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?
A) 7.00 torr
B) 16.7 torr
C) 10.2 torr
D) 7.14 torr
E) 34.0 torr

Question

Quizplus · Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's Law: P_solution = X_solvent * P_pure_solvent. The mole fraction of water (solvent) is 3.50 / (1.50 + 3.50) = 0.7. Therefore, P_solution = 0.7 * 23.8 torr = 16.7 torr.

A Solution Is Prepared by Adding 1