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.

For an ideal gas mixture, the partial pressure $$P_i$$ of any component is given by
$$P_i = x_i \, P_{\text{total}}$$
where $$x_i$$ is the mole fraction of that component. Hence, the ratio of two partial pressures equals the ratio of the corresponding mole fractions.

The composition is provided as mass % at sea level:
nitrogen (N2) : 70.0 %, oxygen (O2) : 27.0 %, argon (Ar) : 3.0 %.

Assume 100 g of this dry air (any convenient mass may be chosen).
  Mass of N2 = 70 g  Mass of O2 = 27 g  Mass of Ar = 3 g

Convert each mass to moles using molar masses $$M$$:
$$n_{\text{N}_2} = \frac{70}{28} = 2.50 \text{ mol}$$
$$n_{\text{O}_2} = \frac{27}{32} = 0.84375 \text{ mol}$$
$$n_{\text{Ar}} = \frac{3}{40} = 0.075 \text{ mol}$$

Total moles
$$n_{\text{total}} = 2.50 + 0.84375 + 0.075 = 3.41875 \text{ mol}$$

Mole fractions
$$x_{\text{N}_2} = \frac{2.50}{3.41875} = 0.7313$$
$$x_{\text{O}_2} = \frac{0.84375}{3.41875} = 0.2467$$
$$x_{\text{Ar}} = \frac{0.075}{3.41875} = 0.02193$$

Because $$\dfrac{P_{\text{A}}}{P_{\text{B}}} = \dfrac{x_{\text{A}}}{x_{\text{B}}}$$ :

Case (i):
$$\frac{P_{\text{N}_2}}{P_{\text{O}_2}} = \frac{x_{\text{N}_2}}{x_{\text{O}_2}} = \frac{0.7313}{0.2467} \approx 2.96$$

Case (ii):
$$\frac{P_{\text{O}_2}}{P_{\text{Ar}}} = \frac{x_{\text{O}_2}}{x_{\text{Ar}}} = \frac{0.2467}{0.02193} \approx 11.2$$

Thus the required ratios are
(i) $$2.96$$ and (ii) $$11.2$$.

Comparing with the options, these correspond to Option D.

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.

Step 1: Find the mass of carbon present in $$220 \text{ mg}$$ of $$CO_2$$.
Molar mass: $$CO_2$$ has $$12 \text{ g}$$ of C in $$44 \text{ g}$$ of $$CO_2$$.
Hence the mass fraction of C in $$CO_2$$ is $$\frac{12}{44} = \frac{3}{11}$$.

Mass of C obtained from combustion = $$220 \text{ mg} \times \frac{3}{11}$$.
Calculate: $$\frac{220}{11} = 20$$, so $$20 \times 3 = 60 \text{ mg}$$.

Step 2: Calculate percentage of carbon in the original compound.
Original sample mass = $$500 \text{ mg}$$.
Percentage of C = $$\frac{60 \text{ mg}}{500 \text{ mg}} \times 100$$.

Simplify: $$\frac{60}{500} = 0.12$$, therefore percentage of C = $$0.12 \times 100 = 12\%$$.

Nearest integer value = $$12$$.

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.

Let us analyze both statements:

Statement I: "Permutit process is more efficient compared to the synthetic resin method for the softening of water."

This is incorrect. The synthetic resin (ion exchange resin) method is more efficient than the permutit process. Synthetic resins can remove both cations and anions, while permutit only exchanges cations.

Statement II: "Synthetic resin method results in the formation of soluble sodium salts."

This is incorrect. The synthetic resin method uses cation exchange resins (H⁺ form) and anion exchange resins (OH⁻ form). The cation exchange resin replaces Ca²⁺, Mg²⁺ etc. with H⁺, and the anion exchange resin replaces anions with OH⁻. The final product is deionized water, not sodium salts. (The permutit process forms sodium salts, not the synthetic resin method.)

Since both statements are incorrect, the answer is Option 1: Both the statements I and II are incorrect.

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$$_3$$.

The main radical responsible for ozone depletion in the presence of UV radiation is the chlorine radical (Cl•).

