Question

State Raoult's law.

The vapour pressure of pure Benzene at a certain temperature is $0.850bar$. A non-volatile, non-electrolyte solid weighing $0.5g$ when added to $39.0g$ of benzene $\left(molarmass78gmo{l}^{-1}\right)$. Vapour pressure of the solution, then is $0.845bar$. What is the molar mass of the solid substance?

Answer 1

• Raoult's law $\to$
  

For volatile solute:

Raoult's law states that in a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that compound in the pure state.

For non-volatile solute:

Raoult's law states that the relative lowering in vapour pressure of a solution containing a nonvolatile solute is equal to the mole fraction of solute in the solution.

• According to Raoult’s law.
  

$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{n_A}{n_A+n_B}.....\left(1\right)$

Given:-

$p^{\circ }=0.850\text{bar}$

$p_s=0.845\text{bar}$

Let $m$ be the molecular mass of the solid substance.

Molecular mass of benzene $=78$

$n_A=\frac{0.5}{m};n_B=\frac{39}{78}=0.5$

Substituting the values in equation $\left(1\right)$, we have.

$\frac{0.850-0.845}{0.850}=\frac{\left(\frac{0.5}{m}\right)}{\left(\frac{0.5}{m}\right)+0.5}=\frac{0.5}{0.5+0.5m}=\frac{1}{1+m}$

$\Rightarrow m=\frac{0.850}{0.005}-1=169g/mol$

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