Question

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

• A. 58
• B. 180
• C. 170
• D. 145

Answer 1

Answer: (C)

170

Answer 2

$P^{\circ}_{A}=0.850\,bar,\, P_{s}=0.845\,bar$ $w=0.5\,g, \,m=?$ weight of solvent (benzene) $= 39.0\, g$ and molecular weight of benzene $= 78\, g$ As we know, $\frac{P^{\circ}_{A}-P_{s}}{P^{\circ}_{A}}=x_{B}=\frac{n_{B}}{n_{A}}$ $\frac{0.850-0.845}{0.850}=\frac{\frac{w_{B}}{M_{B}}}{\frac{w_{A}}{M_{A}}}=\frac{\frac{0.5}{m}}{\frac{39}{78}}$ On solving, we get molecular mass of solid structure $(m) = 170\, g.$