If 651 grams of ethylene glycol, HOCH2CH2OH, is dissolved in 1.50 kg of water at 90.°C, what is the vapor pressure of the water in the solution?

Vapor pressure of solution = 467.9 Torr

Explanation:

This about a colligative property, vapor pressure lowering.

ΔP = P°. Xm

ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

Xm = Mole fraction, for solute.

Let's convert the mass of solute and solvent to moles (mass / molar mass)

651 g / 62.07 g/m = 10.5 moles of ethylene glycol

1500 g / 18 g/m = 83.3 moles of water

Total moles = 10.5 moles of solute + 83.3 moles of solvent ⇒ 93.8 moles

Mole fraction of solute = 10.5 / 93.8 = 0.11

Vapor pressure of water at 90°C is 525.8 Torr

525.8 Torr - Vapor pressure of solution = 525.8 Torr. 011

Vapor pressure of solution = 525.8 Torr. 011 - 525.8 Torr

Vapor pressure of solution = 467.9 Torr