The mechanism of ozone depletion by chlorine radicals:

CFCs (chlorofluorocarbons) release chlorine atoms when exposed to UV radiation in the stratosphere:

$$\text{CF}_2\text{Cl}_2 \xrightarrow{UV} \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet$$

The chlorine radical then catalytically destroys ozone:

$$\text{Cl}^\bullet + \text{O}_3 \to \text{ClO}^\bullet + \text{O}_2$$

$$\text{ClO}^\bullet + \text{O} \to \text{Cl}^\bullet + \text{O}_2$$

A single chlorine radical can destroy thousands of ozone molecules.

The answer is Option 1: Cl•.

For good quality cement, the ratio:

$$\frac{\text{Lime}}{\text{Silica} + \text{Alumina} + \text{Ferric oxide}}$$

should be close to: $${2}$$

This ratio ensures proper strength and setting properties of cement.
Hence, Option C is correct

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.

Greenhouse gases are those that absorb and emit infrared radiation, contributing to the greenhouse effect.

A. Water vapour: The most abundant greenhouse gas. Yes.

B. Ozone (O₃): A greenhouse gas that absorbs infrared radiation. Yes.

C. I₂ (Iodine): Molecular iodine is not a greenhouse gas. It does not significantly absorb infrared radiation in the atmosphere. No.

D. Molecular hydrogen (H₂): H₂ is a homonuclear diatomic molecule with no permanent dipole moment, so it does not absorb infrared radiation. No.

The greenhouse gases are A and B only.

This matches option 3.

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.

BOD measures the amount of dissolved oxygen required by aerobic microorganisms to break down the organic matter present in a given water sample. It serves as a direct indicator of organic water pollution:

• High BOD: Indicates a large amount of organic waste, which consumes significant dissolved oxygen, leaving less for aquatic life (high pollution).
• Low BOD: Indicates very little organic waste or debris, meaning the dissolved oxygen remains abundant and stable.

Standard Water Quality Classification (NCERT Guidelines):

According to environmental chemistry standards, water quality is classified based on its BOD value (measured in ppm or mg/L):

• $$\text{BOD} < 5\text{ ppm}$$: Classified as very clean water.
• $$\text{BOD} \ge 17\text{ ppm}$$: Classified as highly polluted water.

Since the analyzed pond water has a BOD value of 4 (which is less than 5), it falls squarely into the category of unpolluted, healthy water.

Note on the shared web page: The automated green checkmark on the reference page incorrectly labels it as "Highly polluted water." This is factually wrong; a low BOD value of 4 represents clean, unpolluted water.

Conclusion:

A BOD value of 4 indicates minimal organic waste, designating the pond as having clean water.

Answer: Option C — Very clean water

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).

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

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

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

BaCO₃ is a white precipitate.

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

This matches option 2: BaCO₃.

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.

We are given that 1.1 g of a carbonate $$\text{M}_2\text{CO}_3$$ on treatment with excess HCl produces 0.01 mol of $$\text{CO}_2$$.

Write the balanced equation: $$ \text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2 $$

Find moles of $$\text{M}_2\text{CO}_3$$. From the balanced equation, 1 mol of $$\text{M}_2\text{CO}_3$$ produces 1 mol of $$\text{CO}_2$$.

Therefore, moles of $$\text{M}_2\text{CO}_3$$ = moles of $$\text{CO}_2$$ = 0.01 mol.

Calculate molar mass: $$ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{1.1}{0.01} = 110 \text{ g mol}^{-1} $$

Verification: If molar mass of $$\text{M}_2\text{CO}_3$$ = 110, then $$2M + 12 + 48 = 110$$, giving $$M = 25$$. This corresponds to a monovalent metal with atomic mass 25.

The molar mass of $$\text{M}_2\text{CO}_3$$ is 110 g mol$$^{-1}$$.

An organic compound gives 0.220 g of CO$$_2$$ and 0.126 g of H$$_2$$O on complete combustion, and the percentage of carbon is 24%.

$$\text{Mass of C} = \frac{12}{44} \times 0.220 = \frac{12 \times 0.220}{44} = \frac{2.64}{44} = 0.06 \text{ g}$$

$$\text{Mass of H} = \frac{2}{18} \times 0.126 = \frac{0.252}{18} = 0.014 \text{ g}$$

Since the percentage of carbon is 24%:

$$\frac{0.06}{m} \times 100 = 24$$

$$m = \frac{0.06 \times 100}{24} = 0.25 \text{ g}$$

$$\% \text{H} = \frac{0.014}{0.25} \times 100 = 5.6\%$$

The percentage of hydrogen is $$5.6 = 56 \times 10^{-1}$$.

