$$\text{Moles of oxalic acid} = \frac{3.1500}{126} = 0.025 \text{ mol}$$

$$\text{Volume of solution} = 250.0 \text{ mL} = 0.250 \text{ L}$$

$$\text{Molarity} = \frac{\text{moles}}{\text{volume in litres}} = \frac{0.025}{0.250} = 0.1 \text{ M}$$

Therefore, Assertion A is TRUE.

The calculation in Assertion A directly uses the molar mass of 126 g/mol stated in Reason R. Without knowing the molar mass, we cannot compute the molarity. Hence, Reason R is the correct explanation for Assertion A.

The correct answer is Option (3): Both A and R are true and R is the correct explanation of A.

At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.7 L (or 22.4 L at old STP).

Using 22.7 L/mol (new STP at 273.15 K, 1 bar):

$$\text{Moles} = \frac{2.8375}{22.7} = 0.125 \text{ mol}$$

Number of molecules:

$$N = 0.125 \times 6.022 \times 10^{23} = 7.527 \times 10^{22}$$

The correct answer is Option 4: $$7.527 \times 10^{22}$$ and 0.125 mol.

Let us count the significant figures in each number:

(A) 0.00253: Leading zeros are not significant. So significant figures = 2, 5, 3 = 3 significant figures.

(B) 1.0003: All digits are significant (zeros between non-zero digits are significant). So significant figures = 1, 0, 0, 0, 3 = 5 significant figures.

(C) 15.0: The trailing zero after the decimal point is significant. So significant figures = 1, 5, 0 = 3 significant figures.

(D) 163: All non-zero digits are significant. So significant figures = 1, 6, 3 = 3 significant figures.

A, C, and D all have 3 significant figures.

The correct answer is Option 4: A, C and D only.

Match List-I with List-II for mole concept calculations.

A. 16 g of CH$$_4$$(g):

Molar mass of CH$$_4$$ = 16 g/mol, so 16 g = 1 mole.

Electrons in CH$$_4$$: C has 6e, 4H have 4e = 10 electrons per molecule.

Total electrons = $$1 \times 6.02 \times 10^{23} \times 10 = 60.2 \times 10^{23}$$ electrons = II

B. 1 g of H$$_2$$(g):

Molar mass of H$$_2$$ = 2 g/mol, so 1 g = 0.5 mol.

Volume at STP = $$0.5 \times 22.4 = 11.2 \approx 11.4$$ L (using 22.7 L/mol for updated STP) = IV

C. 1 mole of N$$_2$$(g):

Mass = $$1 \times 28 = 28$$ g. Weighs 28 g = I

D. 0.5 mol of SO$$_2$$(g):

Mass = $$0.5 \times 64 = 32$$ g. Weighs 32 g = III

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

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

We need to identify the correct ratio of silica to alumina in good quality cement.

Portland cement is manufactured by heating limestone ($$CaCO_3$$) with clay containing $$SiO_2$$ and $$Al_2O_3$$ in a kiln, and its quality depends on the ratios of these oxides.

For good quality Portland cement, the ratio of silica ($$SiO_2$$) to alumina ($$Al_2O_3$$) should be approximately 2.5 to 4, with an ideal value of about $$3$$, while the ratio of lime ($$CaO$$) to the total acidic oxides ($$SiO_2 + Al_2O_3 + Fe_2O_3$$) is about $$2$$.

When the silica to alumina ratio is too low, the cement sets quickly but lacks strength; conversely, a ratio that is too high causes slow setting.

Hence, the correct answer is Option 4: 3.

We need to identify the minimum dissolved oxygen (DO) for fish growth and the BOD limit for clean water.

For the healthy growth of fish, the concentration of dissolved oxygen in water should be more than 6 ppm. Below this level, fish survival becomes difficult.

Clean water has a Biochemical Oxygen Demand (BOD) of less than 5 ppm. Water with BOD above 5 ppm is considered polluted, and highly polluted water has BOD > 17 ppm.

$$X = 6$$ ppm (dissolved oxygen for fish) and $$Y = 5$$ ppm (BOD for clean water).

The correct answer is Option A: X = 6, Y = 5.

We need to identify the correct statement about smog.

Classical smog:

• Occurs in cool, humid climate
• Contains smoke, fog, and $$\text{SO}_2$$
• $$\text{SO}_2$$ makes it a reducing smog
• Does NOT contain $$\text{NO}_2$$ as a primary component

Photochemical smog:

• Occurs in warm, dry, sunny climate
• Contains $$\text{NO}_2$$, ozone, PAN, formaldehyde
• Has high concentration of oxidizing agents (ozone, PAN, NO₂)

Evaluating each option:

Option A: "$$\text{NO}_2$$ is present in classical smog" — Incorrect. $$\text{NO}_2$$ is in photochemical smog, not classical.

Option B: "Both $$\text{NO}_2$$ and $$\text{SO}_2$$ are present in classical smog" — Incorrect. $$\text{NO}_2$$ is not in classical smog.

Option C: "Photochemical smog has high concentration of oxidizing agents" — Correct. Ozone and PAN are strong oxidizing agents.

Option D: "Classical smog also has high concentration of oxidizing agents" — Incorrect. Classical smog is reducing in nature.

The answer is Option C.

We need to identify A, B, and C in the photochemical smog reactions.

Reaction (i): $$\text{NO}_2 \xrightarrow{h\nu} \text{A} + \text{B}$$

When NO$$_2$$ absorbs UV radiation, it undergoes photodissociation:

$$\text{NO}_2 \xrightarrow{h\nu} \text{NO} + \text{O}$$

So $$A = \text{NO}$$ and $$B = \text{O}$$ (atomic oxygen).

Reaction (ii): $$\text{B} + \text{O}_2 \to \text{C}$$

$$\text{O} + \text{O}_2 \to \text{O}_3$$

So $$C = \text{O}_3$$ (ozone).

Reaction (iii): $$\text{A} + \text{C} \to \text{NO}_2 + \text{O}_2$$

$$\text{NO} + \text{O}_3 \to \text{NO}_2 + \text{O}_2$$ ✓

Therefore, A = NO, B = O, C = O$$_3$$.

The correct answer is Option 4: NO, O and O$$_3$$.

Facts given in NCERT .For a good quality cement, the ratio of lime to the total of the oxides of Si, Al and Fe should be as close as to 4

We need to evaluate two statements about water quality parameters.

Evaluating Statement I: "If BOD is 4 ppm and dissolved oxygen is 8 ppm, then it is a good quality water."

BOD (Biochemical Oxygen Demand) measures the amount of dissolved oxygen consumed by microorganisms to decompose the organic matter present in water. A lower BOD indicates less organic pollution.

- Clean water typically has a BOD of less than 5 ppm.

- Water is considered polluted when BOD exceeds 17 ppm.

Dissolved oxygen (DO) indicates the amount of oxygen available in water for aquatic organisms. Higher DO values indicate better water quality.

- Good quality water has DO around 6-8 ppm or higher.

Given BOD = 4 ppm (which is less than 5 ppm, so acceptable) and DO = 8 ppm (which is in the good range), this water qualifies as good quality water. Statement I is correct.

Evaluating Statement II: "If the concentration of zinc and nitrate salts are 5 ppm each, then it can be good quality water."

The permissible limits for various ions in drinking water as per standard guidelines are:

- Zinc: The permissible limit is approximately 5 ppm. A concentration of 5 ppm of zinc is within the acceptable range.

- Nitrate (NO$$_3^-$$): The permissible limit for nitrate in drinking water is approximately 45-50 ppm. A concentration of 5 ppm of nitrate is well below this limit.

Since both zinc (5 ppm) and nitrate (5 ppm) are within their respective permissible limits, the water can be considered good quality. Statement II is correct.

Both statements are correct. The correct answer is Option 4: Both the statements I and II are correct.

Matching industries with their waste:

(A) Steel plants → Slag (III): Steelmaking produces slag as a byproduct (mixture of metal oxides and silicon dioxide).

(B) Thermal power plants → Fly ash (II): Burning coal in thermal power plants produces fly ash.

(C) Fertilizer Industries → Gypsum (I): Phosphoric acid production in fertilizer industries produces phosphogypsum as waste.

(D) Paper mills → Bio-degradable wastes (IV): Paper mills produce biodegradable organic waste (lignin, cellulose residues).

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

The correct answer is Option 1.

Photochemical smog formation requires:

• Sunlight (UV radiation)
• Warm temperature
• NOₓ and hydrocarbons from vehicular emissions
• Stagnant air conditions

Photochemical smog is minimum in cold conditions with less sunlight. Among the given options:

Srinagar, J&K in January: Very cold (below 0°C), minimal sunlight, low temperatures → least favorable for photochemical smog.

This matches option 1.

We need to identify which process does NOT disturb the delicate balance of $$CO_2$$ and $$O_2$$ in the atmosphere.

Understanding the $$CO_2$$-$$O_2$$ balance: In nature, there is a natural cycle that maintains the balance between $$CO_2$$ and $$O_2$$:

- Photosynthesis (by plants): $$6CO_2 + 6H_2O \xrightarrow{sunlight} C_6H_{12}O_6 + 6O_2$$ (consumes $$CO_2$$, produces $$O_2$$)

- Respiration (by all living organisms): $$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$$ (consumes $$O_2$$, produces $$CO_2$$)

These two processes are complementary and have been occurring for millions of years, maintaining a natural equilibrium.

Analyzing each option:

Option 1: Respiration - Respiration is a natural biological process that is part of the natural carbon cycle. The $$CO_2$$ produced by respiration is consumed by photosynthesis, and the $$O_2$$ consumed by respiration is produced by photosynthesis. This natural cycle does NOT disturb the $$CO_2/O_2$$ balance.

Option 2: Burning of coal - Burning coal is combustion of fossil fuels, which releases large amounts of $$CO_2$$ that were stored underground for millions of years. This adds excess $$CO_2$$ to the atmosphere without a corresponding increase in photosynthesis, thereby disturbing the balance.

Option 3: Deforestation - Cutting down forests reduces the number of trees performing photosynthesis, which means less $$CO_2$$ is absorbed and less $$O_2$$ is produced. This disturbs the balance by shifting it toward higher $$CO_2$$ and lower $$O_2$$.

Option 4: Burning of petroleum - Like coal, burning petroleum releases stored carbon as $$CO_2$$, adding excess $$CO_2$$ to the atmosphere and disturbing the balance.

The correct answer is Option 1: Respiration does NOT disturb the $$CO_2/O_2$$ balance.

Among the given industrial activities, we need to identify the one least responsible for global warming.

Global warming is primarily caused by greenhouse gas emissions, especially CO₂.

Steel manufacturing: Produces large amounts of CO₂ from coke used in blast furnaces.

Electricity generation in thermal power plants: Burns fossil fuels (coal, gas), releasing enormous quantities of CO₂. This is one of the largest contributors to global warming.

Industrial production of urea: Uses CO₂ as a raw material ($$CO_2 + 2NH_3 \rightarrow (NH_2)_2CO + H_2O$$) but the process itself involves significant energy consumption and emissions.

Manufacturing of cement: While cement production does produce CO₂ (from limestone decomposition and fuel burning), its contribution is relatively lower compared to steel and thermal power generation in terms of the overall industrial share.

Therefore, among the given options, manufacturing of cement is held least responsible for global warming.

We need to find the ratio of densities of fcc to bcc lattices for metal M.

Formula: The density of a crystal lattice is given by:

$$\rho = \frac{Z \times M}{N_A \times a^3}$$

where $$Z$$ is the number of atoms per unit cell, $$M$$ is the molar mass, $$N_A$$ is Avogadro's number, and $$a$$ is the edge length.

Number of atoms per unit cell.

For fcc: $$Z_{fcc} = 4$$

For bcc: $$Z_{bcc} = 2$$

Given edge lengths.

$$a_{fcc} = 2.0$$ Angstrom

$$a_{bcc} = 2.5$$ Angstrom

Calculate the ratio of densities.

Since the metal M is the same in both cases, $$M$$ and $$N_A$$ are the same:

$$\frac{\rho_{fcc}}{\rho_{bcc}} = \frac{Z_{fcc}}{Z_{bcc}} \times \frac{a_{bcc}^3}{a_{fcc}^3}$$

$$= \frac{4}{2} \times \frac{(2.5)^3}{(2.0)^3}$$

$$= 2 \times \frac{15.625}{8.0}$$

$$= 2 \times 1.953125$$

$$= 3.90625$$

Rounding to the nearest integer: $$\approx 4$$

The correct answer is $$\mathbf{4}$$ (Option D).

When 0.5 g of the organic compound X containing 60% carbon undergoes complete combustion, the mass of carbon in the compound can be determined.

The mass of carbon in the compound is calculated as $$\text{Mass of C} = 0.5 \times \frac{60}{100} = 0.3$$ g.

