Question

The vapour pressure of pure benzene at a certain temperature is 0.850 bar and 0.5 gm of a non-volatile non-electrolyte solid weighing is added to 39.0 gm of benzene (Molar mass 78 g mol\(^{-1}\)) then the vapour pressure of the solution is 0.845 bar. What is the molar mass of the solid substance?

Hint:

Raoult's law is useful for calculating the molar mass of a non-volatile solute by observing the change in vapour pressure of a solution.

Answer 1

Step 1: Use Raoult's Law.
Raoult's law states that the reduction in vapour pressure is proportional to the mole fraction of the solute in the solution: \[ \Delta P = P_0 - P = P_0 \times X_{\text{solute}} \] where: - \( P_0 \) is the vapour pressure of pure benzene (0.850 bar), - \( P \) is the vapour pressure of the solution (0.845 bar), - \( X_{\text{solute}} \) is the mole fraction of the solute.

Step 2: Calculate the mole fraction of the solute.
The decrease in vapour pressure is: \[ \Delta P = 0.850 - 0.845 = 0.005 \, \text{bar} \] From Raoult's law: \[ X_{\text{solute}} = \frac{\Delta P}{P_0} = \frac{0.005}{0.850} = 0.00588 \]

Step 3: Calculate the moles of benzene.
The number of moles of benzene is: \[ n_{\text{benzene}} = \frac{39.0}{78} = 0.5 \, \text{mol} \]

Step 4: Calculate the moles of solute.
The mole fraction \( X_{\text{solute}} \) is also equal to: \[ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{benzene}}} \] Substitute the values: \[ 0.00588 = \frac{n_{\text{solute}}}{n_{\text{solute}} + 0.5} \] Solving for \( n_{\text{solute}} \): \[ n_{\text{solute}} = \frac{0.00588 \times 0.5}{1 - 0.00588} = 0.00296 \, \text{mol} \]

Step 5: Calculate the molar mass of the solute.
The molar mass of the solute is: \[ M_{\text{solute}} = \frac{\text{mass of solute}}{n_{\text{solute}}} = \frac{0.5}{0.00296} = 169.9 \, \text{g/mol} \]

Step 6: Conclusion.
Thus, the molar mass of the solid substance is approximately 169.9 g/mol.