Question

A KCl solution is prepared by dissolving 25.0 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 23.76mm Hg?

A) 16.5mm Hg

B) 24.9mm Hg

C) 23.2mm Hg

D) 21.6mm Hg

E) 22.7mm Hg

Answer 1

To calculate the vapor pressure of the solution, use Raoult's law. The vapor pressure of the solution is 23.2 mm Hg. So, option C is correct.

Step-by-step explanation:

To calculate the vapor pressure of a solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is determined by multiplying the mole fraction of the solvent by the vapor pressure of the pure solvent. Specifically, in the given scenario, where water serves as the solvent and KCl as the solute, the vapor pressure of the solution is governed by the product of the mole fraction of water and the vapor pressure exhibited by pure water. This law is applicable under ideal conditions, assuming no significant interactions between the solute and solvent molecules, and it provides a theoretical framework for understanding the behavior of solutions in terms of vapor pressures.

First, we need to calculate the mole fraction of water:

Mole fraction of water (X solvent) = moles of water / total moles

Using the given masses and molar masses:

moles of KCl = 25.0 g / 74.55 g/mol = 0.335 mol

moles of water = 250.0 g / 18.015 g/mol = 13.87 mol

total moles = moles of KCl + moles of water = 14.205 mol

Mole fraction of water (X solvent) = 13.87 mol / 14.205 mol = 0.976

Now we can use Raoult's law:

Vapor pressure of the solution = X solvent * Vapor pressure of water

Vapor pressure of the solution = 0.976 * 23.76 mm Hg = 23.2 mm Hg