From this mass, the number of moles of carbon is found by dividing by the molar mass of carbon:

$$n_C = \frac{0.3}{12} = 0.025$$ mol.

Under complete combustion, each mole of carbon produces one mole of carbon dioxide according to the reaction:

$$C + O_2 \rightarrow CO_2$$

Thus, the number of moles of CO₂ formed is $$n_{CO_2} = 0.025$$ mol.

The mass of CO₂ produced is then calculated by multiplying by its molar mass:

$$m_{CO_2} = 0.025 \times 44 = 1.1$$ g $$= 11 \times 10^{-1}$$ g.

The correct answer is 11.

The balanced chemical equation is:

$$3BaCl_2 + 2Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6NaCl$$

We have 5 moles of $$BaCl_2$$ and 2 moles of $$Na_3PO_4$$.

From the stoichiometry, 3 moles of $$BaCl_2$$ react with 2 moles of $$Na_3PO_4$$. For 2 moles of $$Na_3PO_4$$, we need $$\frac{3}{2} \times 2 = 3$$ moles of $$BaCl_2$$. Since we have 5 moles of $$BaCl_2$$ (more than enough), $$Na_3PO_4$$ is the limiting reagent.

Now, from the stoichiometry, 2 moles of $$Na_3PO_4$$ produce 1 mole of $$Ba_3(PO_4)_2$$:

$$\text{Moles of } Ba_3(PO_4)_2 = \frac{2}{2} = 1\;\text{mole}$$

So, the maximum number of moles of $$Ba_3(PO_4)_2$$ formed is $$1$$.

On complete combustion, the organic compound produces CO$$_2$$. The carbon in CO$$_2$$ comes entirely from the organic compound.

Mass of CO$$_2$$ produced = 0.792 g

Mass of carbon in CO$$_2$$:

$$ \text{Mass of C} = \frac{12}{44} \times 0.792 = \frac{12 \times 0.792}{44} = \frac{9.504}{44} = 0.216 \text{ g} $$

Percentage of carbon in the organic compound:

$$ \% C = \frac{0.216}{0.492} \times 100 = 43.9\% \approx 44\% $$

We need to find the energy of one mole of photons of radiation with frequency $$2 \times 10^{12}$$ Hz.

Formula: $$E = N_A h\nu$$

Calculation:

$$E = 6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 2 \times 10^{12}$$

$$= 6.022 \times 6.626 \times 2 \times 10^{23-34+12}$$

$$= 6.022 \times 13.252 \times 10^{1}$$

$$= 79.80 \times 10^{1}$$

$$\approx 798$$ J mol$$^{-1}$$

The energy is $$\boxed{798}$$ J mol$$^{-1}$$.

The reaction: $$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$$

Moles of Mg = $$\frac{2.4}{24} = 0.1$$ mol

Moles of H$$_2$$ = 0.1 mol (1:1 ratio)

Volume at STP = $$0.1 \times 22.4 = 2.24$$ L = $$224 \times 10^{-2}$$ L

The balanced equation for the reaction:

$$ Zn + 2HCl \rightarrow ZnCl_2 + H_2 $$

Moles of zinc:

$$ n_{Zn} = \frac{11.5}{65.4} = 0.1758 \text{ mol} $$

From the stoichiometry, 1 mole of Zn produces 1 mole of H$$_2$$:

$$ n_{H_2} = 0.1758 \text{ mol} $$

Volume of H$$_2$$ at STP:

$$ V = n \times V_m = 0.1758 \times 22.7 = 3.99 \approx 4 \text{ L} $$

Find the volume of monobasic strong acid solution required to neutralize 25 mL of 0.24 M NaOH.

The molarity of the acid solution can be determined from its density and molar mass. With density = 1.21 kg/L = 1210 g/L and molar mass = 24.2 g/mol, the molarity is $$\frac{1210}{24.2} = 50$$ M.

For a monobasic strong acid neutralizing NaOH, the condition $$M_1V_1 = M_2V_2$$ applies. Substituting $$M_1 = 50$$ M, $$M_2 = 0.24$$ M and $$V_2 = 25$$ mL gives $$50 \times V_1 = 0.24 \times 25\,. $$

Solving for $$V_1$$ yields $$V_1 = \frac{0.24 \times 25}{50} = \frac{6}{50} = 0.12\text{ mL}$$ and expressing this in the required form gives $$V_1 = 0.12\text{ mL} = 12 \times 10^{-2}\text{ mL}\,,$$ so the required volume is $$\boxed{12}$$.

We need to find the molar mass of an organic compound containing 60% carbon, given that 0.01 mol produces 4.4 g of CO$$_2$$ on complete combustion.

Molar mass of CO$$_2$$ = 44 g/mol.

$$\text{Moles of CO}_2 = \frac{4.4}{44} = 0.1 \text{ mol}$$

Each mole of CO$$_2$$ contains 1 mole of carbon.

Moles of C = 0.1 mol from 0.01 mol of compound.

So each mole of compound contains $$\frac{0.1}{0.01} = 10$$ moles of carbon.

Mass of carbon per mole of compound = $$10 \times 12 = 120$$ g.

Since the compound is 60% carbon:

$$\frac{120}{M} = 0.60$$

$$M = \frac{120}{0.60} = 200 \text{ g/mol}$$

The correct answer is 200.

We need to find the volume of stock NaOH solution required to prepare 500 mL of 0.1 M NaOH solution.

The molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = $$\frac{5}{40} = 0.125$$ mol, and the volume of the stock solution is 450 mL = 0.450 L, giving a molarity of $$\frac{0.125}{0.450} = \frac{25}{90} = \frac{5}{18}$$ M.

Using the dilution formula $$M_1V_1 = M_2V_2$$, we have $$\frac{5}{18} \times V_1 = 0.1 \times 500$$, which yields $$V_1 = \frac{50 \times 18}{5} = 180$$ mL.

Thus, 180 mL of the stock solution is required.

In a hexagonal close-packed (hcp) structure, for every atom there are 2 tetrahedral voids and 1 octahedral void, giving a total of 3 voids per atom.

The number of atoms in 0.02 mol is $$N = 0.02 \times N_A = 0.02 \times 6.02 \times 10^{23} = 1.204 \times 10^{22}$$.

The total number of voids is $$3N = 3 \times 1.204 \times 10^{22} = 3.612 \times 10^{22} = 36.12 \times 10^{21}$$.

Rounding to the nearest integer, the total number of voids is $$\boxed{36} \times 10^{21}$$.

We are given that $$120$$ g of an organic compound containing only carbon and hydrogen on complete combustion gives $$330$$ g of $$CO_2$$ and $$270$$ g of $$H_2O$$. Since the percentages of carbon and hydrogen are required, we begin by finding the moles of $$CO_2$$ and $$H_2O$$ produced.

Molar mass of $$CO_2 = 44$$ g/mol, so

$$\text{Moles of } CO_2 = \frac{330}{44} = 7.5 \text{ mol}$$

Similarly, molar mass of $$H_2O = 18$$ g/mol, giving

$$\text{Moles of } H_2O = \frac{270}{18} = 15 \text{ mol}$$

Since each mole of $$CO_2$$ contains one mole of carbon (atomic mass = 12 g/mol), the mass of carbon in the original compound is

$$\text{Mass of C} = 7.5 \times 12 = 90 \text{ g}$$

From each mole of $$H_2O$$ there are two moles of hydrogen (atomic mass = 1 g/mol), so the mass of hydrogen becomes

$$\text{Mass of H} = 15 \times 2 \times 1 = 30 \text{ g}$$

Adding these masses gives $$90 + 30 = 120$$ g, which matches the given mass of the compound. Now calculating the percentages:

$$\% \text{ of C} = \frac{90}{120} \times 100 = 75\%$$

$$\% \text{ of H} = \frac{30}{120} \times 100 = 25\%$$

Therefore, the correct option is Option D: 75 and 25.

The composition of the compound is 8.7% hydrogen, 74% carbon, and 17.3% nitrogen, and its molar mass is 162 g/mol. We first calculate the mole ratio of each element:

$$\text{Moles of C} = \frac{74}{12} = 6.167$$

$$\text{Moles of H} = \frac{8.7}{1} = 8.7$$

$$\text{Moles of N} = \frac{17.3}{14} = 1.236$$

Dividing by the smallest value, 1.236, gives

$$\text{C} : \text{H} : \text{N} = \frac{6.167}{1.236} : \frac{8.7}{1.236} : \frac{1.236}{1.236}$$

$$= 4.99 : 7.04 : 1 \approx 5 : 7 : 1$$

Thus the empirical formula is $$C_5H_7N$$, and its formula mass is $$5 \times 12 + 7 \times 1 + 14 = 60 + 7 + 14 = 81$$ g/mol.

The multiplication factor is $$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{162}{81} = 2$$.

Multiplying the subscripts in the empirical formula by 2 yields the molecular formula

$$ (\text{C}_5\text{H}_7\text{N})_2 = \text{C}_{10}\text{H}_{14}\text{N}_2$$

Hence, the correct answer is Option D ($$C_{10}H_{14}N_2$$).

We have the balanced equation: $$4HNO_3(l) + 3KCl(s) \to Cl_2(g) + NOCl(g) + 2H_2O(g) + 3KNO_3(s)$$

We need to find the mass of $$HNO_3$$ required to produce 110.0 g of $$KNO_3$$. The molar mass of $$KNO_3 = 39 + 14 + 48 = 101$$ g/mol. The moles of $$KNO_3$$ produced are $$\frac{110.0}{101} \approx 1.089$$ mol.

From the stoichiometry, 4 moles of $$HNO_3$$ produce 3 moles of $$KNO_3$$. So the moles of $$HNO_3$$ required are: $$n_{HNO_3} = \frac{4}{3} \times 1.089 = 1.452$$ mol.

The molar mass of $$HNO_3 = 1 + 14 + 48 = 63$$ g/mol. The mass of $$HNO_3$$ required is: $$m = 1.452 \times 63 = 91.48 \approx 91.5$$ g.

Hence, the correct answer is Option C.

Since hemoglobin contains $$0.34\%$$ iron by mass and the sample has a mass of $$3.3 \text{ g}$$, with the atomic mass of Fe being $$56 \text{ u}$$ and $$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$$, we first calculate the mass of iron present.

Substituting the given percentage into the formula $$m_{Fe} = \frac{0.34}{100} \times 3.3$$ gives $$m_{Fe} = 0.01122 \text{ g}$$.

Next, we determine the number of moles of iron by using $$n_{Fe} = \frac{m_{Fe}}{M_{Fe}} = \frac{0.01122}{56}$$, which yields $$n_{Fe} = 2.004 \times 10^{-4} \text{ mol}$$.

Finally, multiplying the moles of iron by Avogadro’s number via $$N = n_{Fe} \times N_A = 2.004 \times 10^{-4} \times 6.022 \times 10^{23}$$ results in $$N = 1.21 \times 10^{20}$$ atoms.

Therefore, the number of Fe atoms in 3.3 g of hemoglobin is $$1.21 \times 10^{20}$$, which corresponds to Option C.

The fuel is $$C_{15}H_{30}$$ with density $$0.756$$ g/mL. We need to find the weight of $$O_2$$ required and $$CO_2$$ released for every litre of fuel.

The balanced combustion reaction is:

$$ C_{15}H_{30} + \frac{45}{2}O_2 \rightarrow 15CO_2 + 15H_2O $$

Verification: C: 15 = 15 ✓; H: 30 = 30 ✓; O: 45 = 30 + 15 = 45 ✓

The mass of 1 litre of fuel is: $$ \text{Mass} = 1000 \text{ mL} \times 0.756 \text{ g/mL} = 756 \text{ g} $$ The molar mass of $$C_{15}H_{30}$$ is: $$ M = 15 \times 12 + 30 \times 1 = 180 + 30 = 210 \text{ g/mol} $$ so the number of moles of fuel in 1 litre is: $$ n = \frac{756}{210} = 3.6 \text{ mol} $$

Each mole of fuel requires $$\frac{45}{2} = 22.5$$ moles of $$O_2$$, so the mass of $$O_2$$ required is: $$ \text{Mass of } O_2 = 3.6 \times 22.5 \times 32 = 3.6 \times 720 = 2592 \text{ g} $$

Each mole of fuel produces 15 moles of $$CO_2$$, so the mass of $$CO_2$$ released is: $$ \text{Mass of } CO_2 = 3.6 \times 15 \times 44 = 3.6 \times 660 = 2376 \text{ g} $$

The weight of oxygen required is 2592 g and $$CO_2$$ released is 2376 g. The correct answer is Option C: 2592 g and 2376 g.

We have the reaction $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$. We are given 20 g of $$N_2$$ and 5 g of $$H_2$$.

