JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 42

Answer: (a) 2

Explanation:
1. Electrons lost: FeC₂O₄ gives 1 (Fe²⁺→Fe³⁺) + 2 (C₂O₄²⁻→2CO₂) = 3; Fe₂(C₂O₄)₃ gives 3 × 2 = 6; FeSO₄ gives 1; Fe₂(SO₄)₃ gives 0.
2. Total = 3 + 6 + 1 = 10 mol of electrons.
3. Each MnO₄⁻ accepts 5 electrons (Mn⁷⁺→Mn²⁺), so moles of KMnO₄ = 10/5 = 2.

Teacher's Note:
1. Count the electrons lost by every oxidisable species and divide by the n-factor of the oxidant.
2. Fe³⁺ salts such as Fe₂(SO₄)₃ cannot be oxidised further.

Answer: (a) 10⁻⁶ m

Explanation:
1. Mass of acid in 10 mL = 0.27 × 10/100 = 0.027 g, so its volume = 0.027/0.9 = 0.03 cm³.
2. The monolayer is a cylinder of radius 10 cm: V = πr²h, so h = 0.03/(3 × 100) = 1 × 10⁻⁴ cm.
3. h = 10⁻⁴ cm = 10⁻⁶ m.

Teacher's Note:
1. Convert the mass of acid to volume with density, then equate it to the cylinder volume.
2. Remember to convert cm to m at the end.

Answer: (a) 56 g of N₂ + 10 g of H₂

Explanation:
1. 28 g of N₂ reacts exactly with 6 g of H₂, so H₂ required per gram of N₂ = 6/28.
2. (a) 56 g N₂ needs 12 g H₂ but only 10 g is present, so H₂ is limiting.
3. (b) 35 g N₂ needs 7.5 g H₂ (8 g present); (c) 14 g N₂ needs 3 g H₂ (4 g present); (d) exact ratio. In (b), (c), (d) H₂ is not limiting.

Teacher's Note:
1. Compare the available amount with the amount required by stoichiometry for each reactant.
2. The reactant that runs out first is the limiting reagent.

Answer: (c) 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

Explanation:
1. Oxygen per gram of reactant: (a) 160/44 = 3.64 g; (b) 160/124 = 1.29 g.
2. (c) 96/224 = 0.43 g; (d) 32/48 = 0.67 g.
3. The smallest value is for iron, option (c).

Teacher's Note:
1. Compute grams of O₂ consumed divided by grams of reactant for every option.
2. Heavy reactants with few oxygen atoms per formula unit give small values.

Answer: (d) M_A = 5 × 10⁻³ and M_B = 10 × 10⁻³

Explanation:
1. 5(M_A + 2M_B) = 125 × 10⁻³, so M_A + 2M_B = 25 × 10⁻³.
2. 10(2M_A + 2M_B) = 300 × 10⁻³, so M_A + M_B = 15 × 10⁻³.
3. Subtracting: M_B = 10 × 10⁻³ and M_A = 5 × 10⁻³ kg mol⁻¹.

Teacher's Note:
1. Write mass = moles × molar mass for each compound and solve the two equations.
2. Check by substituting back into both relations.

Answer: (b) 495 g

Explanation:
1. Molar mass of C₅₇H₁₁₀O₆ = 57 × 12 + 110 + 96 = 890 g/mol.
2. 445 g = 0.5 mol, which gives (110/2) × 0.5 = 27.5 mol of H₂O.
3. Mass of H₂O = 27.5 × 18 = 495 g.

Teacher's Note:
1. Convert grams to moles, apply the mole ratio from the balanced equation, convert back to grams.
2. Here 2 mol of the fat gives 110 mol of water.

Answer: (c) 1

Explanation:
1. 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O.
2. Each oxalate (2 carbon atoms, C from +3 to +4) loses 2 electrons and yields 2 CO₂.
3. So one CO₂ molecule corresponds to 1 electron.

Teacher's Note:
1. Carbon changes oxidation state from +3 in oxalate to +4 in CO₂.
2. Each carbon atom therefore loses one electron.

Answer: (a) 100 ppm

Explanation:
1. 10⁻³ mol of CaSO₄ per litre is equivalent to 10⁻³ mol of CaCO₃ per litre.
2. Mass of CaCO₃ = 10⁻³ × 100 = 0.1 g per 1000 g of water.
3. Hardness = 0.1/1000 × 10⁶ = 100 ppm.

Teacher's Note:
1. Hardness always uses CaCO₃ (molar mass 100 g/mol) as the reference.
2. 1 ppm means 1 g per 10⁶ g of solution.

Answer: (d) 50

Explanation:
1. The unit µg/year shows a zero-order reaction.
2. For zero order, t½ = a₀/(2k).
3. t½ = 5 µg / (2 × 0.05 µg/year) = 50 years.

Teacher's Note:
1. The unit of k identifies the order: concentration per time means zero order.
2. Decomposing from 5 µg to 2.5 µg is one half-life.

Answer: (c) 7.5 kg

Explanation:
1. Mass of ¹H in the body = 10% of 75 kg = 7.5 kg.
2. Replacing each ¹H by ²H doubles the mass of the hydrogen, which becomes 15 kg.
3. Gain = 15 − 7.5 = 7.5 kg.

