JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 22

Answer: (a) 180

Explanation:
1. Moles of NaOH in the stock = 5/40 = 0.125 mol in 450 mL.
2. Moles needed for 500 mL of 0.1 M = 0.5 × 0.1 = 0.05 mol.
3. Volume of stock = (0.05/0.125) × 450 = 180 mL.

Teacher's Note:
1. Dilution problems are solved by equating moles of solute before and after dilution.
2. Always work out the molarity (or the moles per mL) of the stock first.

Answer: (d) 25 %

Explanation:
1. Molar mass of C₄H₄N₂O₂ = 48 + 4 + 28 + 32 = 112 g mol⁻¹.
2. Mass of nitrogen = 2 × 14 = 28 g.
3. % N = 28/112 × 100 = 25 %.

Teacher's Note:
1. Count the atoms of every element from the structure before calculating the molar mass.
2. Percentage of an element = (mass of the element in one mole / molar mass) × 100.

Answer: (c) 5

Explanation:
1. Mass percent, mole fraction, molarity, ppm and molality all express concentration.
2. Mole is the unit of amount of substance, not of concentration.
3. So the number of concentration units is 5.

Teacher's Note:
1. Concentration relates the amount of solute to the amount of solution or solvent.
2. A mole is only an amount, so it cannot be a concentration unit by itself.

Answer: (b) 1 L of the marketed solution contains 75 g of H₂O₂

Explanation:
1. '25 volume' means 1 volume of solution releases 25 volumes of O₂ at STP.
2. 2H₂O₂ → 2H₂O + O₂: 68 g of H₂O₂ gives 22.7 L of O₂ at STP.
3. H₂O₂ needed for 25 L of O₂ = 68 × 25/22.7 ≈ 75 g, so 1 L of solution contains about 75 g of H₂O₂.

Teacher's Note:
1. Volume strength = 11.2 × molarity (at 22.4 L molar volume).
2. Here 1 L of solution releases 25 L of oxygen.

Answer: (c) 2 : 1

Explanation:
1. Molal solution: n mol in (500 − 62n) g of water, so n = 0.25 × (500 − 62n)/1000, giving n ≈ 0.123 mol and mass ≈ 7.6 g.
2. Molar solution: 0.25 × 0.250 = 0.0625 mol, mass = 0.0625 × 62 ≈ 3.9 g.
3. Ratio ≈ 7.6 : 3.9 ≈ 2 : 1.

Teacher's Note:
1. Molality uses the mass of solvent, molarity the volume of solution.
2. The two compositions differ because 500 g of solution is not 500 g of solvent.

Answer: (b) 12

Explanation:
1. Mass of carbon in one mole = 0.858 × 84 ≈ 72 g, i.e. 6 mol of C atoms.
2. Mass of hydrogen = 84 − 72 = 12 g, i.e. 12 H atoms.
3. The formula is C₆H₁₂.

Teacher's Note:
1. Find the moles of C first, then get H by difference.
2. 85.8 % carbon with C:H = 1:2 is the signature of an alkene or cycloalkane, CₙH₂ₙ.

Answer: (a) 200

Explanation:
1. Moles of CO₂ = 4.4/44 = 0.1 mol, so 0.1 mol of C came from 0.01 mol of compound.
2. Each mole of compound contains 10 mol C = 120 g C.
3. Molar mass = 120/0.60 = 200 g mol⁻¹.

Teacher's Note:
1. Use the carbon balance: moles of C in the compound equal moles of CO₂ formed.
2. Mass of C per mole divided by the mass fraction of C gives the molar mass.

Answer: (d) 148

Explanation:
1. Mass of chloroform = 671.141 × 1.49 = 1000 g.
2. Moles of CH₂Cl₂ = 2.6 × 10⁻³ × 0.671141 = 1.745 × 10⁻³ mol; mass = 1.745 × 10⁻³ × 85 = 0.148 g.
3. ppm = 0.148/1000.148 × 10⁶ ≈ 148.

Teacher's Note:
1. ppm by mass is grams of solute per 10⁶ g of solution.
2. Convert the molarity to moles using the solution volume, then to mass.

Answer: (c) 150

Explanation:
1. 50 volume means 1 L of solution gives 50 L of O₂ at STP.
2. Moles of O₂ = 50/22.7 = 2.20 mol; from 2H₂O₂ → 2H₂O + O₂, moles of H₂O₂ = 2 × 2.20 = 4.41 mol.
3. Strength = 4.41 × 34 ≈ 150 g L⁻¹.

Teacher's Note:
1. Each mole of O₂ needs 2 mol of H₂O₂.
2. Strength in g/L = 34 × 2 × (volume strength / molar volume).

Answer: (b) 44

Explanation:
1. Carbon in CO₂ = (12/44) × 0.792 = 0.216 g.
2. % C = 0.216/0.492 × 100 ≈ 43.9 %.
3. Nearest integer: 44 %.

