Question

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

• A. 0.01 m
• B. 0.001 m
• C. 0.005 m
• D. 0.02 m

Answer 1

Answer: (A)

Given, $\Delta T_{b}=0.01^{\circ} C =0.01 \,K$

$i =1.92 $

$K_{b} =0.52 \,K \,kg \,mol ^{-1}$

As we know that,

$\Delta T_{b} =i K_{b} \times m $

$\therefore m =\frac{0.01}{1.92 \times 0.52} mol / kg $

$m$ (molality) $=0.01 \,mol / kg$ or $ 0.01 \,m$