Question

A solution is prepared by adding 1.50 mol glucose, which is not volatile, to 3.50 mol...

A solution is prepared by adding 1.50 mol glucose, which is not volatile, to 3.50 mol water. What is the vapor pressure of this solution at 25 C given that the vapor pressure of pure water is 23.8 torr?

Homework Answers

Answer #1

Mole fraction of water in glucose solution can be calculated as follows:

Mole fraction of water = moles of water/(total moles of glucose and water)

Mole fraction of water = 3.50 mol / (1.50 + 3.50) mol

Mole fraction of water = 0.700

Vapor pressure of glucose solution can be calculated by using Raoult’s Law expression as follows:

Vapor pressure = Vapor pressure of pure water x mole fraction of water

Vapor pressure = 23.8 torr x 0.700

Vapor pressure = 16.7 torr

Therefore, vapor pressure of glucose solution is 16.7 torr.

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