Question

The vapour pressure of pure benzene at a particular temperature is $260$ bar. At that temperature, the vapour pressure of a solution containing $2$ g non-volatile non-electrolytic solid in $100$ g of benzene is $250$ bar. What is the molecular mass of the solute?

• A. $60.4$ g/mole
• B. $40.6$ g/mole
• C. $70.5$ g/mole
• D. $30.5$ g/mole

Answer 1

Answer: (B)

$40.6$ g/mole

Vapor pressure of pure benzene that is $p_A^o$ is $260$ bar
Vapor pressure of the solution when a solute is added is $250$ bar, that is $p_A$ is $250$ bar.
Lowering of vapor pressure is $p_A^o-p_A$
Thus relative lowering of vapor pressure is $\frac{p_A^o-p_A}{p_A^o}$$=$$X_B$, where $X_B$ is the mole fraction of the solute.
$X_B$ can also be written as $\frac{n_B}{n_A}$ where $n_B$ and $n_A$ are the number of moles of solute and solvent and $n_B$ is much smaller compared to $n_A$.

Thus $\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{n_B}{n_A}$
$\frac{n_B}{n_A}$ can also be written as $\frac{W_B}{M_B}\times \frac{M_A}{W_A}$ where $M_B$ and $M_A$ are the molecular masses of solute and solvent. Whereas, $W_B$ and $W_A$ are the amount of solute and solvent present in the solution.
Putting all the values in the equation:

$\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{W_B}{M_B}\times \frac{M_A}{W_A}$

$\frac{260-250}{260}$$=$$\frac{2}{M_B}\times \frac{78}{100}$, where $78$ g/mole is the molecular weight of benzene.
From this the value of $M_B$ obtained is almost $40.6$ g/mol.
Thus molecular weight of the solute is $40.6$ g/mol.