Question

The vapour pressure of pure benzene, $C_6{H}_6$ at ${50}^{o}C$ is 268 mm. Number of moles of non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of 167.0 mm at ${50}^{o}C$ is (multiply answer with 10) :

Answer 1

The expression for the relative lowering of vapor pressure is $\frac{P^0-P}{P^0}=\frac{n}{n+N}$
Substitute values in the above expression.
$\frac{268-167}{268}=\frac{n}{n+1}$
$0.3769=\frac{n}{n+1}$
$0.3769n+0.3769=n$
$0.6230n=0.3769$
$n=0.605$