Question

A solution is prepared by dissolving 5 g of a non-volatile solute in 200 g of water. It has a vapor pressure of 31.84 mm Hg at 300 K. Calculate the molar mass of the solute.
(Vapor pressure of pure water at 300 K = 32 mm Hg)

Hint:

Raoult's Law allows the calculation of molar mass of non-volatile solutes by using vapor pressure lowering. Ensure to use proper unit conversions and balance the mole fraction equations carefully.

Answer 1

We can use Raoult's Law to find the molar mass of the solute. Raoult's Law states: \[ \frac{P_{\text{solvent}}}{P_{\text{solvent, pure}}} = \frac{n_{\text{solute}}}{n_{\text{solution}}} \] Where: - \( P_{\text{solvent}} \) is the vapor pressure of the solution, - \( P_{\text{solvent, pure}} \) is the vapor pressure of the pure solvent, - \( n_{\text{solute}} \) is the number of moles of solute, - \( n_{\text{solution}} \) is the number of moles of the solution. Step 1: Calculate the mole fraction of the solute: \[ \frac{P_{\text{solvent}}}{P_{\text{solvent, pure}}} = 1 - \frac{31.84}{32} = 0.005 \] Step 2: Use this value to find the moles of solute: \[ \frac{n_{\text{solute}}}{n_{\text{solution}}} = 0.005 \quad \Rightarrow \quad n_{\text{solute}} = 0.005 \times \left( \frac{200}{18} \right) \] This gives: \[ n_{\text{solute}} = 0.005 \times 11.11 = 0.0555 \text{ mol} \] Step 3: Calculate the molar mass: \[ \text{Molar mass of solute} = \frac{\text{mass of solute}}{n_{\text{solute}}} = \frac{5 \, \text{g}}{0.0555 \, \text{mol}} = 90.05 \, \text{g/mol} \] So, the molar mass of the solute is 90.05 g/mol.