The answer is $$\mathbf{56}$$.

We are given the reaction $$2C(s) + O_2(g) \rightarrow 2CO(g)$$ and that 12 g of carbon is burnt in 48 g of oxygen.

To determine the amount of carbon monoxide formed, we first calculate the moles of each reactant. Moles of C = $$\frac{12}{12} = 1$$ mol and moles of O$$_2$$ = $$\frac{48}{32} = 1.5$$ mol.

From the stoichiometric relationship 2 mol C requires 1 mol O$$_2$$, it follows that 1 mol C requires 0.5 mol O$$_2$$. Since there are 1.5 mol O$$_2$$ available, oxygen is in excess and carbon is the limiting reagent.

Because 2 mol C produce 2 mol CO, 1 mol C will produce 1 mol CO, so the amount of CO produced is 1 mol.

At STP, the molar volume is 22.7 L/mol, giving a volume of 1 mol × 22.7 L/mol = 22.7 L, which can be expressed as 227 × 10$$^{-1}$$ L.

The correct answer is 227.

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 use the density formula for a crystal unit cell:

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

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

We have $$d = 4.0$$ g cm$$^{-3}$$, $$M = 56 + 16 = 72$$ g mol$$^{-1}$$ (for FeO), $$a = 5.0$$ $$\text{\AA}$$ $$= 5.0 \times 10^{-8}$$ cm, and $$N_A = 6.0 \times 10^{23}$$ mol$$^{-1}$$.

Now, $$a^3 = (5.0 \times 10^{-8})^3 = 125 \times 10^{-24} = 1.25 \times 10^{-22}$$ cm$$^3$$.

So $$N_A \times a^3 = 6.0 \times 10^{23} \times 1.25 \times 10^{-22} = 75$$.

From the density formula:

$$Z = \frac{d \times N_A \times a^3}{M} = \frac{4.0 \times 75}{72} = \frac{300}{72} \approx 4.17$$

Rounding to the nearest integer gives $$Z = 4$$.

So, the answer is $$4$$.

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.

We need to find the molarity of a commercially sold concentrated HCl solution that is 35% HCl by mass with a density of 1.46 g/mL.

The formula for molarity in terms of mass percentage and density is given by:

$$\text{Molarity} = \frac{\text{Mass percentage} \times \text{Density} \times 10}{\text{Molar mass of solute}}$$

The molar mass of HCl is calculated as:

$$M_{HCl} = 1 + 35.5 = 36.5$$ g/mol

Substituting the known values into the formula yields:

$$\text{Molarity} = \frac{35 \times 1.46 \times 10}{36.5}$$

$$= \frac{35 \times 14.6}{36.5}$$

$$= \frac{511}{36.5}$$

$$= 14.0$$ M

To verify this result, consider 1 L (1000 mL) of the solution, for which the mass of solution is:

$$1000 \times 1.46 = 1460$$ g.

The mass of HCl present is 35% of this, namely:

$$0.35 \times 1460 = 511$$ g.

The number of moles of HCl is then:

$$\frac{511}{36.5} = 14.0$$ mol,

confirming a molarity of 14.0 M $$\checkmark$$

Therefore, the correct answer is Option B: 14.0 M.

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 balanced equation for the reduction of magnetite by carbon monoxide is given by $$Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g)$$. The molar mass of $$Fe_3O_4$$ is calculated as $$3(56) + 4(16) = 168 + 64 = 232\ \text{g/mol}$$; hence, the number of moles of $$Fe_3O_4$$ in a $$4640\ \text{g}$$ sample is $$\frac{4640}{232} = 20\ \text{mol}$$.

Similarly, the molar mass of carbon monoxide is $$12 + 16 = 28\ \text{g/mol}$$, so the number of moles in $$2520\ \text{g}$$ of CO is $$\frac{2520}{28} = 90\ \text{mol}$$. According to the stoichiometry, each mole of $$Fe_3O_4$$ requires 4 moles of CO, so for $$20\ \text{mol}$$ of $$Fe_3O_4$$ the required amount is $$20 \times 4 = 80\ \text{mol CO}$$. Since $$90\ \text{mol CO}$$ are available, carbon monoxide is in excess and $$Fe_3O_4$$ is the limiting reagent.

