Question

A solution of a non-volatile solute in water freezes at $-{0.30}^{o}C$. The vapour pressure of pure water at 298 K is 23.51 mm Hg and $K_f$ for water is 1.86 degree/molal. The vapour pressure of this solution at 298 K in cm of Hg to the nearest integer is________.

Answer 1

We have; $\Delta T=K_f\times molality$

Also from Raoult's law

$\frac{P^0-P_S}{P_S}=\frac{w\times M}{m\times W}=\frac{w\times 1000\times M}{m\times W\times 1000}$

$\frac{P^0-P_S}{P_S}=molality\times \frac{M}{1000}$

$\frac{P^0-P_S}{P_S}=\frac{\Delta T}{K_f}\times \frac{M}{1000}$

Given, $P^0=23.51$ mm of Hg; $M=18$

$\Delta T=0.3,K_f=1.86Kmolalit{y}^{-1}$

$\therefore \frac{23.51-P_S}{P_S}=\frac{0.3}{1.86}\times \frac{18}{1000}$

$\therefore P_S=23.44mmHg=2.344cmHg$