Question

The vapour pressure of pure benzene at any temperature is 0.850 bar. 0.5 g non-volatile non-electrolytic solid is dissolved in 39.0 g benzene (molar mass 78 g mol$^{-1}$). The vapour pressure of the solution is 0.845 bar. Calculate the molar mass of the solid.

Hint:

For non-volatile solutes, use \(\dfrac{\Delta P}{P^\circ} = X_{\text{solute}}\).
If the solute amount is very small, \(X_{\text{solute}} \approx \dfrac{n_{\text{solute}}}{n_{\text{solvent}}}\) is a quick approximation; here we used the exact expression.

Answer 1

For a non–volatile solute, Raoult's law gives relative lowering of vapour pressure:
\[ \frac{\Delta P}{P^\circ} \;=\; X_{\text{solute}} \;=\; \frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}} \]
Given:
\(P^\circ = 0.850\) bar,\; \(P = 0.845\) bar \(\Rightarrow \Delta P = 0.005\) bar.
Mass of benzene \(= 39.0\) g, \(M_{\text{benzene}} = 78\) g mol\(^{-1}\) \(\Rightarrow n_{\text{solvent}} = \frac{39.0}{78} = 0.50\) mol.
Mass of solute \(= 0.50\) g, molar mass \(= M\) \(\Rightarrow n_{\text{solute}} = \frac{0.50}{M}\) mol.

Step 1: Compute relative lowering.
\[ \frac{\Delta P}{P^\circ} = \frac{0.005}{0.850} = 0.005882\ (\approx 5.882\times 10^{-3}) \]

Step 2: Apply Raoult's relation.
\[ 0.005882 = \frac{\dfrac{0.50}{M}}{\dfrac{0.50}{M} + 0.50} \]
Multiply numerator and denominator by \(M\):
\[ 0.005882 = \frac{0.50}{0.50 + 0.50M} \]
\[ 0.005882(0.50 + 0.50M) = 0.50 \]
\[ 0.002941 + 0.002941\,M = 0.50 $\Rightarrow$ 0.002941\,M = 0.497059 \]
\[ M = \frac{0.497059}{0.002941} \approx 169\ \text{g mol}^{-1} \]
\[ \boxed{\text{Molar mass of the solid } \approx 1.69 \times 10^{2}\ \text{g mol}^{-1}} \]