The molar mass of $$N_2$$ is 28 g/mol, so the moles of $$N_2$$ are $$\dfrac{20}{28} = \dfrac{5}{7} \approx 0.714$$ mol. The molar mass of $$H_2$$ is 2 g/mol, so the moles of $$H_2$$ are $$\dfrac{5}{2} = 2.5$$ mol.

From the stoichiometry, 1 mole of $$N_2$$ requires 3 moles of $$H_2$$. So $$\dfrac{5}{7}$$ moles of $$N_2$$ would require $$3 \times \dfrac{5}{7} = \dfrac{15}{7} \approx 2.14$$ moles of $$H_2$$. We have 2.5 moles of $$H_2$$ available, which is more than enough. Therefore, $$N_2$$ is the limiting reagent because it will be consumed first.

Now, from the balanced equation, 1 mole of $$N_2$$ produces 2 moles of $$NH_3$$. So $$\dfrac{5}{7}$$ moles of $$N_2$$ will produce $$2 \times \dfrac{5}{7} = \dfrac{10}{7} \approx 1.42$$ moles of $$NH_3$$.

Hence, the correct answer is Option C.

The reaction of $$SO_2Cl_2$$ with excess water is:

$$SO_2Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl$$

So each mole of $$SO_2Cl_2$$ produces 1 mole of $$H_2SO_4$$ and 2 moles of $$HCl$$.

Neutralization reactions:

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

$$HCl + NaOH \rightarrow NaCl + H_2O$$

For each mole of $$SO_2Cl_2$$:

- 1 mole $$H_2SO_4$$ requires 2 moles of $$NaOH$$

- 2 moles $$HCl$$ require 2 moles of $$NaOH$$

Total $$NaOH$$ required per mole of $$SO_2Cl_2 = 2 + 2 = 4$$ moles.

Given that 16 moles of $$NaOH$$ are required:

$$\text{Moles of } SO_2Cl_2 = \dfrac{16}{4} = 4$$

Therefore, the correct answer is Option C: $$4$$.

We need to find the answer to the expression using significant figures rules. The expression is $$\frac{0.02858 \times 0.112}{0.5702}$$. The numerator is $$0.02858 \times 0.112 = 0.00320096$$, and dividing by 0.5702 gives $$\frac{0.00320096}{0.5702} = 0.005613...$$

In multiplication and division, the result should have the same number of significant figures as the quantity with the least significant figures. 0.02858 has 4 significant figures, 0.112 has 3 significant figures, and 0.5702 has 4 significant figures, so the minimum is 3 significant figures.

Rounding the answer to 3 significant figures gives $$0.005613... \approx 0.00561$$ (3 significant figures). The answer is Option B: 0.00561.

The question asks about the highest industrial consumption of molecular hydrogen ($$H_2$$).

Key Industrial Uses of Hydrogen:

1. Haber-Bosch Process (Ammonia synthesis): $$N_2 + 3H_2 \rightarrow 2NH_3$$ — This is by far the largest industrial consumer of hydrogen gas worldwide. Approximately 50-60% of all industrially produced hydrogen is used to manufacture ammonia.

2. Hydrogenation of oils and fats (carbon compounds) — uses hydrogen but in much smaller quantities.

3. Production of methanol ($$CO + 2H_2 \rightarrow CH_3OH$$) — significant but smaller than ammonia production.

4. Production of HCl (hydrogen + chlorine) — relatively small scale.

Since ammonia ($$NH_3$$) is a compound of nitrogen, the element whose compound consumes the most molecular hydrogen industrially is Nitrogen.

The correct answer is Option D: Nitrogen.

Clark's method of water softening involves adding slaked lime (calcium hydroxide, $$Ca(OH)_2$$) to hard water that contains temporary hardness (bicarbonates of calcium and magnesium).

Reaction with calcium bicarbonate:

Calcium bicarbonate reacts with calcium hydroxide to form insoluble calcium carbonate ($$CaCO_3$$), which precipitates out.

Reaction with magnesium bicarbonate:

Magnesium bicarbonate reacts with calcium hydroxide to form insoluble calcium carbonate and insoluble magnesium hydroxide ($$Mg(OH)_2$$), both of which precipitate out.

Therefore, the metal salts formed during softening of hard water using Clark's method are $$CaCO_3$$ and $$Mg(OH)_2$$.

The correct answer is Option B: $$CaCO_3$$ and $$Mg(OH)_2$$.

PV=nRT is standard ideal gas equation, assuming no. of moles as constant, on increasing T ,pressure and volume would increase represented by plot B

We need to identify which of the given substances is not a pesticide.

What is a Pesticide?

A pesticide is any substance used to kill, repel, or control pests. Pesticides include several categories:

- Insecticides: Used to kill insects (e.g., DDT, BHC, Malathion)

- Herbicides: Used to kill weeds (e.g., 2,4-D, Atrazine)

- Fungicides: Used to kill fungi

- Rodenticides: Used to kill rodents

In the NCERT syllabus for JEE, the standard examples of "pesticides" focus primarily on insecticides, particularly organochlorine and organophosphate compounds.

Now let us evaluate each option:

Option A: DDT (Dichlorodiphenyltrichloroethane)

DDT is the most well-known example of an organochlorine insecticide. It was widely used to control mosquitoes and agricultural pests. DDT is a standard NCERT example of a pesticide.

Option B: Dieldrin

Dieldrin is another organochlorine insecticide. It was extensively used against soil-dwelling insects and crop pests. It is listed as a pesticide in NCERT.

Option C: Organophosphate

Organophosphates are an entire class of insecticide pesticides. Common examples include malathion and parathion. These are standard NCERT examples of pesticides that work by inhibiting the enzyme acetylcholinesterase.

Option D: Sodium Arsenite

Sodium arsenite ($$NaAsO_2$$) is an inorganic arsenic compound. It is primarily used as a herbicide (weedicide) and rodenticide (rat poison). While herbicides and rodenticides are technically types of pesticides, sodium arsenite is not classified among the standard NCERT pesticide examples. In the NCERT framework, the term "pesticide" is used specifically for insecticides like DDT, Dieldrin, and organophosphates. Sodium arsenite does not belong to the organochlorine or organophosphate classes and is not listed as a pesticide in NCERT.

Hence, the correct answer is Option D: Sodium Arsenite.

Clark's method is used for removing temporary hardness from water. Temporary hardness is caused by the presence of dissolved bicarbonates of calcium and magnesium, i.e., $$Ca(HCO_3)_2$$ and $$Mg(HCO_3)_2$$.

In Clark's method, a calculated amount of slaked lime, $$Ca(OH)_2$$, is added to the hard water. The reactions that occur are:

For calcium bicarbonate: $$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$$. Here, calcium carbonate ($$CaCO_3$$) precipitates out since it is insoluble in water.

For magnesium bicarbonate: $$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$$. In this case, magnesium hydroxide ($$Mg(OH)_2$$) precipitates out because $$MgCO_3$$ is slightly soluble in water. The excess $$Ca(OH)_2$$ converts the magnesium compound to the insoluble $$Mg(OH)_2$$.

So the products obtained during the treatment of hard water using Clark's method are $$CaCO_3$$ and $$Mg(OH)_2$$.

Hence, the correct answer is Option C.

We are asked to identify substance 'X' in Portland cement that enhances (controls) the setting time.

Portland cement is manufactured by heating a mixture of limestone and clay in a rotary kiln. The calcium silicates and aluminates formed during this process would normally set (harden) very quickly when mixed with water. To regulate and slow down this setting process, a small amount of gypsum is added during the grinding stage.

Gypsum has the chemical formula $$CaSO_4 \cdot 2H_2O$$ (calcium sulphate dihydrate). It reacts with tricalcium aluminate ($$C_3A$$) to form ettringite, which coats the aluminate particles and prevents them from reacting too fast with water. This gives the mason adequate working time before the cement hardens.

Now let us examine the options. Option A, $$CaSO_4 \cdot \frac{1}{2}H_2O$$, is Plaster of Paris — it is not added to Portland cement. Option C, anhydrous $$CaSO_4$$, and Option D, $$CaCO_3$$ (limestone), are also not the standard additives used to control setting time. The correct additive is gypsum, $$CaSO_4 \cdot 2H_2O$$.

Hence, the correct answer is Option B.

We need to evaluate the Assertion and Reason about BOD (Biochemical Oxygen Demand).

Evaluating the Assertion:

BOD of clean water is typically around 1-5 ppm. Polluted water can have a BOD value of 17 ppm or higher, as it contains a large amount of organic matter that requires more oxygen for decomposition.

Therefore, the Assertion is true.

Evaluating the Reason:

BOD (Biochemical Oxygen Demand) is defined as the amount of oxygen required by bacteria to break down only the biodegradable organic matter in water. It does NOT measure the oxygen required for non-biodegradable organic material.

The Reason states that BOD measures oxygen required to oxidise both biodegradable and non-biodegradable organic material, which is incorrect.

Therefore, the Reason is false.

Conclusion:

Assertion is true but Reason is false.

The correct answer is Option C.

Let us verify each statement individually.

Statement I: Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide.

This is correct. Classical smog (also called London smog or reducing smog) occurs in cool, humid climates. It contains smoke, fog, and $$SO_2$$. The $$SO_2$$ acts as a reducing agent, hence it is called reducing smog.

Statement II: Photochemical smog has components: ozone, nitric oxide, acrolein, formaldehyde, PAN, etc.

This is also correct. Photochemical smog (also called Los Angeles smog) is formed by the action of sunlight on nitrogen oxides and hydrocarbons. Its components include ozone ($$O_3$$), nitric oxide (NO), acrolein ($$CH_2=CHCHO$$), formaldehyde ($$HCHO$$), and peroxyacetyl nitrate (PAN).

Since both Statement I and Statement II are correct, the answer is Option A.

We need to evaluate both statements about industrial waste management.

Statement I: The non bio-degradable fly ash and slag from steel industry can be used by cement industry.

This is CORRECT. Fly ash (from thermal power plants) and blast furnace slag (from steel industry) are commonly used as supplementary cementitious materials in the cement industry. Fly ash is used in Portland Pozzolana Cement (PPC), and slag is used in Portland Slag Cement (PSC). This is a well-established practice of industrial waste utilization.

Statement II: The fuel obtained from plastic waste is lead free.

This is CORRECT. When plastic waste is converted into fuel through pyrolysis, the resulting fuel is free from lead. Unlike conventional leaded fuels (which contained tetraethyl lead as an additive), the fuel derived from plastics does not contain lead compounds, making it a cleaner alternative in that regard.

Since both statements are correct:

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

We need to identify the pair in which both substances are herbicides (chemicals used to kill unwanted plants/weeds).

We recall the classification of common pesticides. Aldrin and Dieldrin are well-known insecticides — they belong to the class of chlorinated hydrocarbon insecticides used to kill insects, not weeds.

On the other hand, sodium chlorate ($$NaClO_3$$) and sodium arsenite ($$Na_3AsO_3$$) are both classified as herbicides. Sodium chlorate is a non-selective herbicide that kills plants by oxidative damage, and sodium arsenite has historically been used as a weed killer.

Now let us examine the other options. Option B pairs sodium chlorate (herbicide) with Aldrin (insecticide) — not both herbicides. Option C pairs sodium arsenite (herbicide, noting the question writes "arsinate" but means arsenite) with Dieldrin (insecticide) — again, not both herbicides.

Hence, the only pair where both substances are herbicides is sodium chlorate and sodium arsenite.

Hence, the correct answer is Option D: Sodium chlorate and sodium arsenite.

Photochemical smog, also called Los Angeles smog, is formed in warm, dry, and sunny climates (not humid). It is an oxidizing smog, in contrast to reducing smog (London smog), which contains SO$$_2$$ and occurs in humid, cool climates.

It is formed by the action of sunlight on nitrogen oxides and unsaturated hydrocarbons (from automobile exhausts). The key reactions involve NO$$_2$$, O$$_3$$, and hydrocarbons producing formaldehyde, acrolein, and PAN (peroxyacetyl nitrate).

Among the options provided, A states it occurs in a humid climate — incorrect, as it occurs in warm, dry, sunny climates; B describes it as a mixture of smoke, fog, and SO$$_2$$ — incorrect, as that is classical (London/reducing) smog; C claims it is reducing smog — incorrect, since it is an oxidizing smog; D attributes it to the reaction of unsaturated hydrocarbons — correct, as unsaturated hydrocarbons react with NO$$_x$$ in the presence of sunlight to form photochemical smog.

Option D is the correct answer.

We need to match the water pollutants/chemicals in List-I with their effects in List-II.

Excess sulphate in drinking water causes a laxative effect (diarrhoea); therefore, A - Sulphate → (III) Laxative effect.

Excess fluoride in water causes fluorosis, resulting in bending of bones and mottling of teeth; hence, B - Fluoride → (II) Bending of bones.

