Question

Vapour pressure of pure benzene is $640mm$ and its mol. wt. is $78$. When $2.175g$ of a non-volatile substance is dissolved in $39g$ benzene the vapour pressure of the solution becomes $600mm$. Calculate mol. wt. of solute.

Answer 1

vapour pressure of pure Benzene = $P_A^{\circ }=640mm$

Mol. wt of Benzene = ${78}^o$

mole of non volatile substance = n

mole of Benzene = $\frac{39}{78}=0.5$

mole fraction = $\frac{n}{n+0.5}$

$P_A^o-P_A^o.\frac{n}{n+0.5}=600$

$640-600=640\frac{n}{n+0.5}$

$\frac{40}{640}=\frac{n}{n+0.5}$

$n+0.5=16n$

$15.5n=0.5$

$n=\frac{0.5}{15.5}$

$\frac{2.175}{M}=\frac{0.5}{15.5}$
$M=\frac{2.175\times 15.5}{0.5}$

$=67.425$