Each mole of $$Fe_3O_4$$ yields 3 moles of iron, so $$20\ \text{mol}$$ of $$Fe_3O_4$$ produces $$20 \times 3 = 60\ \text{mol Fe}$$; the mass of this iron is $$60 \times 56 = 3360\ \text{g}$$. Therefore, the mass of iron obtained is 3360 g, corresponding to 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.

An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with liquid (density 0.95 g/mL), and 40.5 g when filled with an ideal gas at 0.82 atm and 250 K. We need to find the molar mass of the gas.

First, the mass of liquid introduced into the vessel is calculated as $$135.0 - 40.0 = 95.0$$ g. The volume of the vessel is then given by the ratio of this mass to the liquid’s density, namely $$\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{95.0}{0.95} = 100\text{ mL} = 0.1\text{ L}.$$

The mass of the gas is found from the difference between the filled and empty vessel masses, which yields $$40.5 - 40.0 = 0.5$$ g.

Since the gas behaves ideally, we apply the ideal gas law: $$PV = nRT.$$ Solving for the number of moles gives $$n = \frac{PV}{RT} = \frac{0.82 \times 0.1}{0.082 \times 250} = \frac{0.082}{20.5} = 0.004\text{ mol}.$$

The molar mass of the gas is then determined by dividing the mass by the number of moles: $$M = \frac{\text{mass}}{n} = \frac{0.5}{0.004} = 125\text{ g/mol}.$$

The correct answer is Option D: 125 g/mol.

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 $$Mg(OH)_2$$.

the standard Ideal Gas Equation:

$$pV = nRT$$

Where:

• $$p$$ = Pressure
• $$V$$ = Volume
• $$n$$ = Number of moles
• $$R$$ = Gas constant
• $$T$$ = Temperature

We know that the number of moles ($$n$$) is equal to the mass of the gas ($$m$$) divided by its molar mass ($$M$$):

$$n = \frac{m}{M}$$

Substituting this back into the ideal gas equation:

$$pV = \left(\frac{m}{M}\right)RT$$

Rearranging the formula to isolate pressure ($$p$$):

$$p = \left(\frac{m}{V}\right) \frac{RT}{M}$$

Since density ($$d$$) is defined as mass per unit volume ($$d = \frac{m}{V}$$), we can substitute $$d$$ into the equation:

$$p = d \cdot \frac{RT}{M}$$
$$\text{Slope } (m) = \frac{RT}{M}$$

The slope of the line is directly proportional to the temperature ($$\text{Slope} \propto T$$).

• Higher Temperature $$\rightarrow$$ Steeper Slope (closer to the $$y$$-axis).
• Lower Temperature $$\rightarrow$$ Gentler Slope (closer to the $$x$$-axis).
  
  

Hence, Option B is correct.

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

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. 

Option B: Dieldrin

Dieldrin is another organochlorine insecticide. It was extensively used against soil-dwelling insects and crop pests. 

Option C: Organophosphate

Organophosphates are an entire class of insecticide pesticides. Common examples include malathion and parathion. These are 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.

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 acid mainly responsible for the damage to the Taj Mahal. The yellowing and deterioration of the Taj Mahal’s white marble is primarily caused by acid rain. Located in Agra, the monument is surrounded by several industries, including the Mathura oil refinery, which emit gases responsible for this phenomenon.

The burning of fossil fuels releases $$SO_2$$ and $$SO_3$$ into the atmosphere. These gases react with water vapour to form sulfuric acid $$H_2SO_4$$ through a series of reactions:

$$SO_2 + H_2O \rightarrow H_2SO_3$$

$$2SO_2 + O_2 \rightarrow 2SO_3$$

$$SO_3 + H_2O \rightarrow H_2SO_4$$

Marble is primarily composed of $$CaCO_3$$, and sulfuric acid reacts with it according to the equation $$CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2$$.

This process, often referred to as “Marble cancer” or “stone leprosy,” produces powdery $$CaSO_4$$ that flakes off, leading to the deterioration of the monument.

Option A: sulfuric acid.

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 $$\boxed{\text{Option A}}$$.