Nicotine has historically been used as a pesticide (insecticide) to kill insects in agriculture, so C - Nicotine → (I) Pesticide.

Sodium arsenite functions as a herbicide (weed killer), giving D - Sodium arsenite → (IV) Herbicide.

This leads to the matching A - III, B - II, C - I, and D - IV, corresponding to Option B.

Eutrophication is the process where a water body becomes excessively enriched with nutrients (especially nitrogen and phosphorus), leading to excessive plant and algal growth.

Understand the eutrophication process: When excess nutrients enter a water body, they promote rapid growth of algae (algal bloom). This dense algal growth covers the surface, blocking sunlight from reaching deeper water.

Consequences of eutrophication: When the algae die, their decomposition by aerobic bacteria consumes large amounts of dissolved oxygen. This leads to:

$$\bullet$$ Depletion of dissolved oxygen (hypoxia or anoxia)

$$\bullet$$ Death of aquatic organisms (fish, invertebrates) that depend on oxygen

$$\bullet$$ Overall loss in biodiversity

Evaluate the options: Option A: Increase in biodiversity — Incorrect; eutrophication kills aquatic organisms due to oxygen depletion.

Option B: Loss in biodiversity — Correct; oxygen depletion causes death of fish and other aquatic life.

Option C: Break down of organic matter — While decomposition does occur, this is part of the process, not the main result of eutrophication.

Option D: Decrease in BOD — Incorrect; BOD actually increases due to the large amount of organic matter from dead algae.

The correct answer is Option B: Loss in biodiversity.

We need to match each pollutant in List-I with its correct source in List-II.

(A) Microorganisms

Microorganisms such as bacteria, viruses, and other pathogens are primarily found in domestic sewage (II). Human waste and household water contain large amounts of harmful microorganisms.

So, A matches with II.

(B) Plant nutrients

Plant nutrients like nitrates, phosphates, and potassium compounds come from chemical fertilizers (III). When excess fertilizers run off into water bodies, they cause eutrophication.

So, B matches with III.

(C) Toxic heavy metals

Toxic heavy metals such as mercury (Hg), lead (Pb), cadmium (Cd), and chromium (Cr) are released from chemical factories (IV) as industrial waste.

So, C matches with IV.

(D) Sediment

Sediment (soil, sand, and rock particles) is produced during strip mining (I). Strip mining removes the surface layer of earth, leading to large amounts of sediment runoff.

So, D matches with I.

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

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

We need to match pollutants in water with the diseases they cause at the given concentration levels.

(A) Sulphate (>500 ppm): Excess sulphate in water causes a laxative effect (diarrhoea). This matches with (III).

(B) Nitrate (>50 ppm): Excess nitrate causes methemoglobinemia (also called "blue baby syndrome"). Nitrate gets reduced to nitrite in the body, which converts haemoglobin to methemoglobin, reducing oxygen-carrying capacity. This matches with (I).

(C) Lead (>50 ppb): Lead is a toxic heavy metal that causes kidney damage and also damages the nervous system and liver. This matches with (IV).

(D) Fluoride (>2 ppm): Excess fluoride causes brown mottling of teeth (dental fluorosis). It also causes bone disease (skeletal fluorosis) at higher concentrations. This matches with (II).

A - III, B - I, C - IV, D - II

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

We need to identify which gas does NOT contribute to the greenhouse effect.

The greenhouse effect occurs when certain gases in the atmosphere absorb and re-emit infrared radiation, trapping heat. A gas must be able to change its dipole moment during molecular vibration to absorb infrared radiation.

Option A: $$H_2O$$ vapour - Water vapour is the most abundant greenhouse gas. It has a bent molecular geometry and a permanent dipole moment, allowing it to absorb infrared radiation effectively. This IS a greenhouse gas.

Option B: $$O_3$$ - Ozone has a bent structure and absorbs infrared radiation. This IS a greenhouse gas.

Option C: $$N_2$$ - Nitrogen is a homonuclear diatomic molecule. It has no permanent dipole moment and its symmetric stretching vibration does not produce a change in dipole moment. Therefore, it does NOT absorb infrared radiation and is NOT a greenhouse gas.

Option D: $$CH_4$$ - Methane has vibrational modes that change the dipole moment, making it a potent greenhouse gas. This IS a greenhouse gas.

The correct answer is Option C: $$N_2$$.

We need to identify the correct and complete reaction representing acid rain caused by SO₂. SO₂ in the atmosphere undergoes oxidation and reacts with water to form sulphuric acid (H₂SO₄), which falls as acid rain.

Option A: $$2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$$ — This only shows the oxidation of SO₂ to SO₃, not the complete acid rain reaction. Option B: $$\text{SO}_2 + \text{O}_3 \rightarrow \text{SO}_3 + \text{O}_2$$ — This uses ozone as oxidizer and still does not produce acid rain (no H₂SO₄). Option C: $$\text{SO}_2 + \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{SO}_4$$ — This produces H₂SO₄ but uses H₂O₂, which is not the primary pathway. Option D: $$2\text{SO}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4$$ — This is the complete reaction showing SO₂ being oxidized and combined with water to form sulphuric acid.

Option D represents the correct and complete reaction for acid rain caused by SO₂, so the answer is Option D: $$2\text{SO}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4$$.

We need to identify the water samples that are highly polluted with organic wastes based on their BOD values. The BOD values for the samples are A = 3 ppm, B = 18 ppm, C = 21 ppm, and D = 4 ppm.

Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen needed by aerobic organisms to break down organic material in water. A higher BOD indicates more organic pollution.

Water with BOD < 5 ppm is considered clean, while water with BOD > 17 ppm is considered highly polluted.

A = 3 ppm — Clean water
B = 18 ppm — Highly polluted (BOD > 17 ppm)
C = 21 ppm — Highly polluted (BOD > 17 ppm)
D = 4 ppm — Clean water

The highly polluted samples are B and C.

The correct answer is Option C: B & C.

We need to identify which compound is generally NOT found in photochemical smog.

Photochemical smog is formed by the action of sunlight on nitrogen oxides ($$NO_x$$) and volatile organic compounds (VOCs) released mainly by automobile exhaust.

The key components of photochemical smog include:

- Nitrogen oxides: $$NO$$ and $$NO_2$$

- Ozone ($$O_3$$)

- Formaldehyde ($$HCHO$$) and other aldehydes

- Peroxyacetyl nitrate (PAN)

- Various organic peroxides

$$SO_2$$ (sulphur dioxide) is a component of classical smog (also called London smog or reducing smog), which is caused by burning of coal and is associated with cool, humid conditions. It is NOT a typical component of photochemical smog.

Hence, the correct answer is Option C.

We need to evaluate both statements about water pollution.

Statement I: "In polluted water, values of both dissolved oxygen and BOD are very low."

Polluted water contains a large amount of organic matter. Microorganisms consume dissolved oxygen to decompose this organic matter. Therefore:

• Dissolved oxygen (DO) is low in polluted water — this part is correct.
• Biochemical Oxygen Demand (BOD) is high in polluted water, because a large quantity of oxygen is needed to break down the organic waste — this part is incorrect.

So Statement I is false.

Statement II: "Eutrophication results in decrease in the amount of dissolved oxygen."

Eutrophication is the excessive growth of algae and aquatic plants due to nutrient enrichment (phosphates, nitrates). When these organisms die, their decomposition by bacteria consumes large amounts of dissolved oxygen, leading to oxygen depletion in the water body.

So Statement II is true.

Therefore, the correct answer is Option D: Statement I is false but Statement II is true.

On the surface of polar stratospheric clouds, chlorine nitrate ($$ClONO_2$$) undergoes hydrolysis: $$ClONO_2 + H_2O \rightarrow HOCl + HNO_3$$, so A = HOCl and B = $$HNO_3$$. The reaction with HCl is $$ClONO_2 + HCl \rightarrow Cl_2 + HNO_3$$, producing B = $$HNO_3$$ and C = $$Cl_2$$. Thus A = HOCl, B = $$HNO_3$$, C = $$Cl_2$$, which matches Option A.

The answer is $$\boxed{\text{Option A}}$$.

First extract the data from the table given in the question.

Compound 1 → $$\%P = 50,\; \%Q = 50$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{50}{50}=1$$
Compound 2 → $$\%P = 44.4,\; \%Q = 55.6$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{44.4}{55.6}=0.8=\dfrac45$$
Compound 3 → $$\%P = 40,\; \%Q = 60$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{40}{60}= \dfrac23$$

Case 1: Checking Option A - “If empirical formula of compound 3 is $$P_3Q_4$$, then empirical formula of compound 2 is $$P_3Q_5$$.”

For compound 3 assume $$P_3Q_4$$.
Weight ratio $$\dfrac{P}{Q}= \dfrac{3M_P}{4M_Q}= \dfrac23$$   $$\Rightarrow 9M_P = 8M_Q$$   $$\Rightarrow \dfrac{M_P}{M_Q}= \dfrac89$$ $$-(1)$$
With the same atomic masses, the ratio for $$P_3Q_5$$ (suggested for compound 2) would be
$$\dfrac{P}{Q}= \dfrac{3M_P}{5M_Q}= \dfrac{3\left(\tfrac89 M_Q\right)}{5M_Q}= \dfrac{24}{45}=0.533$$
The experimental ratio for compound 2 is $$0.8$$, not $$0.533$$. Hence Option A is WRONG.

Case 2: Checking Option B - “If empirical formula of compound 3 is $$P_3Q_2$$ and atomic weight of P is 20, atomic weight of Q is 45.”

Assume $$M_P = 20$$ and empirical formula $$P_3Q_2$$ for compound 3.
Mass of P in one empirical unit = $$3 \times 20 = 60$$
Let $$M_Q = x$$. Mass of Q in one empirical unit = $$2x$$.
Percentage of P: $$\dfrac{60}{60+2x}\times100 = 40\%$$ (from data of compound 3)
$$\dfrac{60}{60+2x}=0.40 \;\Rightarrow\; 60 = 0.40(60+2x)=24+0.8x$$
$$60-24 = 0.8x \;\Rightarrow\; 36 = 0.8x \;\Rightarrow\; x = 45$$
Thus $$M_Q = 45\;\text{u}$$. Option B is CORRECT.

Case 3: Checking Option C - “If empirical formula of compound 2 is $$PQ$$, then empirical formula of compound 1 is $$P_5Q_4$$.”

For compound 2 let empirical formula be $$PQ$$.
Weight ratio $$\dfrac{P}{Q}= \dfrac{M_P}{M_Q}=0.8=\dfrac45$$ ⇒ take $$M_P = 4k,\; M_Q = 5k$$ $$-(2)$$.
With these atomic masses, the ratio for $$P_5Q_4$$ (suggested for compound 1) becomes
$$\dfrac{P}{Q}= \dfrac{5M_P}{4M_Q}= \dfrac{5\,(4k)}{4\,(5k)}= \dfrac{20k}{20k}=1$$
This matches the experimental ratio of compound 1 (50 % : 50 % ⇒ 1 : 1). Hence Option C is CORRECT.

Case 4: Checking Option D - “If atomic weights of P and Q are 70 and 35 respectively, empirical formula of compound 1 is $$P_2Q$$.”

Given $$M_P = 70,\; M_Q = 35$$, compound 1 has equal masses of P and Q.
To get equal masses: number of moles $$n_P : n_Q = \dfrac{1}{70} : \dfrac{1}{35} = 1 : 2$$.
Thus the simplest integer formula is $$PQ_2$$, not $$P_2Q$$. Option D is WRONG.

Therefore the correct choices are:
Option B (atomic weight of Q is 45) and Option C (empirical formula $$P_5Q_4$$).

Final answer: Option B, Option C.

We need to identify which substance enhances the lathering property of soap.

Lathering refers to the formation of foam when soap is mixed with water. To improve lathering, substances called builders or lather boosters are added to soap.

Considering the given options:

- Sodium stearate: This is itself a soap (sodium salt of stearic acid). It does not specifically enhance lathering.

- Sodium carbonate: This is a water softener (washing soda). It helps soap work in hard water but does not directly enhance lathering.

- Sodium rosinate: This is the sodium salt of rosin (abietic acid), obtained from pine trees. It is specifically added to soap to increase lather formation. Rosin soap produces rich, stable foam.

- Trisodium phosphate: This is a cleaning agent and water softener but does not enhance lathering.

Sodium rosinate is added to soaps as a lather-enhancing agent. It improves the foaming ability of the soap. Hence, the correct answer is Option C.

We are given that 56.0 L of nitrogen gas is mixed with excess hydrogen gas and 20 L of ammonia is produced. We need to find the volume of unused nitrogen gas.

Write the balanced equation

$$N_2 + 3H_2 \rightarrow 2NH_3$$

Use the volume ratio (at same T and P, volume ratio = mole ratio)

From the equation: 1 volume of $$N_2$$ produces 2 volumes of $$NH_3$$.

