Question

i). State Raoult's Law.

ii). 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 $mo{l}^{-1}$). Vapour pressure of the solution, then is $0.845$ bar. What is the molar mass of the solid substance?

Answer 1

i). Raoult's law: The partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.

$p=p^{o}X$ ----a

Where p is the vapour pressure of a component of the solution, $p^o$ is the vapour pressure of pure component and X is the mole the fraction of that component in the solution.

ii). Let the molar mass of the solid substance be M g/mol.
The number of moles of solid substance $=\frac{0.5g}{Mg/mol}$

39.0 g of benzene (molar mass 78 g /mol) corresponds to the number of moles $\frac{39.0}{78}=0.5$ moles.

The mole fraction of solid substance is $X=\frac{\frac{0.5}{M}}{0.5+\frac{0.5}{M}}$
The relative lowering in vapour pressure is equal to its mole fraction.

$\frac{P^o-P}{P^o}=X$

$\frac{0.850-0.845}{0.850}=\frac{\frac{0.5}{M}}{0.5+\frac{0.5}{M}}$

$0.00588=\frac{\frac{0.5}{M}}{0.5+\frac{0.5}{M}}$

$0.5+\frac{0.5}{M}=\frac{\frac{0.5}{M}}{0.00588}$

$0.5+\frac{0.5}{M}=\frac{85}{M}$

Multiply both sides with M

$0.5M+0.5=85$

$0.5M=84.5$

$M=169g/mol$

Hence, the molar mass of the solid substance is $169$ g/mol.