A KCl solution is prepared by dissolving 60.0 g KCl in 250.0 g of water at 25°C.What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?
A)20.5 mm Hg
B)21.3 mm Hg
C)22.5 mm Hg
D)25.5 mm Hg

Question

Quizplus · Accepted Answer

The vapor pressure of a solution is given by Raoult's law, which is expressed as P = X_water P_water, where P is the vapor pressure of the solution, X_water is the mole fraction of water, and P_water is the vapor pressure of pure water. To calculate the mole fraction of water, we need to know the moles of KCl and water in the solution.

Moles of KCl = 60.0 g / 74.55 g/mol = 0.804 mol
Moles of water = 250.0 g / 18.015 g/mol = 13.874 mol

Total moles of solute and solvent = 0.804 mol + 13.874 mol = 14.678 mol

Mole fraction of water = 13.874 mol / 14.678 mol = 0.944

Substituting the values in Raoult's law, we get:
P = 0.944 x 23.76 mm Hg = 22.44 mm Hg, which is approximately 21.3 mm Hg.

Therefore, the correct answer is B) 21.3 mm Hg.

A KCl Solution Is Prepared by Dissolving 60