Calculate the volume of $$N_2$$ used

Volume of $$NH_3$$ produced = 20 L

$$\text{Volume of } N_2 \text{ used} = \frac{20}{2} = 10 \text{ L}$$

Calculate unused $$N_2$$

$$\text{Unused } N_2 = 56.0 - 10 = 46 \text{ L}$$

The answer is 46.

We need to find the volume at which all liquid water evaporates while maintaining equilibrium vapour pressure. The mass of water is 0.90 g and its molar mass is 18 g/mol, so $$n = \frac{0.90}{18} = 0.05 \text{ mol}$$. The vapour pressure is 32.0 Torr, which is $$P = 32.0 \text{ Torr} = \frac{32.0}{760} \text{ atm} = 0.04211 \text{ atm}$$.

When all liquid water evaporates, all 0.05 mol exists as vapour at the equilibrium pressure. By the ideal gas law, $$PV = nRT$$, so $$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{32/760}$$, which gives $$= \frac{0.05 \times 0.082 \times 300}{0.04211}$$ and then $$= \frac{1.23}{0.04211} = 29.2 \approx 29 \text{ litres}$$.

The answer is 29 litres.

We are given that at the same pressure, 3.0 g of gas A at 300 K occupies the same volume as 0.2 g of hydrogen at 200 K.

For an ideal gas: $$PV = nRT$$. Since both gases occupy the same volume at the same pressure: $$V_A = V_{H_2}$$, which implies $$\frac{n_A R T_A}{P} = \frac{n_{H_2} R T_{H_2}}{P}$$ and hence $$n_A \times T_A = n_{H_2} \times T_{H_2}$$.

Expressing moles in terms of mass and molar mass, $$n_A = \frac{3.0}{M_A}$$ and $$n_{H_2} = \frac{0.2}{2.0} = 0.1$$ mol. Substituting into the relation gives $$\frac{3.0}{M_A} \times 300 = 0.1 \times 200$$, so $$\frac{900}{M_A} = 20$$ and thus $$M_A = \frac{900}{20} = 45 \text{ g mol}^{-1}$$.

The molar mass of gas A is 45 g mol$$^{-1}$$.

We need to find the value of $$x$$ such that the number of Mg atoms in the solution is $$x \times 10^{20}$$.

The concentration of Mg is $$48 \text{ ppm}$$. For aqueous solutions, ppm means mg of solute per L of solution (since density of dilute solution $$\approx 1 \text{ g/mL}$$).

$$48 \text{ ppm} = 48 \text{ mg of Mg per litre of solution}$$

Volume of solution = $$2 \text{ L}$$

$$\text{Mass of Mg} = 48 \times 2 = 96 \text{ mg} = 0.096 \text{ g}$$

Atomic mass of Mg = $$24 \text{ g/mol}$$

$$\text{Moles of Mg} = \frac{0.096}{24} = 0.004 \text{ mol} = 4 \times 10^{-3} \text{ mol}$$

$$\text{Number of Mg atoms} = \text{Moles} \times N_A = 4 \times 10^{-3} \times 6.02 \times 10^{23}$$

$$= 4 \times 6.02 \times 10^{-3+23}$$

$$= 24.08 \times 10^{20}$$

$$\text{Number of Mg atoms} = 24.08 \times 10^{20} \approx 24 \times 10^{20}$$

So $$x = 24$$ (nearest integer).

Hence, the value of $$x$$ is 24.

We need to find the mass of $$CO_2$$ produced when 100 g of methane (CNG) is mixed with 208 g of oxygen.

The balanced chemical equation for the combustion of methane is $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$.

The number of moles of methane present is calculated by dividing its mass by its molar mass: $$\frac{100}{16} = 6.25$$ mol.

The number of moles of oxygen available is determined similarly: $$\frac{208}{32} = 6.5$$ mol.

According to the balanced equation, one mole of methane requires two moles of oxygen. To fully react 6.25 moles of methane, $$6.25 \times 2 = 12.5$$ moles of oxygen would be needed, but only 6.5 moles are available.

Since oxygen is insufficient to react with all the methane, it is the limiting reagent.

The stoichiometry of the reaction indicates that two moles of oxygen produce one mole of carbon dioxide. Therefore, the moles of carbon dioxide formed are $$\frac{6.5}{2} = 3.25$$ mol.

Multiplying the moles of carbon dioxide by its molar mass gives the mass produced: $$3.25 \times 44 = 143$$ g.

Therefore, the amount of carbon dioxide produced is 143 g.

The reaction is given by $$X + Y + 3Z \rightleftharpoons XYZ_3$$ and the initial amounts are 1 mol of X, 1 mol of Y, and 0.05 mol of Z, while the atomic masses are 10 amu for X, 20 amu for Y, and 30 amu for Z.

Since the reaction requires X, Y, and Z in the molar ratio 1 : 1 : 3 and only 0.05 mol of Z is available, the maximum moles of $$XYZ_3$$ that can be formed is $$\frac{0.05}{3}$$ mol, indicating that Z is the limiting reagent.

Substituting the atomic masses into the molar mass formula gives $$M_{XYZ_3} = 10 + 20 + 3 \times 30 = 10 + 20 + 90 = 120 \text{ g/mol}$$.

From the moles of product and its molar mass, the mass formed is $$\text{Mass} = \frac{0.05}{3} \times 120 = \frac{6}{3} = 2 \text{ g}$$.

Therefore, the yield of $$XYZ_3$$ is 2 g.

We need to find the percentage of oxygen in an organic compound containing C, H, and O. The mass of carbon is determined from CO$$2$$: Mass of CO$$2$$ = 0.793 g, its molar mass is 44 g/mol, and each mole contains 12 g of carbon, giving $$\text{Mass of C} = \frac{12}{44} \times 0.793 = \frac{9.516}{44} = 0.2163 \text{ g}$$. The mass of hydrogen is found from H$$2$$O: Mass of H$$2$$O = 0.442 g, molar mass 18 g/mol, each mole contains 2 g of hydrogen, hence $$\text{Mass of H} = \frac{2}{18} \times 0.442 = \frac{0.884}{18} = 0.04911 \text{ g}$$.

Subtracting these from the total mass of the compound (0.492 g) gives the mass of oxygen as $$\text{Mass of O} = 0.492 - 0.2163 - 0.04911 = 0.2266 \text{ g}$$. The percentage of oxygen is then $$\% \text{ O} = \frac{0.2266}{0.492} \times 100 = 46.06\%$$. Rounding to the nearest integer yields 46, which is the required percentage of oxygen.

We need to find the number of nitrogen atoms in 681 g of $$C_7H_5N_3O_6$$.

The molar mass of $$C_7H_5N_3O_6$$ is determined by $$M = 7(12) + 5(1) + 3(14) + 6(16) = 84 + 5 + 42 + 96 = 227 \text{ g/mol}$$, so the number of moles in 681 g is $$n = \frac{681}{227} = 3 \text{ mol}$$.

Each molecule contains 3 nitrogen atoms, hence the total number of nitrogen atoms is $$\text{Total N atoms} = 3 \text{ mol} \times 3 \times 6.02 \times 10^{23} \text{ mol}^{-1} = 9 \times 6.02 \times 10^{23} = 54.18 \times 10^{23} = 5418 \times 10^{21}$$.

Expressing this as $$x \times 10^{21}$$ yields $$x = 5418$$, so the correct answer is 5418.

Initially, a 1 L aqueous solution contains 0.02 mmol of $$H_2SO_4$$. To determine the total millimoles of $$H_2SO_4$$ in the final solution, 50 % of this solution (0.5 L) is taken. Because 0.5 L is half of 1 L, the amount of $$H_2SO_4$$ in this aliquot is half of 0.02 mmol, which equals 0.01 mmol.

Next, this 0.5 L aliquot is diluted with deionized water to give 1 L of solution A; since dilution does not change the amount of solute, the quantity of $$H_2SO_4$$ remains 0.01 mmol.

Afterward, an additional 0.01 mmol of $$H_2SO_4$$ is added to solution A, yielding
$$\text{Total } H_2SO_4 = 0.01 + 0.01 = 0.02 \text{ mmol}$$

Finally, the result can be expressed in the required format as
$$0.02 \text{ mmol} = 20 \times 10^{-3} \text{ mmol}$$

The correct answer is 20.

We need to find the molar mass of an ideal gas using the ideal gas equation. The mass of gas (w) is 100 g, the volume (V) is 416 L, the temperature (T) is 27°C = 300 K, the pressure (P) is 1.5 bar, and R = 0.083 L bar K$$^{-1}$$ mol$$^{-1}$$.

Using the ideal gas equation $$PV = nRT$$ with $$n = \frac{w}{M}$$ (w = mass, M = molar mass) gives $$PV = \frac{w}{M} \times RT$$. Solving for the molar mass yields $$M = \frac{wRT}{PV}$$, so $$M = \frac{100 \times 0.083 \times 300}{1.5 \times 416}$$.

The numerator is $$100 \times 0.083 \times 300 = 2490$$ and the denominator is $$1.5 \times 416 = 624$$, hence $$M = \frac{2490}{624} = 3.99 \approx 4 \text{ g mol}^{-1}$$. The molar mass is approximately 4 g mol$$^{-1}$$, which corresponds to helium (He).

Hence, the answer is 4.

We need to find the moles of methane required to produce 81 g of water after complete combustion.

The balanced equation for the combustion of methane is $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$.

The moles of $$H_2O$$ produced are $$\frac{81}{18} = 4.5$$ mol.

According to the balanced equation, 1 mol of $$CH_4$$ yields 2 mol of $$H_2O$$, so the moles of $$CH_4$$ required are $$\frac{4.5}{2} = 2.25$$ mol.

Writing this in the specified format gives $$2.25$$ mol = $$225 \times 10^{-2}$$ mol.

Therefore, the answer is 225.

We need to find the pressure of nitrogen gas at 45°C in a rigid tank, given the initial conditions.

Since the tank is rigid (constant volume) and the amount of gas is constant, we use Gay-Lussac's Law:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

$$T_1 = 27°C + 273 = 300 \text{ K}$$

$$T_2 = 45°C + 273 = 318 \text{ K}$$

$$\frac{30}{300} = \frac{P_2}{318}$$

$$P_2 = \frac{30 \times 318}{300}$$

$$P_2 = \frac{9540}{300}$$

$$P_2 = 31.8 \text{ atm}$$

$$P_2 \approx 32 \text{ atm}$$

The pressure in the tank at 3:00 pm will be 32 atm.

We have 'x' g of $$O_2$$ mixed with 200 g of Ne in a cylinder. The total pressure is 25 bar and the partial pressure of Ne is 20 bar. We need to find x.

By Dalton's law of partial pressures, the total pressure equals the sum of partial pressures:

$$P_{total} = P_{Ne} + P_{O_2}$$

$$25 = 20 + P_{O_2}$$

$$P_{O_2} = 5 \text{ bar}$$

Now, the partial pressure of a gas is related to its mole fraction by $$P_i = \chi_i \times P_{total}$$. We can also write the ratio of partial pressures:

$$\frac{P_{O_2}}{P_{Ne}} = \frac{n_{O_2}}{n_{Ne}}$$

We first find the moles of Ne. With 200 g of Ne and molar mass 20 g/mol:

$$n_{Ne} = \frac{200}{20} = 10 \text{ mol}$$

Now using the pressure ratio:

$$\frac{5}{20} = \frac{n_{O_2}}{10}$$

$$n_{O_2} = \frac{5 \times 10}{20} = 2.5 \text{ mol}$$

The mass of $$O_2$$ is:

$$x = n_{O_2} \times M_{O_2} = 2.5 \times 32 = 80 \text{ g}$$

Hence, the correct answer is 80.

To determine the basicity of acid 'A', we apply the law of equivalence, which states that in a complete neutralization reaction, the milliequivalents (mEq) of the acid must equal the milliequivalents of the base. The formula for calculating milliequivalents is Molarity × Volume × n-factor. First, we calculate the milliequivalents for the base, $$M(OH)_2$$. Its volume is 30 mL, its molarity is 0.05 M, and its acidity (n-factor) is 2 because it can donate two hydroxide ions. Multiplying these together gives $$30 \times 0.05 \times 2 = 3$$ mEq. Next, we set up the expression for acid 'A', which has a volume of 10 mL, a molarity of 0.1 M, and an unknown basicity (n-factor) that we will call '$$x$$'. This gives us $$10 \times 0.1 \times x = 1x$$ mEq. By setting the milliequivalents equal to each other ($$1x = 3$$), we find that $x$ equals 3. Therefore, the basicity of the acid is 3.

