Question

The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5g$ is added to $39.0g$ of benzene (molar mass $78g/mol$). The vapour pressure of the solution then is $0.845bar$. What is the molecular mass of the solid substance?

• A. $58$
• B. $180$
• C. $170$
• D. $145$

Answer 1

The correct option is C $170$

$P_{benzene}^{\circ }=0.850bar;W_{benzene}=39.0g$.
$M.W.=78g/mol$
$P_{solution}^{\circ }=0.845bar;w=0.5g$
Let n be the no. of moles of non volatile solid; w and m be the weight of solid and molecular mass respectively.
Using, $\frac{P_{benzene}^{\circ }-P_{solution}^{\circ }}{P_{benzene}^{\circ }}=\frac{n_{solid}}{n_{benzene}}$
$\frac{0.850-0.845}{0.850}=\frac{\frac{0.5}{m}}{\frac{39}{78}}$
$\frac{0.005}{0.850}=\frac{0.5}{m}\times \frac{78}{39}$
$m=\frac{100\times 0.850}{0.005}=170$