A KCl solution is prepared by dissolving 40.0 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?

22.0 mmHg

Explanation:

The vapor pressure of a solution is a colligative property, which means that it is determined by the number of particles (molecules or ions) of solute present in a solution.

Raoult's law permits the calculations of the change of the vapor pressure of a solvent when a solute is added.

The equation is:

P solvent - P solution = ΔP = X solute * P solven

Where:

P solvent = vapor pressure of the pure solvent. P solution = vapor pressure of the solution X solute = molar fraction of the solute

In the case of ionic solutes, you must take into account the number of ions that result from the ionization.

Calculating the molar fraction:

number of moles = mass in grams / molar mass number of moles of KCl: 40.0 g / 74.5513 g/mol = 0.567 mol moles of ions = 2 * number of moles of KCl = 1.134 mol moles of water: 250.0g / 18.015 g/mol = 13.877 mol

total moles = 1.134 mol + 13.877 mol = 15.011 mol

X solute = moles of ions / total moles = 1.134 mol / 15.011 mol = 0.0755

Calculating the change in the vapor pressure of the solution:

ΔP = X solute * P solvent = 0.0755 * 23.76 mmHg = 1.78 mmHg

Vapor pressure of the solution:

P solution = P solvent + ΔP = 23.76 mmHg - 1.79 mm Hg = 21.97mmHg

Rounding to three significant figures (because 40.0g has three significant figures) : 22.0 mmHg ← answer.