Question

Latent heat of fusion and latent heat of vaporization of a solvent are in the ratio of 2:5 respectively. Also, freezing point and boiling point (in kelvin scale) are in the ratio of 1:2 respectively.
Thus, the ratio of elevation in boiling point to depression in freezing point of 1 molal solution is:

• A. 1:2
• B. 8:5
• C. 3:4
• D. 3:8

Answer 1

The correct option is B 8:5

$K_b$ (molal elevation constant) is related to latent heat of vaporization $\Delta H_w$ and boiling point $T_b$ of solvent by relation.

$K_b=\frac{R{T}_b^2}{1000\Delta H_v}$

Similarly, $K_f$ (molal depression constant) is related to latent heat of fusion $\Delta H_f$ and boiling point $T_f$ of solvent by relation.

$K_f=\frac{R{T}_f^2}{1000\Delta H_f}$

Thus, $\frac{K_b}{K_f}={\left(\frac{T_b}{T_f}\right)}^2\left(\frac{\Delta H_f}{\Delta H_v}\right)={\left(\frac{2}{1}\right)}^2\left(\frac{2}{5}\right)=\frac{8}{5}$

$\Delta T_b=m{K}_b,\Delta T_f=m{K}_f$

$\therefore \frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}=\frac{8}{5}$