The vapour pressure of pure benzene at a certain temperature is $0.850 \ bar$. $A$ non-volatile,non-electrolyte solid weighing $0.5 \ g$ when added to $39.0 \ g$ of benzene (molar mass $78 \ g \ mol^{-1}$),the vapour pressure of the solution becomes $0.845 \ bar$. What is the molar mass of the solid substance?

$(170 \ g \ mol^{-1}$ The given values are:
$p_{1}^{0} = 0.850 \ bar$; $p = 0.845 \ bar$; $M_{1} = 78 \ g \ mol^{-1}$; $w_{2} = 0.5 \ g$; $w_{1} = 39 \ g$
Using the relative lowering of vapour pressure formula:
$\frac{p_{1}^{0} - p}{p_{1}^{0}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
Substituting the values:
$\frac{0.850 - 0.845}{0.850} = \frac{0.5 \times 78}{M_{2} \times 39}$
$\frac{0.005}{0.850} = \frac{39}{M_{2} \times 39}$
$\frac{1}{170} = \frac{1}{M_{2}}$
$M_{2} = 170 \ g \ mol^{-1}$