Question

Freezing point of an aqueous solution is $-0.186{}^{\circ }C$. If the values of $K_b$ and $K_f$ of water are respectively $0.52Kkg$ $mo{l}^{-1}$ and $1.86Kkgmo{l}^{-1}$, 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

Answer: (E)

$\because$ Freezing point of the aqueous solution

$=-0.186^{\circ }C$

$\therefore$ Depression in freezing point,

$\Delta T_f=0^{\circ }C-\left(-0.186^{\circ }C\right)$

$=+0.186^{\circ }C=+0.186K$

We know that

$\Delta T_f=K_f\cdot m....(i)$

and elevation in boiling point,

$\Delta T_b=K_b\cdot m.....(ii)$

Eq. (i) Eq. (ii) gives

$\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}$

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

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

$=0.052K$