Question

The molal lowering of vapour pressure( in cm to nearest integer) for $H_{2}O$ at ${100}^{o}C$ is_________.

Answer 1

$\because \frac{P^0-P_S}{P_S}=\frac{n}{n+N}$
$\therefore P^0-P_S=P^0\left[n/\right(n+N\left)\right]$
Let 1 mole of solute is dissolved in 1000 g solvent water.
Thus, $n=1$ and $N=1000/18$
Also, $P^0=760mm$ for water at ${100}^{o}C$
$\therefore P^0-P_S=\left[\frac{1}{1+\left(1000/18\right)}\right]\times 760=13.44mm=1.344cm$