Question

The ratio of the boiling point (in K) of ethanol to its freezing point (in K) is 9:4, while the ratio of its latent heat of vaporization to its latent heat of fusion is 9:1. If the boiling point of pure ethanol is 360 K and freezing point of a solution of a compound P in ethanol is 152 K, then what is boiling point of same solution?

• A. 355.5 K
• B. 364.5 K
• C. 368 K
• D. 352 K

Answer 1

The correct option is B 364.5 K

Given, $\frac{T_b}{T_f}=\frac{9}{4}and\frac{l_v}{l_f}=\frac{9}{1}$
So, $\frac{k_b}{k_f}=\frac{\frac{R{T}_b^2}{1000l_v}}{\frac{R{T}_f^2}{1000l_f}}=\frac{T_b^2}{T_f^2}\times \frac{l_f}{l_b}=\frac{9^2}{4^2}\times \frac{1}{9}=\frac{9}{16}$
Since boiling point of pure ethanol = 360 K (given), so freezing point of pure ethanol = $T_f=\frac{4}{9}\times T_b=\frac{4}{9}\times 360=160K$
So, $△T_f=T_f^0-T_f^S=160-152=8K$ $\left(Freezingpointofsolution\right(T_f^S)=152K,given)$
So,$\frac{△T_b}{△T_f}=\frac{k_{b}m}{k_{f}m}=\frac{9}{16}\Rightarrow △T_b=\frac{9}{16}\times 8=4.5K$
So, $T_b^S=T_b^0+△T_b=360+4.5=364.5K$