JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 20

Answer: (c) 23.3

Explanation:
1. Take 100 g of iron. Mass gained = 10 g of oxygen = 0.625 mol of O.
2. \( \ce{Fe2O3} \) has 2 Fe per 3 O: Fe rusted = 0.625 x 2/3 = 0.4167 mol = 23.3 g.
3. Percentage rusted = 23.3%.

Teacher's Note:
1. The mass gained is oxygen only. Link it to the iron through the formula.

Answer: (a) 7.75 g

Explanation:
1. Ag = 10/108 = 0.0926 mol; S = 1/32 = 0.03125 mol.
2. \( \ce{2Ag + S -> Ag2S} \): Ag/2 = 0.0463 > S = 0.03125, so sulphur is limiting.
3. \( \ce{Ag2S} \) = 0.03125 mol x 248 = 7.75 g.

Teacher's Note:
1. Divide by the coefficients to find the limiting reagent.

Answer: (d) y/x must be greater than or equal to 1.33

Explanation:
1. Carbon burning to CO needs 16 g of O per 12 g of C, so y/x = 1.33.
2. Carbon burning to \( \ce{CO2} \) needs 32 g per 12 g, so y/x = 2.67.
3. No solid carbon remains only if all carbon is oxidised, so y/x must be at least 1.33 (more oxygen is simply left over).

Teacher's Note:
1. The minimum oxygen needed to consume all the carbon corresponds to forming CO.

Answer: (b) 180

Explanation:
1. Balanced: \( \ce{3I2 + 7XeF6 -> 6IF7 + 7Xe} \).
2. IF7 = 210 x 6/7 = 180 mmol.

Teacher's Note:
1. Balance the fluorine: seven XeF6 supply 42 F atoms, which make six IF7.

Answer: (d) 500

Explanation:
1. 1 L has mass 1800 g and holds 18 x 98 = 1764 g of acid.
2. Solvent = 1800 - 1764 = 36 g = 0.036 kg.
3. Molality = 18/0.036 = 500 m.

Teacher's Note:
1. Molality needs the mass of the solvent, which is found from the density.

Answer: (a) mole fraction x and molality m

Explanation:
1. Mole fraction and molality depend only on masses or moles, not on volume.
2. They do not change with temperature.

Teacher's Note:
1. Only volume-based units such as molarity change with temperature.

Answer: (d) 46%

Explanation:
1. Take 1 mol of solution: ethanol = 0.25 mol = 11.5 g; water = 0.75 mol = 13.5 g.
2. Percentage = 11.5/25 x 100 = 46%.

Teacher's Note:
1. Mole fraction needs conversion to mass using molar masses.

Answer: (a) \( 5.8\times10^{-4} \) mol

Explanation:
1. 1 L of water = 1000 g, containing \( 10^{-3}/100\times1000 = 0.01 \) g of \( \ce{NH3} \).
2. Moles = 0.01/17 = \( 5.8\times10^{-4} \) mol.

Teacher's Note:
1. Use the density to turn 1 L into 1000 g, then take the percentage.

Answer: (a) 27 g

Explanation:
1. Methanol = 16/32 = 0.5 mol.
2. 0.5/(0.5 + n) = 0.25, so n = 1.5 mol of water.
3. Mass of water = 1.5 x 18 = 27 g.

Teacher's Note:
1. Solve the mole fraction expression for the moles of water.

Answer: (a) 80.8%

Explanation:
1. Mass of solution = 196 g, with 156.8 g of \( \ce{H2SO4} \).
2. 4 g \( \ce{SO3} \) = 0.05 mol gives 0.05 mol \( \ce{H2SO4} \) = 4.9 g.
3. Total \( \ce{H2SO4} \) = 161.7 g in 200 g, so 80.85%, about 80.8%.

Teacher's Note:
1. Sulphur trioxide adds both its own mass and its acid content to the solution.

Answer: (c) 3%

Explanation:
1. Mass = 15 x 0.792 + 15 x 1 = 11.88 + 15 = 26.88 g.
2. Final volume = 26.88/0.924 = 29.09 mL.
3. Contraction = (30 - 29.09)/30 x 100 = 3%.

Teacher's Note:
1. Mass is conserved on mixing but volume is not.

Answer: (a) 3 and 4 only

Explanation:
1. Equal masses give equal moles only if the molar masses are equal.
2. \( \ce{N2} \) = CO = 28 and \( \ce{N2O} \) = \( \ce{CO2} \) = 44.
3. The pairs in 3 and 4 qualify; 1 (44, 28) and 2 (28, 68) do not.

Teacher's Note:
1. Equal masses and equal numbers of molecules need equal molar masses.

Answer: (a) 1, 3 and 4 only

Explanation:
1. Ag = 100/108 = 0.926 mol.
2. KCN = 2 x 0.926 = 1.85 mol = 120 g (true).
3. \( \ce{O2} \) = 0.926/4 = 0.2315 mol = 7.4 g (true); volume = 0.2315 x 22.4 = 5.19 L, about 5.20 L (true). 0.742 g is wrong.

