Question

Freezing point of an aqueous solution is ${-0.186}^{o}C$. If the values of $K_b$ and $K_f$ of water are respectively $0.52$ $K$ $kg$ ${mol}^{-1}$ and $1.86$ $K$ $kg$ ${mol}^{-1}$ then the elevation of boiling point of the solution in $K$ is

• A. $0.52$
• B. $1.04$
• C. $1.34$
• D. $0.134$
• E. $0.052$

Answer 1

The correct option is E $0.052$

$\because$ Freezing point of the aqueous solution $=-{0.186}^{o}C$

$\therefore$ Depression in freezing point,
$\Delta T_f=0^{o}C-\left({-0.186}^{o}C\right)$
$=+{0.186}^{o}C=+0.186K$

We know that,
$\Delta T_f=K_f\cdot m$ ....(i)

and elevation in boiling point,
$\Delta T_b=K_b\cdot m$ ....(ii)

Equation (i) Equation (ii) gives
$\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}$

$\Rightarrow \frac{0.186K}{\Delta T_b}=\frac{1.86Kkg{mol}^{-1}}{0.52Kkg{mol}^{-1}}$

$\Delta T_b=\frac{0.186\times 0.52}{1.86}=0.052K$