Question

When 1 g each of compounds AB and $\mathrm{AB}_2$ 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____________ $\times 10^{-1}$ (Nearest integer) (Given : Molal boiling point elevation constant is $0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Hint:

Use the elevation of boiling point for AB and AB2 to get their molar masses, then subtract.

Answer 1

Correct Answer: 25

$\Delta T_b = K_b \times m$ with $m = \frac{w/M}{W_{kg}}$ and $W = 15$ g $= 0.015$ kg.
For AB: $2.7 = 0.5 \times \frac{1/M_{AB}}{0.015}$, so $M_{AB} = \frac{0.5}{2.7 \times 0.015} = 12.35$ g mol$^{-1}$.
For $AB_2$: $1.5 = 0.5 \times \frac{1/M_{AB_2}}{0.015}$, so $M_{AB_2} = \frac{0.5}{1.5 \times 0.015} = 22.22$ g mol$^{-1}$.
Mass of B $= M_{AB_2} - M_{AB} = 9.88$, so the atomic mass of A $= 12.35 - 9.88 = 2.47$ amu $= 24.7 \times 10^{-1} \approx 25 \times 10^{-1}$.
The answer is 25.