Question

Calculate the mole fraction of solute, if the vapour pressure of pure benzene at certain temperature is 640 mmHg and vapour pressure of solution of a solute in benzene is 600 mmHg.

Hint:

To find the mole fraction of the solute, use the relationship between vapour pressures and mole fractions given by Raoult’s Law.

Answer 1

Step 1: According to Raoult’s Law, the vapour pressure of a solution is given by: \[ P_{\text{solution}} = P_{\text{solvent}} \times X_{\text{solvent}} + P_{\text{solute}} \times X_{\text{solute}} \] Where: - \( P_{\text{solution}} \) is the vapour pressure of the solution. - \( P_{\text{solvent}} \) is the vapour pressure of the pure solvent. - \( X_{\text{solvent}} \) and \( X_{\text{solute}} \) are the mole fractions of the solvent and solute, respectively. Step 2: The equation can be rewritten as: \[ P_{\text{solution}} = P_{\text{solvent}} \times X_{\text{solvent}} \] Since the solute has no vapour pressure, we can ignore the second term. Step 3: Rearranging the equation to solve for the mole fraction of the solute \( X_{\text{solute}} \): \[ P_{\text{solution}} = P_{\text{solvent}} (1 - X_{\text{solute}}) \] \[ X_{\text{solute}} = 1 - \frac{P_{\text{solution}}}{P_{\text{solvent}}} \] Step 4: Substitute the given values: \[ X_{\text{solute}} = 1 - \frac{600 \, \text{mmHg}}{640 \, \text{mmHg}} = 1 - 0.9375 = 0.0625 \] Thus, the mole fraction of the solute is: \[ \boxed{0.0625} \]