When 1 g each of compounds AB and AB₂ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in a m u ) is____________ × 10⁻¹ (Nearest integer)
(Given : Molal boiling point elevation constant is 0.5 K kg mol⁻¹ )

Solution

Chemistry ma’am · 4 steps

Get the molar mass of AB and of AB₂ from their boiling point rise, then subtract to find B.

1. Molar mass formula. The boiling point rise is Δ T_b = K_b × m, where the molality is m = w/M × 1000/W. Rearranging gives the molar mass of the solute.

   M = (K_b × w × 1000)/(Δ T_b × W)

2. Put in numbers. For AB: M_(AB) = (0.5 × 1 × 1000)/(2.7 × 15) = 500/40.5 = 1000/81 ≈ 12.35. For AB₂: M_(AB₂) = 500/(1.5 × 15) = 500/22.5 = 200/9 ≈ 22.22.
3. Find B, then A. AB₂ has one more B than AB, so B = 22.22 - 12.35 = 9.88 (exactly 800/81). Then A = 12.35 - 9.88 = 2.47 amu (exactly 200/81).
4. So. A = 2.47 amu = 24.7 × 10⁻¹ amu, which rounds to 25.

The common slip: 24. Rounding the molar masses early to 12.3 and 22.2 gives A = 2.4, which leads to 24, not 25. Keep the fractions until the last step.

The idea to keep. For two compounds of the same elements, find each molar mass from the colligative data. The difference in formula then gives one atom's mass directly.

Answer from the NTA final answer key. Solution by PrepWiser.