Teacher's Note:
1. Percentage of carbon = (12/44) × (mass of CO₂ / mass of compound) × 100.
2. Never forget that only 12 of every 44 g of CO₂ is carbon.

Answer: (a) 4 L

Explanation:
1. Zn + 2HCl → ZnCl₂ + H₂, so 1 mol Zn gives 1 mol H₂.
2. Moles of Zn = 11.5/65.4 = 0.176 mol.
3. Volume of H₂ = 0.176 × 22.7 ≈ 4.0 L.

Teacher's Note:
1. The mole ratio Zn : H₂ is 1 : 1 here.
2. Convert moles of gas to volume with the molar volume at STP.

Answer: (d) 22.7 L

Explanation:
1. Moles of C = 12/12 = 1 mol; moles of O₂ = 48/32 = 1.5 mol.
2. 1 mol of C needs only 0.5 mol of O₂, so carbon is the limiting reagent.
3. Moles of CO = 1 mol, volume = 22.7 L (that is 227 × 10⁻¹ L).

Teacher's Note:
1. Find the limiting reagent before calculating the product.
2. Moles of product follow the limiting reagent only.

Answer: (c) 3.64 m

Explanation:
1. 1 L of solution has mass 1000 g and contains 3 mol NaCl = 175.5 g.
2. Mass of water = 1000 − 175.5 = 824.5 g = 0.8245 kg.
3. Molality = 3/0.8245 = 3.64 m (that is 364 × 10⁻² m).

Teacher's Note:
1. Molality needs the mass of solvent, so subtract the mass of solute from the mass of solution.
2. Molality is larger than molarity here because the solvent mass is less than 1 kg.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. KOH absorbs CO₂ from air: 2KOH + CO₂ → K₂CO₃ + H₂O, so its effective concentration changes with time.
2. Hence the solution must be standardised before use. Both statements are true and R explains A.

Teacher's Note:
1. Alkali solutions are not primary standards because they absorb CO₂ and moisture.
2. Standardise NaOH or KOH with a primary standard such as oxalic acid before using it.

Answer: (b) 1.39 m

Explanation:
1. In 100 mL of solution: 10 mL Br₂ = 32 g = 0.2 mol; 90 mL CCl₄ = 144 g = 0.144 kg.
2. Molality = 0.2/0.144 = 1.39 m (139 × 10⁻²).

Teacher's Note:
1. Take 100 mL of solution as the basis for volume-percent problems.
2. Convert volumes to masses with the densities before computing molality.

Answer: (a) 1362 mL

Explanation:
1. Moles of CO₂ = 0.2 × 0.300 = 0.06 mol.
2. Volume at STP = 0.06 × 22.7 = 1.362 L = 1362 mL.

Teacher's Note:
1. Moles = molarity × volume in litres.
2. Multiply the moles by the molar volume to get the gas volume.

Answer: (d) 555 mm Hg

Explanation:
1. n(X) = 0.6/20 = 0.03 mol; n(Y) = 0.45/45 = 0.01 mol.
2. Mole fraction of X = 0.03/0.04 = 0.75.
3. p(X) = 0.75 × 740 = 555 mm Hg.

Teacher's Note:
1. Partial pressure = mole fraction × total pressure (Dalton's law).
2. Always convert grams to moles before finding mole fractions.

Answer: (a) C₈H₆

Explanation:
1. For CₓHy: H₂O formed = y/2 = 3, so y = 6.
2. O₂ needed = x + y/4 = 9.5, so x = 9.5 − 1.5 = 8.
3. Formula = C₈H₆.

Teacher's Note:
1. For CₓHy, oxygen needed is x + y/4 and water formed is y/2.
2. Solve the two equations for x and y.

Answer: (a) 2NaOCl + H₂O₂ → 2NaCl + H₂O + O₂

Explanation:
1. A reducing agent is itself oxidised. In H₂O₂ the oxidation state of oxygen is −1.
2. In reaction (a) oxygen goes from −1 in H₂O₂ to 0 in O₂ (oxidation), so H₂O₂ is the reductant.
3. In (b), (c) and (d) oxygen goes from −1 to −2, so H₂O₂ is the oxidant.

Teacher's Note:
1. H₂O₂ can act as both an oxidant and a reductant because −1 lies between −2 and 0.
2. It reduces strong oxidants such as hypochlorite, permanganate and dichromate.

Answer: (a) C₉H₈

Explanation:
1. H₂O formed = y/2 = 4, so y = 8.
2. O₂ needed = x + y/4 = 11, so x = 11 − 2 = 9.
3. Formula = C₉H₈.

Teacher's Note:
1. Use the same two equations: y/2 equals moles of water and x + y/4 equals moles of oxygen.
2. Check by balancing the final equation.