Question

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

• A. $125g$
• B. $105g$
• C. $80.5g$
• D. $65.25g$

Answer 1

The correct option is D $65.25g$

According to Raoult's law,
$\frac{P_0-P_s}{P_0}=\frac{n}{N}$
Where,
$P_0=\text{Vapour pressure of benzne}{P}_s=\text{Vapour pressure of solution}n=\text{number of moles of solute}N=\text{number of moles of benzene}$

$\frac{w_{solute}\times M_{solvent}}{W_{solvent}\times M_{solute}}=\frac{P_0-P_s}{P_0}$

$M_{solute}=\frac{w_{solute}\times M_{solvent}}{W_{solvent}}\times \frac{P_0}{P_0-P_s}$
$w_{solute}=2.175g$
$W_{benzene}=39.0g$

$M_{benzene}=78g/mol$

$P_0=640mmHg$

$P_s=600mmHg$

$M_{solute}=\frac{2.175\times 78}{39.0}\times \frac{600}{640-600}$

$M_{solute}=\frac{2.175\times 78}{39.0}\times \frac{600}{40}=65.25g$