Question

The vapour pressure of pure benzene at a certain temperature is $640$mm Hg. A non-volatile solute weighing $2.175$g is added to $39.0$g of benzene. The vapour pressure of solution is $600$ mm Hg. What is the molar mass of the solute?

Answer 1

Relative lowering of vapour pressure of solution is given as:

$\frac{P_0-p_{solvent}}{P_0}={\chi }_{solute}=\frac{n_{solute}}{n_{solute}+n_{solvent}}$

where $p_{solvent}$=vapour pressure of solution

$P_0$=vapour pressure of pure solvent

${\chi }_{solute}$=molefraction of non-volatile solute

for benzene as given $P_0=640$mm of Hg

Given that $p_{solvent}=600$ mm of Hg

$M_{solvent}=6\times 12+6\times 1=78g/mol$

$m_{solvent}=39g$

$m_{solute}=2.175g$

as $n=mass/Molarmass$

$\frac{n_{solute}}{n_{solute}+n_{solvent}}=\frac{2.175/M}{2.175/M+39/78}$

$\frac{n_{solute}}{n_{solute}+n_{solvent}}=\frac{2.175}{2.175+0.5M}$

thus upon substitution we ger:

$\frac{640-600}{640}=\frac{2.175}{2.175+0.5M}$

$\frac{1}{16}\times (2.175+0.5M)=2.175$

$0.5M=2.175\times 16-2.175$

$0.5M=32.625

$M=65.25$

Molar mass of solute =65.25 g/mol.