Question

An aqueous solution freezes at -${0.186}^{o}C$ ($K_f$ $={1.86}^{o}Ckgmo{l}^{-1}$ and $K_b$ $={0.512}^{o}Ckgmo{l}^{-1}$). What is the elevation in boiling point?

• A. ${0.186}^{o}C$
• B. ${0.512}^{o}C$
• C. ${0.86}^{o}C$
• D. ${0.0512}^{o}C$

Answer 1

The correct option is D ${0.0512}^{o}C$

Depression in freezing point is given by $\Delta T_f=-K_f\times m$, where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.

Similarly, elevation in boiling point is given by $\Delta T_b=K_b\times m$, where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.

Taking molality as a constant for the solution, we have

$-\frac{\Delta T_f}{K_f}=\frac{\Delta T_b}{K_b}$ $⟹$ $\frac{0.186}{1.86}=\frac{x}{0.512}$

Therefore, $x=0.0512$

Hence, the elevation in boiling point is ${0.0512}^{o}C$.