We consider a body-centred cubic (BCC) unit cell with edge length $$a = 300$$ pm $$= 300 \times 10^{-10}$$ cm $$= 3 \times 10^{-8}$$ cm and density $$\rho = 6.0$$ g cm$$^{-3}$$. For BCC, the number of atoms per unit cell is $$Z = 2$$.

Using the density relation $$\rho = \frac{Z \times M}{N_A \times a^3}$$ and solving for $$M$$ gives $$M = \frac{\rho \times N_A \times a^3}{Z}$$. Substituting the known values yields $$M = \frac{6.0 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{2}$$.

Since $$a^3 = (3 \times 10^{-8})^3 = 27 \times 10^{-24} = 2.7 \times 10^{-23}$$ cm$$^3$$, substituting this into the expression for $$M$$ gives $$M = \frac{6.0 \times 6.022 \times 10^{23} \times 2.7 \times 10^{-23}}{2} = \frac{6.0 \times 6.022 \times 2.7}{2} = \frac{97.56}{2} = 48.78$$ g/mol.

To find the number of atoms in 180 g of the element, the number of moles is calculated as $$\text{Number of moles} = \frac{180}{48.78} = 3.69$$ mol. Therefore, the number of atoms is $$\text{Number of atoms} = 3.69 \times 6.022 \times 10^{23} = 22.22 \times 10^{23}$$.

Therefore, the number of atoms present in 180 g is $$22 \times 10^{23}$$ (nearest integer).

We are given that anion $$B^-$$ forms a cubic close packing (CCP, i.e., FCC) structure, with cation $$A^+$$ occupying all octahedral voids. The ionic radii are $$r_{A^+} = 102$$ pm and $$r_{B^-} = 181$$ pm.

In an FCC unit cell, there are 4 anions per unit cell (8 corner atoms contributing $$\frac{1}{8}$$ each and 6 face-centred atoms contributing $$\frac{1}{2}$$ each: $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$). The number of octahedral voids in an FCC unit cell equals the number of atoms, which is 4. Since $$A^+$$ occupies all octahedral voids, we have 4 cations per unit cell, giving the stoichiometry AB. This is the well-known rock salt (NaCl) structure.

In the rock salt structure, the octahedral voids are located at the edge centres and at the body centre of the cube. Consider one edge of the unit cell: at one end sits a $$B^-$$ anion (at a corner), at the midpoint of the edge sits an $$A^+$$ cation (in the octahedral void), and at the other end sits another $$B^-$$ anion (at the adjacent corner). The cation and anions touch along the edge.

The edge length $$a$$ therefore equals the distance from the centre of one anion through the cation to the centre of the next anion:

$$a = r_{B^-} + 2r_{A^+} + r_{B^-} = 2r_{A^+} + 2r_{B^-}$$

This can also be written as $$a = 2(r_{A^+} + r_{B^-})$$. The factor of 2 appears because both a cation radius and an anion radius span from the edge centre to each corner, and there are two such spans along the full edge.

Now we substitute the given values:

$$a = 2(102 + 181) = 2 \times 283 = 566 \text{ pm}$$

Hence, the correct answer is 566.

Let the atomic weight of A be $$M_A$$ and the atomic weight of B be $$M_B$$.

The molecular weight of $$A_2B = 2M_A + M_B$$

The molecular weight of $$AB_3 = M_A + 3M_B$$

Given: 0.15 moles of $$A_2B$$ and 0.15 moles of $$AB_3$$ weigh equally.

Weight of $$A_2B = 0.15 \times (2M_A + M_B)$$

Weight of $$AB_3 = 0.15 \times (M_A + 3M_B)$$

Since they weigh equally:

$$0.15(2M_A + M_B) = 0.15(M_A + 3M_B)$$

Dividing both sides by 0.15:

$$2M_A + M_B = M_A + 3M_B$$

$$2M_A - M_A = 3M_B - M_B$$

$$M_A = 2M_B$$

Therefore, the atomic weight of A is 2 times the atomic weight of B.

We are given that metal M crystallizes in an FCC lattice with edge length $$a = 4.0 \times 10^{-8}$$ cm, and the density $$\rho = 9.03$$ g/cm$$^3$$. We need to find the atomic mass.

For an FCC unit cell, the number of atoms per unit cell is $$Z = 4$$ (since there are 8 corner atoms each contributing $$\frac{1}{8}$$ and 6 face-centered atoms each contributing $$\frac{1}{2}$$, giving $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$).

The relation between density and the unit cell parameters is given by:

where $$M$$ is the molar mass, $$a$$ is the edge length, and $$N_A$$ is Avogadro's number.

Rearranging for $$M$$:

Now substituting the values:

So we get:

Rounding to the nearest integer, we get $$M \approx 87$$ g/mol.

Hence, the correct answer is 87.

The compound has the formula $$C_xH_yO_z$$ and weighs 1.80 g. On complete combustion, it produces 2.64 g of $$CO_2$$ and 1.08 g of $$H_2O$$.

Moles of $$CO_2 = \frac{2.64}{44} = 0.06\,\text{mol}$$. Since each mole of $$CO_2$$ contains one mole of carbon, moles of C $$= 0.06$$. Mass of C $$= 0.06 \times 12 = 0.72\,\text{g}$$.

Moles of $$H_2O = \frac{1.08}{18} = 0.06\,\text{mol}$$. Since each mole of $$H_2O$$ contains 2 moles of hydrogen, moles of H $$= 0.12$$. Mass of H $$= 0.12 \times 1 = 0.12\,\text{g}$$.

Mass of oxygen in the compound $$= 1.80 - 0.72 - 0.12 = 0.96\,\text{g}$$.

Percentage of oxygen $$= \frac{0.96}{1.80} \times 100 = 53.33\%$$.

Therefore, the percentage of oxygen in the organic compound is $$53.33\%$$, which corresponds to option (1).

We begin with the van der Waals equation for n moles of a real gas

$$\left(P + \frac{a n^{2}}{V^{2}}\right)\,(V - nb) \;=\; nRT$$

In this expression the pressure-correction term $$\frac{a n^{2}}{V^{2}}$$ is added directly to the pressure $$P$$. Because only like quantities can be added, the whole fraction $$\frac{a n^{2}}{V^{2}}$$ must possess exactly the same unit as $$P$$.

Let us denote the unit of pressure by $$[P]$$, the unit of volume by $$[V]$$ and the unit of amount of substance by $$[n]$$. Correspondingly, let $$[a]$$ be the unknown unit of the parameter $$a$$ which we have to determine.

Since the dimensions of the two addends are equal, we can write

$$\Bigl[\frac{a n^{2}}{V^{2}}\Bigr] \;=\; [P]$$

Substituting the separate units we get

$$[a] \,[n]^{2}\,[V]^{-2} \;=\; [P]$$

Rearranging for $$[a]$$ gives

$$[a] \;=\; [P]\,[V]^{2}\,[n]^{-2}$$

Now we replace each symbol by the commonly used practical units in which the van der Waals constants are usually expressed:

• The convenient laboratory unit for pressure is the atmosphere, so $$[P] = \text{atm}$$.
• The convenient laboratory unit for volume is the cubic decimetre, so $$[V] = \text{dm}^3$$.
• The SI-derived unit for the amount of substance is the mole, so $$[n] = \text{mol}$$.

Substituting these explicit units we arrive at

$$[a] \;=\; (\text{atm})\,(\text{dm}^{3})^{2}\,(\text{mol})^{-2}$$

Simplifying the exponent on the volume unit, we finally have

$$[a] = \text{atm}\,\text{dm}^{6}\,\text{mol}^{-2}$$

This exactly matches Option C in the given list.

Hence, the correct answer is Option C.

Heavy water is $$\text{D}_2\text{O}$$ in which the ordinary hydrogen $$({^1\text{H}})$$ is replaced by its heavier isotope deuterium $$({^2\text{H}} = \text{D})$$.

Case 1: Checking Assertion (A).
Because deuterium behaves chemically like protium yet can be detected separately, replacing $$\text{H}$$ with $$\text{D}$$ in a reactant makes it possible to trace which bonds break and which intermediates form. Such isotopic labelling studies help chemists determine detailed reaction mechanisms. Hence, Assertion (A) is true.

Case 2: Checking Reason (R).
Bond strength increases with atomic mass since the zero-point vibrational energy decreases. Therefore the $$\text{O-D}$$ bond is stronger than the $$\text{O-H}$$ bond. Because a stronger bond is harder to break, reactions that involve cleavage of $$\text{O-D}$$ proceed more slowly than those that cleave $$\text{O-H}$$. Mathematically this is the primary kinetic isotope effect: $$k_{\text{O-H}} \gt k_{\text{O-D}}$$. The statement given in (R) says “the rate of reaction for the cleavage of O-H bond is slower than that of O-D bond,” which is the opposite of the correct relationship. Thus, Reason (R) is false.

Case 3: Relation between (A) and (R).
Since (A) is true and (R) is false, (R) cannot be the correct explanation of (A).

Therefore the correct option is Option B: (A) is true but (R) is false.

Excerpt from NCERT:
Important ingredients present in Portland cement are dicalcium silicate ($$Ca_2SiO_4$$) 26%, tricalcium silicate ($$Ca_3SiO_5$$) 51% and tricalcium aluminate ($$Ca_3Al_2O_6$$) 11%.

Answer:
The major constituent is tricalium silicate, thus the right option is D.

We start by recalling the basic principle of dry-cleaning. In dry-cleaning, clothes are not washed with water; instead, an organic solvent (a liquid that can dissolve grease and oil but is itself almost non-polar) is used. These solvents lift dirt without wetting the fabric with water. Commonly employed dry-cleaning solvents in the chemical industry include chlorinated hydrocarbons and, more recently, super-critical or liquid $$\mathrm{CO_2}$$ because of its low environmental impact.

Let us now look at every option one by one and match it with the list of known dry-cleaning agents.

First option: $$\mathrm{H_2O_2}$$ is hydrogen peroxide. It is widely recognised as an oxidising and bleaching agent. It liberates nascent oxygen according to the reaction $$\mathrm{H_2O_2 \;\rightarrow\; H_2O + \tfrac12\,O_2\uparrow}$$ and therefore is mainly used for bleaching hair, paper pulp, cotton etc. Because it is an aqueous solution and works by oxidation in water, it is not suitable for the non-aqueous process called dry-cleaning.

Second option: $$\mathrm{CCl_4}$$ is carbon tetrachloride. This compound is a non-polar liquid, so it dissolves greasy dirt and has historically been used as a dry-cleaning solvent (although now phased out for toxicity reasons). Hence it is a recognised dry-cleaning agent.

Third option: liquid $$\mathrm{CO_2}$$. Under high pressure, carbon dioxide can exist as a liquid or super-critical fluid. Super-critical $$\mathrm{CO_2}$$ has become a modern, eco-friendly alternative to chlorinated solvents in dry-cleaning. Therefore liquid $$\mathrm{CO_2}$$ also fits the definition of a dry-cleaning agent.

Fourth option: $$\mathrm{Cl_2C{=}CCl_2}$$ is perchloroethylene (also called tetrachloroethylene). Its non-polarity and high density make it the most widely used commercial dry-cleaning solvent. Thus it clearly belongs to the family of dry-cleaning agents.

From the discussion above, the only substance that does not qualify as a dry-cleaning solvent is hydrogen peroxide, $$\mathrm{H_2O_2}$$, which is actually a water-based bleach.

Hence, the correct answer is Option A.

Green chemistry aims to reduce or eliminate the use of hazardous substances in chemical processes. The correct answer is option 4: liquified $$\text{CO}_2$$ for dry cleaning of clothes.

Let us evaluate each option. Chlorine used for bleaching paper is harmful as it produces toxic byproducts such as dioxins and chlorinated compounds, making it environmentally unfriendly. Using a large amount of water alone for washing clothes is wasteful of a precious natural resource, which contradicts green chemistry principles. Tetrachloroethene (perchloroethylene) used in dry cleaning is a volatile organic compound that is toxic and a suspected carcinogen, and it contaminates groundwater — exactly the kind of substance green chemistry seeks to replace.

Liquified $$\text{CO}_2$$ as a solvent for dry cleaning is the green chemistry alternative. $$\text{CO}_2$$ is non-toxic, non-flammable, and the carbon dioxide used is often a byproduct of existing industrial processes, so its use does not add net carbon to the atmosphere. It is an excellent solvent that leaves no harmful residues and is easily recycled in a closed-loop system. This makes it the best example of green chemistry in day-to-day life among the given options.

The correct answer is option 4.

We need to identify the environmental consequence of gas released during anaerobic degradation of vegetation.

Anaerobic degradation means decomposition of organic matter (vegetation) in the absence of oxygen. This process is carried out by methanogenic bacteria. The primary gas released is methane ($$CH_4$$).

