Some Basic Concepts of Chemistry: JEE Advanced PYQ, 2007–2026

Answer: (B)

One litre holds 1 mol of H₂O₂, which gives 0.5 mol of O₂: 0.5 × 22.4 = 11.2 L. So 1 L of solution gives 11.2 L of oxygen: volume strength 11.2.

Answer: (C)

Count the oxygen atoms per formula unit: the three SO₄ groups give 3 × 4 = 12, and the 18 water molecules give 18, a total of 30. Moles of O atoms = 0.10 × 30 = 3.0 mol. Number of atoms = 3.0 × 6.022 × 10²³ ≈ 1.81 × 10²⁴.

Answer: (A)

By Gay-Lussac's law, gases react in simple volume ratios at the same temperature and pressure. The reactions are 2CO + O2 → 2CO2 and 2H2 + O2 → 2H2O(l). Each volume of either fuel needs 1/2 volume of O2, so 20 mL of mixture uses 10 mL of O2 whatever the composition, and 10 mL of O2 is left. Liquid water has negligible volume, so the only product gas is CO2, whose volume equals the volume of CO. The remaining gas is CO2 + leftover O2 = a + 10 = 22, so a = 12 mL.

Answer: (C)

Molar mass of C₆H₅Cl = 72 + 5 + 35.5 = 112.5 g/mol, so the sample is 0.001 mol. It gives 0.001 mol of AgCl, of molar mass 143.5 g/mol: 0.1435 g.

Answer: (C)

The reaction is Ag⁺ + Cl⁻ → AgCl, with 1 : 1 stoichiometry. Moles of Ag⁺ = 0.20 × 0.050 = 0.010. Moles of CaCl₂ = 0.10 × 0.040 = 0.0040, which gives 0.0080 mol of Cl⁻. Cl⁻ is limiting, so 0.0080 mol of AgCl forms. Mass = 0.0080 × 143.5 = 1.148 g ≈ 1.15 g.

Answer: (C)

By the law of definite proportions the compound always has X : Y = 3 : 8 by mass, so the reactants combine only in that ratio. 6.0 g of X would need 6 × 8/3 = 16 g of Y, but only 10.0 g is available. Y is therefore the limiting reactant. 10.0 g of Y combines with 10 × 3/8 = 3.75 g of X. Compound formed = 3.75 + 10.0 = 13.75 g. X left over = 6.0 − 3.75 = 2.25 g. Check: 13.75 + 2.25 = 16.0 g, which equals the total starting mass (conservation of mass).

Answer: (D)

The mass of oxygen combined is 2.55 − 1.35 = 1.20 g. Moles of Al = 1.35/27 = 0.050 and moles of O = 1.20/16 = 0.075. The ratio Al:O = 0.050:0.075 = 2:3, so the empirical formula is Al₂O₃.

Answer: (D)

The law of multiple proportions says that if two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. Take 16 g of oxygen as the fixed mass. For P (63.64% N, 36.36% O), N = 16 × 63.64/36.36 = 28 g. For Q (46.67% N, 53.33% O), N = 16 × 46.67/53.33 = 14 g. For R (30.43% N, 69.57% O), N = 16 × 30.43/69.57 = 7 g. The ratio is 28 : 14 : 7 = 4 : 2 : 1. Raw percentages must first be converted to masses combining with the same mass of one element before they can be compared.

Answer: (D)

Moles per 100 g: Na = 29.1/23 ≈ 1.27, S = 40.5/32 ≈ 1.27, O = 30.4/16 = 1.90. Dividing by 1.27 gives Na:S:O = 1:1:1.5. Because 1.5 is a half-integer, multiply every term by 2 to give 2:2:3. The empirical formula is Na₂S₂O₃. Rounding 1.5 to a whole number changes the formula, and multiplication must be applied once to all the terms.

Answer: (B)

By Gay-Lussac's law and Avogadro's law, volumes of gases at the same T and P are proportional to their moles. The reaction is CxHy + (x + y/4) O₂ → x CO₂ + (y/2) H₂O(l). Step 1: water is liquid after cooling, so the 35 mL is CO₂ plus unreacted O₂. Step 2: KOH absorbs only CO₂, so CO₂ = 35 − 15 = 20 mL, and 10 mL of hydrocarbon giving 20 mL of CO₂ means x = 2. Step 3: the 15 mL left is unreacted O₂, so O₂ used = 50 − 15 = 35 mL, which for 10 mL of hydrocarbon is 3.5 per mL. Step 4: x + y/4 = 3.5, so 2 + y/4 = 3.5 and y = 6. The hydrocarbon is C₂H₆. Check: 10 mL C₂H₆ needs 35 mL O₂ and gives 20 mL CO₂, leaving 15 + 20 = 35 mL of gas.