Calculate the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass. (Assume complete dissociation of the solute.)

Vapor pressure of solution = 25.1 Torr

Explanation:

Vapor pressure lowering is one of the colligative property. Its formula is:

ΔP = P°. Xm. i

ΔP = Vapor pressure pure solvent - Vapor pressure solution

P° is Vapor pressure pure solvent

Xm is the mole fraction of solute (mol of solute / total moles)

i is the Van't Hoff factor, the number of ions particles dissolved.

NaCl → Na⁺ + Cl⁻ (For this dissociation, i = 2)

Let's determine the mole fraction of this solution.

5.50% by mass is 5.50 g of solute in 100 g of solution.

Total mass = 100 g → Mass of solute + Mass of solvent

100 g - 5.5g = 94.5 g (mass of solvent)

Moles of solute → Solute mass / molar mass

5.5 g / 36.45 g/mol = 0.151mol

Moles of solvent → Solvent mass / molar mass

94.5 g / 18 g/mol = 5.25mol

Total moles = Moles of solute + Moles of solvent

0.151 mol + 5.25mol = 5.401 total moles

Xm solute = 0.151 / 5.401 → 0.028

Vapor pressure of water at 25°C → 23.8 Torr

Let's replace the data

Vapor pressure of solution - 23.8 Torr = 23.8 Torr. 0.028. 2

Vapor pressure of solution = 23.8 Torr. 0.028. 2 + 23.8 Torr

Vapor pressure of solution = 25.1 Torr