Methane is a greenhouse gas. Greenhouse gases trap infrared radiation emitted by the Earth's surface and prevent it from escaping into space. This leads to an increase in the average temperature of the Earth's atmosphere, a phenomenon known as global warming.

Methane is approximately 25 times more effective than carbon dioxide ($$CO_2$$) at trapping heat over a 100-year period. Therefore, even small amounts of methane can contribute significantly to global warming.

Additionally, methane in the atmosphere undergoes photochemical reactions that produce ground-level ozone and other secondary pollutants. Prolonged exposure to such pollutants has been associated with increased cancer risk in living organisms.

The other options do not correctly describe the primary effect of methane. Corrosion of metals is caused by electrochemical reactions, not methane. The ozone hole is caused by chlorofluorocarbons (CFCs), not methane. Acid rain is caused by $$SO_2$$ and $$NO_x$$ emissions, not methane.

Hence, the gas released during anaerobic degradation of vegetation may lead to global warming and cancer.

Hence, the correct answer is Option 3.

Exerpt from NCERT: "Some of the heat is trapped by gases such as carbondioxide, methane, ozone, chlorofluorocarbon compounds (CFCs) and water vapour in the atmosphere. Thus, they add to the heating of the atmosphere. This causes global warming."

Thus the right option is C (A, C, D).

First, let us recall what photochemical smog is. It is the brown hazy mixture that is produced when primary pollutants such as $$\text{NO}$$ and $$\text{NO}_2$$, emitted from vehicular exhaust, react in the presence of strong sunlight with hydrocarbons present in air. The major secondary products that are finally present in this smog are ozone $$\left(\text{O}_3\right)$$, nitric oxide $$\left(\text{NO}\right)$$, acrolein $$\left(\text{CH}_2\!\!=\!\text{CH}-\text{CHO}\right)$$, formaldehyde $$\left(\text{HCHO}\right)$$ and peroxyacetyl nitrate $$\left(\text{PAN},\ \text{CH}_3\text{COOONO}_2\right)$$.

Now, all these substances—especially ozone and PAN—are very powerful oxidising agents. An oxidising agent has the tendency to remove electrons from, that is, oxidise, the substances it comes in contact with. Natural as well as synthetic rubber contains unsaturated $$\text{C=C}$$ bonds in its long-chain molecules. Oxidising agents attack these double bonds, break the chains and weaken the material. This chemical attack is popularly called “cracking of rubber”.

So we have:

$$\text{O}_3 + \text{Rubber} \;\longrightarrow\; \text{Cracked\ rubber\ fragments}$$

Because photochemical smog supplies the oxidants—$$\text{O}_3,\ \text{PAN},\ \text{NO}$$ etc.—it definitely causes cracking of rubber that is exposed to the atmosphere. Thus the statement in the Assertion (A) is correct.

Reason (R) says that the presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in photochemical smog makes it oxidising. We have just seen that each of these species is indeed an oxidant (ozone and PAN are very strong; acrolein and formaldehyde are milder but still oxidising). Therefore the Reason (R) is also correct.

Furthermore, we notice that the reason directly supplies the chemical cause behind the cracking mentioned in the assertion: because the smog is oxidising, it attacks the $$\text{C=C}$$ bonds of rubber and cracks it. Therefore (R) is the true and sufficient explanation of (A).

Hence, the correct answer is Option 3.

Statement I says that non-biodegradable wastes are generated by thermal power plants. Thermal power plants burn coal and produce fly ash, bottom ash, and flue gas pollutants. These solid wastes are non-biodegradable. Statement I is true.

Statement II says that biodegradable detergents lead to eutrophication. Even biodegradable detergents contain phosphates, and when these phosphates enter water bodies, they serve as nutrients promoting excessive algal growth, leading to eutrophication. Statement II is true.

Since both statements are true, the correct answer is option (4).

Statement I states that the pH of rain water is normally ~5.6. This is true because atmospheric $$CO_2$$ dissolves in rainwater to form carbonic acid: $$CO_2 + H_2O \rightarrow H_2CO_3$$. This weak acid lowers the pH of pure rainwater from 7.0 to approximately 5.6.

Statement II states that if the pH of rain water drops below 5.6, it is called acid rain. This is also true. Acid rain is defined as rain with a pH below the normal value of 5.6, typically caused by pollutants such as $$SO_2$$ and $$NO_x$$ that form sulphuric acid and nitric acid in the atmosphere.

Since both statements are true, the correct answer is option (1): Both Statement I and Statement II are true.

Statement I says that the value of the parameter "Biochemical Oxygen Demand (BOD)" is important for survival of aquatic life. This is true. BOD measures the amount of dissolved oxygen consumed by microorganisms in the biological decomposition of organic matter in water. A high BOD indicates a large amount of organic pollutants, which depletes dissolved oxygen and threatens aquatic life. Therefore, monitoring BOD is indeed important for the survival of aquatic organisms.

Statement II says that the optimum value of BOD is 6.5 ppm. This is false. The value of 6.5 ppm is actually associated with the optimum pH of water or is sometimes confused with dissolved oxygen levels, not BOD. Clean water typically has a BOD of less than 5 ppm. A BOD value of 6.5 ppm would indicate moderately polluted water. The lower the BOD, the cleaner the water and the better conditions for aquatic life.

Therefore, the correct answer is Option (1): Statement I is true but Statement II is false.

We begin with compound (a) carbon monoxide, whose most well-known toxic action is its strong affinity for the blood pigment haemoglobin. The reaction is written as $$\mathrm{CO + Hb \;\longrightarrow\; HbCO}$$ and the product, called carboxy-haemoglobin, blocks the normal transport of oxygen. Thus the only entry in List-II that fits carbon monoxide is “haemoglobin (iii).”

Next, compound (b) sulphur dioxide is an acidic, hygroscopic gas that injures green vegetation; one characteristic injury noted in horticulture is the abnormal hardening or stiffness of developing flower buds. Therefore sulphur dioxide must be matched with “stiffness of flower buds (iv).”

For compound (c) polychlorinated biphenyls (PCBs), textbooks of environmental chemistry state that these chlorinated aromatic compounds are potent human and animal carcinogens. Hence PCBs correspond to “carcinogenic (i).”

Finally, compound (d) the oxides of nitrogen (NO and NO2) can act as a supplementary nitrogen source for certain higher plants; pear trees belonging to the genus Pyrus are specifically reported to absorb and metabolise these oxides. Consequently, oxides of nitrogen are best paired with “metabolized by pyrus plants (ii).”

Collecting all the matches we have found:

$$\begin{aligned} (a) &\;\;\text{Carbon monoxide} \;\longrightarrow\; (iii)\;\text{Haemoglobin} \\ (b) &\;\;\text{Sulphur dioxide} \;\longrightarrow\; (iv)\;\text{Stiffness of flower buds} \\ (c) &\;\;\text{Polychlorinated biphenyls} \;\longrightarrow\; (i)\;\text{Carcinogenic} \\ (d) &\;\;\text{Oxides of nitrogen} \;\longrightarrow\; (ii)\;\text{Metabolized by pyrus plants} \end{aligned}$$

This sequence corresponds exactly to Option A.

Hence, the correct answer is Option A.

Photochemical smog (also called oxidising smog or Los Angeles smog) is formed during the daytime in the presence of sunlight. It involves the photochemical reaction of nitrogen oxides ($$NO_x$$) and volatile organic compounds (VOCs) in the presence of sunlight, producing ozone ($$O_3$$), peroxyacetyl nitrate (PAN), and other secondary pollutants.

The presence of $$O_3$$ is a characteristic feature of photochemical or oxidising smog. Since ozone and other oxidants are the primary components, this type of smog is called oxidising smog. It increases during the daytime because sunlight drives the photochemical reactions that produce these oxidising pollutants.

Reducing smog (London smog), by contrast, occurs in cooler, humid conditions and contains $$SO_2$$ and soot particles. It is not associated with ozone formation.

Therefore, the correct answer is Option (2): Oxidising smog.

We begin by recalling what eutrophication means. Eutrophication is the enrichment of a water body with nutrients, chiefly nitrates ($$\text{NO}_3^-$$) and phosphates ($$\text{PO}_4^{3-}$$). This nutrient loading stimulates excessive algal and plant growth.

Because of the dense algal bloom, sunlight penetration into deeper layers of water is hindered. Aquatic plants in the deeper zone therefore cannot photosynthesise efficiently, so their production of oxygen decreases. Later, when the algal mass dies, its decomposition by bacteria consumes the dissolved oxygen (DO) present in water.

The stoichiometric oxygen demand of the decomposition can be expressed as

$$\text{Organic Matter} + O_2 \longrightarrow \text{CO}_2 + \text{H}_2O + \text{Energy}.$$

The more biomass there is, the more $$O_2$$ is required for complete oxidation. Hence, eutrophication always tends to lower the dissolved oxygen level.

Now we check each statement one by one.

Option A says, “Eutrophication indicates that the water body is polluted.” We have just seen that eutrophication arises from an overload of nutrients, which is indeed a form of water pollution. So this statement is correct.

Option B says, “The dissolved oxygen concentration below 6 ppm inhibits fish growth.” The thumb rule used by environmental engineers is that fish need at least about $$5 \text{-} 6 \text{ mg L}^{-1}$$ (ppm) of DO for normal growth; if $$\text{DO} \lt 6 \text{ ppm},$$ fish become stressed and growth is hampered. Hence Option B is correct.

Option C says, “Eutrophication leads to increase in the oxygen level in water.” We have already explained that decomposition of the excess biomass consumes oxygen according to

$$\text{BOD} = \frac{\text{Mass of } O_2 \text{ consumed}}{\text{Volume of water}},$$

and a high biochemical oxygen demand (BOD) means lower residual $$O_2$$. Therefore, eutrophication does not increase oxygen; it depletes it. Thus Option C is an incorrect statement.

Option D says, “Eutrophication leads to anaerobic conditions.” As the DO falls far below $$1 \text{ ppm},$$ aerobic microorganisms die, and anaerobic bacteria begin to dominate, producing gases like $$\text{H}_2S$$ and $$\text{CH}_4$$. So Option D is a correct statement.

Only Option C contradicts the scientific facts. Hence, the correct answer is Option C.

We first examine Statement I. Chlorofluorocarbons (CFCs) present in the stratosphere break down only when they absorb photons whose energy is high enough to cleave the $$\mathrm{C\!-\!Cl}$$ bond. The bond dissociation energy of a typical $$\mathrm{C\!-\!Cl}$$ bond is about $$328\;\text{kJ mol}^{-1}$$. The wavelength $$\lambda$$ corresponding to this energy is obtained from the relation $$E=\dfrac{hc}{\lambda}$$, where $$h$$ is Planck’s constant and $$c$$ is the speed of light. Substituting $$E=328\times10^{3}\ \text{J mol}^{-1}$$ and using $$h=6.626\times10^{-34}\ \text{J s},\;c=3.0\times10^{8}\ \text{m s}^{-1},\;N_A=6.022\times10^{23}\ \text{mol}^{-1}$$, we obtain

$$\lambda=\dfrac{hcN_A}{E}=\dfrac{(6.626\times10^{-34})(3.0\times10^{8})(6.022\times10^{23})}{328\times10^{3}}$$ $$\lambda\approx2.2\times10^{-7}\ \text{m}=220\ \text{nm}$$

This wavelength lies in the ultraviolet region, not in the visible region (visible light spans roughly $$400\;\text{nm}$$ to $$700\;\text{nm}$$). Hence CFCs are photodissociated by UV radiation, not by visible light. Furthermore, the primary product is a chlorine radical $$\mathrm{Cl\cdot}$$, not the molecular gas $$\mathrm{Cl_2}$$. Therefore Statement I is false.

Now we analyse Statement II. Stratospheric ozone does interact with nitric oxide, but the reaction is

$$\mathrm{O_3 + NO \;\longrightarrow\; NO_2 + O_2}$$

No molecular nitrogen $$\mathrm{N_2}$$ is generated; instead nitric oxide is converted to nitrogen dioxide, and only one molecule of oxygen gas is produced. Thus the claim that the reaction produces “nitrogen and oxygen gases” is incorrect. Consequently Statement II is also false.

Since both statements have been shown to be incorrect, we select the option that says both statements are false. Looking at the given list, Option B corresponds to this description.

Hence, the correct answer is Option B.

Smog is classified into two types: reducing smog (classical smog) and photochemical smog (oxidizing smog).

Reducing smog, also called London smog or classical smog, occurs in cool, humid climates and is caused by the burning of coal and other fossil fuels. It is a mixture of smoke, fog, and sulfur dioxide ($$SO_2$$). The $$SO_2$$ in reducing smog can further react with moisture to form sulfuric acid, which is why it has a reducing character and causes acid rain.

Photochemical smog, on the other hand, contains oxidants like ozone ($$O_3$$) and is formed in warm, sunny conditions from nitrogen oxides and hydrocarbons.