Teacher's Note:
1. Deuterium has twice the mass of protium.
2. The gain equals the original mass of hydrogen.

Answer: (b) 84.3

Explanation:
1. M₂CO₃ + 2HCl → 2MCl + H₂O + CO₂, so moles of carbonate = moles of CO₂ = 0.01186.
2. Molar mass = 1 g / 0.01186 mol = 84.3 g/mol.

Teacher's Note:
1. The mole ratio of carbonate to CO₂ is 1 : 1.
2. Molar mass equals mass divided by moles.

Answer: (d) 1/412

Explanation:
1. 2C₈H₇SO₃Na + Ca²⁺ → (C₈H₇SO₃)₂Ca + 2Na⁺.
2. 1 g of resin = 1/206 mol of resin.
3. Two moles of resin take up one mole of Ca²⁺, so uptake = 1/(2 × 206) = 1/412 mol per gram.

Teacher's Note:
1. Two monovalent resin anions are needed for each Ca²⁺ ion.
2. Do not forget the factor 2.

Answer: (b) 7 : 32

Explanation:
1. Moles of O₂ ∝ 1/32 and moles of N₂ ∝ 4/28.
2. Ratio = (1/32) × (28/4) = 28/128 = 7/32.
3. The ratio of the number of molecules is 7 : 32.

Teacher's Note:
1. Number of molecules is proportional to number of moles.
2. Divide each mass by the molar mass of the gas.

Answer: (a) 0.875 M

Explanation:
1. Total moles of HCl = 0.750 × 0.5 + 0.250 × 2 = 0.375 + 0.5 = 0.875 mol.
2. Total volume = 1 L.
3. Molarity = 0.875 M.

Teacher's Note:
1. Add moles, not molarities.
2. M = (M₁V₁ + M₂V₂)/(V₁ + V₂).

Answer: (d) 2.05 M

Explanation:
1. Moles of urea = 120/60 = 2 mol.
2. Mass of solution = 1120 g, so volume = 1120/1.15 = 973.9 mL = 0.974 L.
3. Molarity = 2/0.974 = 2.05 M.

Teacher's Note:
1. Molarity needs the volume of the solution, which is obtained from its mass and density.
2. Do not use the volume of water alone.

Answer: (c) 22.05 g

Explanation:
1. Moles of Fe²⁺ = 0.750 × 0.6 = 0.45 mol.
2. Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O, so moles of K₂Cr₂O₇ = 0.45/6 = 0.075.
3. Mass = 0.075 × 294 = 22.05 g.

Teacher's Note:
1. Mohr's salt supplies Fe²⁺, which is the only oxidisable part.
2. One dichromate accepts 6 electrons.

Answer: (c) reduces permanganate to Mn²⁺

Explanation:
1. HCl is a better reducing agent than oxalic acid under these conditions.
2. Cl⁻ is oxidised to Cl₂ by MnO₄⁻, which consumes extra permanganate and gives a wrong end point.

Teacher's Note:
1. Only non-oxidisable acids such as dilute H₂SO₄ are suitable for permanganate titrations.
2. Extra H⁺ does not affect the end point.

Answer: (d) 11.2 L H₂(g) at STP is produced for every mole HCl(aq) consumed

Explanation:
1. 6 mol HCl give 3 mol H₂, so 1 mol HCl gives 0.5 mol H₂.
2. At STP, 0.5 mol of H₂ occupies 0.5 × 22.4 = 11.2 L.
3. 1 mol Al gives 1.5 mol H₂ = 33.6 L at STP only, so (b) and (c) are wrong; HCl is an aqueous solution, so (a) is meaningless.

Teacher's Note:
1. Use mole ratios, then apply 22.4 L/mol only at STP.
2. Gas volume depends on temperature and pressure.

Answer: (b) remain unchanged

Explanation:
1. The mass of one mole of a substance is its actual mass in grams of 6.022 × 10²³ particles, independent of the unit chosen.
2. Only the numerical values of relative atomic masses and the value of the Avogadro constant would change.

Teacher's Note:
1. A change of reference unit changes relative masses, not actual masses.
2. The mole is defined by the number of particles, not by the mass scale.

Answer: (b) 67.2 L

Explanation:
1. 2BCl₃ + 3H₂ → 2B + 6HCl.
2. 21.6 g of B = 2 mol, which needs 3 mol of H₂.
3. Volume at STP = 3 × 22.4 = 67.2 L.

Teacher's Note:
1. Convert mass to moles, use the mole ratio, then multiply by 22.4 L/mol.
2. The given 273 K and 1 atm is STP in this older convention.

Answer: (a) C₆H₈N₂

Explanation:
1. Total parts = 13.5, so mass of C = 9/13.5 × 108 = 72 g (6 atoms).
2. Mass of H = 8 g (8 atoms); mass of N = 28 g (2 atoms).
3. Molecular formula = C₆H₈N₂.

Teacher's Note:
1. Split the molar mass in the given weight ratio.
2. Divide each mass by the atomic mass to count atoms.