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 $\circ$ C given that the vapor pressure of pure water is 23.8 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 ∘\circ∘ C Given That the Vapor Pressure of Pure Water Is

A Solution Is Prepared by Adding 1 $\circ$ C Given That the Vapor Pressure of Pure Water Is