Q22 / 160 · AP EAMCET 2019 (21 Apr Shift 2) · Chemistry

The elevation in the boiling point of an aqueous solution of $NaCl$ is $0.01 ° C$ . If its van't Hoff factor is $1.92,$ the molality of $NaCl$ solution is ___ $(K_{b}$ for water $= 0.52 K kg {mol}^{- 1})$

Solution

The elevation in boiling point is given by:

$∆ T_{b} = i \times K_{b} \times m m = \frac{∆ T_{b}}{i \times K_{b}} m = \frac{0.01}{1.92 \times 0.52} = 0.1 molal$

Hence, the correct answer is option: 'A'.