Therefore, reducing smog is a mixture of smoke, fog, and $$SO_2$$, which is option (B).

The reduction of a metal oxide by coke (carbon) depends on the relative positions of the metal and carbon on the Ellingham diagram. Carbon can reduce a metal oxide only if the Gibbs free energy line for the formation of $$CO/CO_2$$ lies below the line for the metal oxide at the operating temperature.

Aluminium oxide ($$Al_2O_3$$) is extremely stable with a very large negative $$\Delta G_f°$$. On the Ellingham diagram, the line for $$Al_2O_3$$ lies well below the line for $$CO$$ at all practical temperatures. This means carbon cannot reduce $$Al_2O_3$$ to aluminium. This is why aluminium is extracted industrially by electrolysis of molten alumina (Hall-Heroult process) rather than by carbon reduction.

In contrast, $$ZnO$$, $$Fe_2O_3$$, and $$Cu_2O$$ can all be reduced by coke at sufficiently high temperatures, as their Ellingham diagram lines cross above the carbon line. Zinc is extracted by carbon reduction at about 1400°C, iron is produced in blast furnaces using coke, and copper oxide is easily reduced by carbon.

Therefore, the reduction that cannot be carried out with coke is $$Al_2O_3 \to Al$$, which is option (1).

The question asks which statement is an incorrect reason for eutrophication. Eutrophication is the excessive enrichment of water bodies with nutrients (especially nitrogen and phosphorus), which triggers rapid algal and plant growth, oxygen depletion, and ecosystem damage.

(A) Excess usage of fertilisers causes nutrient runoff (nitrogen and phosphorus compounds) into rivers and lakes, directly contributing to eutrophication. This is a correct reason.

(B) Excess usage of detergents introduces phosphates into water bodies, which promotes excessive algal growth. This is a correct reason for eutrophication.

(C) Dense plant population in water bodies is directly linked to excess nutrients already present, and such overgrowth further depletes oxygen, worsening eutrophication. This is a valid reason associated with the eutrophication process.

(D) Lack of nutrients in water bodies that prevent plant growth is the exact opposite of what causes eutrophication. Eutrophication occurs due to an excess of nutrients. A nutrient deficiency leads to oligotrophic (nutrient-poor) conditions, not eutrophication. This is clearly an incorrect reason.

Therefore, the incorrect reason for eutrophication is (D) only.

Let us analyze both statements carefully.

Statement I says: "An allotrope of oxygen is an important intermediate in the formation of reducing smog." The allotrope of oxygen referred to here is ozone ($$O_3$$). However, ozone is an important intermediate in the formation of photochemical smog, not reducing smog. Reducing smog (also called London smog or classical smog) is formed from smoke and fog in cold, humid conditions due to combustion of coal, and involves $$SO_2$$ and particulate matter — not ozone. Therefore, Statement I is false.

Statement II says: "Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog." Photochemical smog is primarily caused by oxides of nitrogen ($$NO_x$$) and volatile organic compounds (VOCs) reacting in the presence of sunlight. Oxides of sulphur ($$SO_x$$) are associated with reducing smog (classical smog), not photochemical smog. Therefore, Statement II is also false.

Since both statements are false, the correct answer is option (1): Both Statement I and Statement II are false.

The removal of sulphur from sulphide ores is achieved by the process of roasting. In roasting, the sulphide ore is heated strongly in the presence of excess air (oxygen) below its melting point. During this process, the sulphide ore is converted to the corresponding metal oxide, and sulphur is released as sulphur dioxide ($$SO_2$$) gas.

For example, when zinc sulphide ore is roasted: $$2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$$. Similarly, for copper: $$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$.

Smelting involves melting the ore with a reducing agent, leaching is a chemical washing process to dissolve the ore, and refining is the purification of the crude metal. None of these specifically target sulphur removal.

Therefore, the correct answer is Option (2): Roasting.

We match each ore with its correct chemical formula.

(a) Haematite is an iron ore with the formula $$\text{Fe}_2\text{O}_3$$, which matches (ii).

(b) Bauxite is an aluminium ore with the formula $$\text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O}$$, which matches (i).

(c) Magnetite is an iron ore with the formula $$\text{Fe}_3\text{O}_4$$, which matches (iv).

(d) Malachite is a copper ore with the formula $$\text{CuCO}_3 \cdot \text{Cu(OH)}_2$$, which matches (iii).

Therefore, the correct matching is (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii).

Among the given gases, $$\text{NO}_2$$ (nitrogen dioxide) is reported to retard photosynthesis. Nitrogen dioxide reacts with water to form nitrous acid and nitric acid. In the atmosphere, elevated concentrations of $$\text{NO}_2$$ interfere with the photosynthetic process in plants by damaging leaf tissue and inhibiting chlorophyll function.

$$\text{CO}$$ is primarily toxic by binding to hemoglobin. $$\text{CO}_2$$ is actually a reactant in photosynthesis and does not retard it. CFCs (chlorofluorocarbons) deplete the ozone layer but are not directly known to retard photosynthesis. $$\text{NO}_2$$ is specifically documented as a photosynthesis inhibitor because it generates reactive oxygen species and causes oxidative stress in plant cells.

Therefore, the correct answer is $$\text{NO}_2$$.

We have been supplied with 100 g of propane, whose molecular formula is $$\mathrm{C_3H_8}$$. The molar mass of propane is obtained by adding the atomic masses: $$3(12\ \text{g mol}^{-1}) + 8(1\ \text{g mol}^{-1}) = 44\ \text{g mol}^{-1}.$$

The general formula for finding moles is first stated:

$$\text{Number of moles} = \dfrac{\text{Given mass}}{\text{Molar mass}}.$$

Using this, the number of moles of propane present is

$$n_{\mathrm{C_3H_8}} = \dfrac{100\ \text{g}}{44\ \text{g mol}^{-1}} = 2.2727\ \text{mol}.$$

Next, the oxygen. The molecular mass of $$\mathrm{O_2}$$ is $$32\ \text{g mol}^{-1}.$$ Hence,

$$n_{\mathrm{O_2,\ given}} = \dfrac{1000\ \text{g}}{32\ \text{g mol}^{-1}} = 31.25\ \text{mol}.$$

The balanced combustion equation for propane is written and will now be used for stoichiometry:

$$\mathrm{C_3H_8 + 5\,O_2 \;\rightarrow\; 3\,CO_2 + 4\,H_2O}.$$

From the coefficients we see that 1 mol of propane needs 5 mol of oxygen. Therefore, the oxygen required for the full consumption of the available propane is

$$n_{\mathrm{O_2,\ reqd}} = 5 \times n_{\mathrm{C_3H_8}} = 5 \times 2.2727 = 11.3636\ \text{mol}.$$

Because $$31.25\ \text{mol} > 11.3636\ \text{mol},$$ oxygen is in excess and propane is the limiting reagent. Hence all propane will be consumed.

By direct proportion from the balanced equation, the amounts of products formed are:

• Carbon dioxide: $$n_{\mathrm{CO_2}} = 3 \times n_{\mathrm{C_3H_8}} = 3 \times 2.2727 = 6.8182\ \text{mol}.$$
• Water vapour: $$n_{\mathrm{H_2O}} = 4 \times n_{\mathrm{C_3H_8}} = 4 \times 2.2727 = 9.0909\ \text{mol}.$$

The oxygen actually consumed equals $$11.3636\ \text{mol},$$ so the oxygen remaining after reaction is

$$n_{\mathrm{O_2,\ left}} = 31.25 - 11.3636 = 19.8864\ \text{mol}.$$

The gaseous mixture after completion therefore contains:

$$\begin{aligned} n_{\mathrm{CO_2}} &= 6.8182\ \text{mol},\\ n_{\mathrm{H_2O}} &= 9.0909\ \text{mol},\\ n_{\mathrm{O_2}} &= 19.8864\ \text{mol}. \end{aligned}$$

The total number of moles in the mixture is obtained by simple addition:

$$n_{\text{total}} = 6.8182 + 9.0909 + 19.8864 = 35.7955\ \text{mol}.$$

We now recall the definition of mole fraction:

$$\chi_i = \dfrac{n_i}{n_{\text{total}}}.$$

Applying this to carbon dioxide,

$$\chi_{\mathrm{CO_2}} = \dfrac{6.8182}{35.7955} = 0.1904.$$

The question states that this value is written as $$x \times 10^{-2},$$ so

$$x = 0.1904 \times 10^{2} = 19.04.$$

Rounding to the nearest integer gives $$x = 19.$$

So, the answer is $$19$$.

We are given that complete combustion of 750 g of an organic compound produces 420 g of $$CO_2$$ and 210 g of $$H_2O$$. We need to find the percentage of hydrogen in the compound.

First, let us find the mass of carbon. The molar mass of $$CO_2$$ is 44 g/mol, and each mole contains 12 g of carbon. So the mass of carbon in 420 g of $$CO_2$$ is $$\frac{12}{44} \times 420 = \frac{5040}{44} = 114.545$$ g.

The percentage of carbon is $$\frac{114.545}{750} \times 100 = 15.27\%$$, which rounds to approximately 15.3%, consistent with the given information.

Now, let us find the mass of hydrogen. The molar mass of $$H_2O$$ is 18 g/mol, and each mole contains 2 g of hydrogen. So the mass of hydrogen in 210 g of $$H_2O$$ is $$\frac{2}{18} \times 210 = \frac{420}{18} = 23.33$$ g.

The percentage of hydrogen is $$\frac{23.33}{750} \times 100 = 3.11\%$$.

Rounding off to the nearest integer, the percentage of hydrogen is $$\mathbf{3}$$.

For the complete combustion of butane, we first recall and write the balanced chemical equation:

$$2\,\mathrm{C_4H_{10}} \;+\; 13\,\mathrm{O_2}\;\longrightarrow\; 8\,\mathrm{CO_2}\;+\; 10\,\mathrm{H_2O}$$

From this equation we see that:

$$2 \text{ mol } \mathrm{C_4H_{10}} \;\xrightarrow{\text{combustion}}\; 10 \text{ mol } \mathrm{H_2O}$$

Now, we are told that the actual amount of water obtained is $$72.0\ \text{g}$$. Let us convert this mass of water into moles. The molar mass of water $$\mathrm{H_2O}$$ is

$$M(\mathrm{H_2O}) = (2 \times 1) + 16 = 18\ \text{g mol}^{-1}.$$

Therefore, the number of moles of water produced is

$$n(\mathrm{H_2O}) \;=\;\frac{72.0\ \text{g}}{18\ \text{g mol}^{-1}} \;=\; 4.0\ \text{mol}.$$

According to the stoichiometric ratio obtained from the balanced equation, $$2$$ moles of butane give $$10$$ moles of water. Hence, to find the moles of butane required for $$4.0$$ moles of water we write the proportion

$$\frac{2\ \text{mol }\mathrm{C_4H_{10}}}{10\ \text{mol }\mathrm{H_2O}} \;=\;\frac{x\ \text{mol }\mathrm{C_4H_{10}}}{4.0\ \text{mol }\mathrm{H_2O}}.$$

Cross-multiplying gives

$$10\,x = 2 \times 4.0,$$

so

$$x = \frac{2 \times 4.0}{10} = 0.8\ \text{mol}.$$

Now we convert the required moles of butane to mass. The molar mass of butane ($$\mathrm{C_4H_{10}}$$) is

$$M(\mathrm{C_4H_{10}}) = (4 \times 12) + (10 \times 1) = 48 + 10 = 58\ \text{g mol}^{-1}.$$

Hence, the mass of butane consumed is

$$m(\mathrm{C_4H_{10}}) = n \times M = 0.8\ \text{mol} \times 58\ \text{g mol}^{-1} = 46.4\ \text{g}.$$

The question asks us to express this answer in the form $$\_\_\_\_\_\times 10^{-1}\ \text{g}$$ and then give the blank as the nearest integer. We rewrite

$$46.4\ \text{g} = 464 \times 10^{-1}\ \text{g}.$$

So the integer that fills the blank is $$464$$.

Hence, the correct answer is Option 464.

Let the gaseous hydrocarbon be $$C_xH_y$$. The balanced combustion reaction is:

$$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

Since all volumes are measured under the same conditions of temperature and pressure, by Avogadro's law, the ratio of volumes equals the ratio of moles for gaseous species.

From the given information, 1 volume of $$C_xH_y$$ requires 6 volumes of $$O_2$$. So:

$$x + \frac{y}{4} = 6 \quad \cdots (i)$$

Also, 1 volume of $$C_xH_y$$ produces 4 volumes of $$CO_2$$. So:

$$x = 4 \quad \cdots (ii)$$

Substituting $$x = 4$$ in equation (i):

$$4 + \frac{y}{4} = 6$$