Teacher's Note:
1. Use the ratio 4 Ag : 8 KCN : 1 \( \ce{O2} \).

Answer: (a) 2, 3 and 4 only

Explanation:
1. C = 2.25 mol; \( \ce{O2} \) = 2.75 mol. Carbon is limiting, so (1) is false.
2. \( \ce{CO2} \) = 2.25 mol = 50.4 L (true). C : O = 12 : 32 = 3 : 8 (true).
3. \( \ce{O2} \) left = 0.5 mol = 11.2 L (true).

Teacher's Note:
1. Find the limiting reagent first; then check each statement.

Answer: (a) 2 and 4 only

Explanation:
1. Balanced: \( \ce{3A + 2B -> A3B2} \); \( \ce{A3B2 + 2C -> A3B2C2} \).
2. 3 mol A and 3 mol B give 1 mol of \( \ce{A3B2} \) (A limiting).
3. 1 mol C reacts with 0.5 mol \( \ce{A3B2} \) to give 0.5 mol \( \ce{A3B2C2} \), leaving 0.5 mol \( \ce{A3B2} \).

Teacher's Note:
1. Balance both reactions and carry the amounts through step by step.

Answer: (b) (p) and (s) only

Explanation:
1. C = 0.44/44 = 0.01 mol (0.12 g). H = 2 x 0.18/18 = 0.02 mol (0.02 g). O = 0.3 - 0.14 = 0.16 g = 0.01 mol.
2. Empirical formula \( \ce{CH2O} \); with two O atoms the molecular formula is \( \ce{C2H4O2} \), so (q) is false.
3. \( \ce{C2H4O2} \): 4 H atoms per molecule (p), \( \ce{CO2} \) = \( \ce{H2O} \) = 2 mol (r false), 0.25 mol gives 0.5 mol \( \ce{CO2} \) = 11.2 L (s true).

Teacher's Note:
1. Find the empirical formula from the combustion data, then use the O count to fix the molecular formula.

Answer: (d) (q) and (r) only

Explanation:
1. C = 10.5/12 = 0.875 mol per 1 mol H, so C : H = 7 : 8. Empirical formula \( \ce{C7H8} \) (92 g/mol).
2. Molar mass = 2 x 46 = 92, so the molecular formula is \( \ce{C7H8} \) (q true). It has 8 H atoms (p false).
3. Combustion gives 7 \( \ce{CO2} \) and 4 \( \ce{H2O} \) (r true). 0.25 mol gives 1.75 mol \( \ce{CO2} \) (s false).

Teacher's Note:
1. Vapour density is half the molar mass.

Answer: (c) (p), (q), (r) and (s)

Explanation:
1. HCl is limiting: 2 mol HCl reacts with 1 mol Zn (s).
2. Zn left = 1 mol of the 2 mol = 50% (p).
3. \( \ce{H2} \) = 1 mol = 22.4 L (q). \( \ce{ZnCl2} \) = 1 mol (r).

Teacher's Note:
1. Identify the limiting reagent first. All four statements follow from that.

Answer: (a) (p) and (s) only

Explanation:
1. \( \ce{AgNO3} \) = 170/170 = 1 mol; HCl = 18.25/36.5 = 0.5 mol. HCl is limiting (s).
2. AgCl = 0.5 mol (r false). \( \ce{AgNO3} \) left = 0.5 mol, 50% of the excess reagent (p).
3. HNO3 here is formed in solution/gas as 0.5 mol, not 1 mol, so (q) is false.

Teacher's Note:
1. The limiting reagent controls the amount of every product.

Answer: (c) (q) and (r) only

Explanation:
1. \( \ce{CaCO3} \) = 1 mol. It gives 1 mol CaO (solid) (r) and 1 mol \( \ce{CO2} \) = 22.4 L (q).
2. There is only one reactant, so there is no excess reagent and no HCl.

Teacher's Note:
1. 100 g of \( \ce{CaCO3} \) is exactly one mole.

Answer: (c) (q) only

Explanation:
1. \( \ce{KCl} \) = 2/3 mol (not 1 mol), so (r) is false.
2. \( \ce{O2} \) = 3/2 x 2/3 = 1 mol = 22.4 L (q true).
3. There is no excess reagent and no HCl.

Teacher's Note:
1. Use the ratio 2 : 2 : 3 for \( \ce{KClO3} \) : KCl : \( \ce{O2} \).

Answer: (a) (p) and (s) only

Explanation:
1. Total volume = 500 mL. \( Al^{3+} \) = 0.02 mol, \( H^+ \) = 0.04 mol: cations = 0.06 mol, so 0.12 M (p).
2. Cl- = 0.06 + 0.04 = 0.10 mol, so 0.2 M (s). There is no sulphate.

Teacher's Note:
1. Add the moles of each ion from both solutions and divide by the final volume.

Answer: (a) (s) only

Explanation:
1. KCl = 0.02 mol in 100 mL = 0.2 M, so [Cl-] = 0.2 M (s) and [K+] = 0.2 M (not 0.12 M).
2. There is no sulphate.

Teacher's Note:
1. Doubling the volume halves the concentration.