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

Raoult's Law can be applied to determine the molar mass of a solute. The law is expressed as:\[\frac{P_{\text{solvent}}}{P_{\text{solvent, pure}}} = \frac{n_{\text{solute}}}{n_{\text{solution}}}\]Definitions:- \( P_{\text{solvent}} \) represents the vapor pressure of the solution.- \( P_{\text{solvent, pure}} \) is the vapor pressure of the pure solvent.- \( n_{\text{solute}} \) denotes the number of moles of solute.- \( n_{\text{solution}} \) indicates the number of moles of the solution.Procedure:Step 1: Compute the mole fraction of the solute.\[\frac{P_{\text{solvent}}}{P_{\text{solvent, pure}}} = 1 - \frac{31.84}{32} = 0.005\]Step 2: Utilize this value to ascertain the moles of solute.\[\frac{n_{\text{solute}}}{n_{\text{solution}}} = 0.005 \quad \Rightarrow \quad n_{\text{solute}} = 0.005 \times \left( \frac{200}{18} \right)\]Resulting in:\[n_{\text{solute}} = 0.005 \times 11.11 = 0.0555 \text{ mol}\]Step 3: Calculate the molar mass.\[\text{Molar mass of solute} = \frac{\text{mass of solute}}{n_{\text{solute}}} = \frac{5 \, \text{g}}{0.0555 \, \text{mol}} = 90.05 \, \text{g/mol}\]Consequently, the molar mass of the solute is 90.05 g/mol.