A solution is prepared by adding 1.50 mol of glucose, which is not volatile, to 3.50 mol of water.What is the vapor pressure of this solution at 25 $\degree$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, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. The mole fraction of water ($X_{\text{water}}$) is calculated as the moles of water divided by the total moles in the solution: $X_{\text{water}} = \frac{3.50}{3.50 + 1.50} = \frac{3.50}{5.00} = 0.70$. The vapor pressure of the solution ($P_{\text{solution}}$) is then $0.70 \times 23.8 \, \text{torr} = 16.66 \, \text{torr}$, which rounds to 16.7 torr.

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

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