# Determine the following properties of the solution?

A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at 25°C. Given that the molal freezing-point-depression constant (Kf) for water is 1.86°C/m and vapor pressure of pure water at 25°C is 23.8 torr, determine the following properties of the solution:

a. Mole fraction of H2O, X waterb. Vapour pressure (torr), P solnc. Freezing point (°C), Tf

Relevance
• Assume you have 1000 mL of the solution. That will contain 0.944 moles of glucose. The mass of glucose in the solution = 0.944 mol X 180.16 g/mol = 170.1 g

The mass of that is 1000 mL X 1.0624 g/mL = 1062.4 g. So, the mass of the solution that is water = 1062.4 - 170.1 g = 892.3 g H2O. Moles H2O = 892.3 g / 18.02 g/mol = 49.52 mol H2O

Mole fraction H2O = 49.52 mol / (49.52+0.944) = 0.981

Vapor pressure = 23.8 torr X (0.981) = 23.3 torr

Molality of the solution = 0.944 mol / 0.892 kg H2O = 1.06 m

Change in freezing point = 1.86°C/m X 1.06 m = 1.97 °C

The solution will freeze at -1.97 °C