The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 35 g of urea (a nonvolatile, non-electrolyte, molar mass = 60.0 g/mol) in 75 g of water.

Question

Chemistry

Lucian Schultz

3 July, 05:40

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 35 g of urea (a nonvolatile, non-electrolyte, molar mass = 60.0 g/mol) in 75 g of water.

+1

Answers (1)

Know the Answer?

Pollard · Answer 1

Vapor pressure of solution = 20.8 Torr

Explanation:

This is about vapor pressure lowering

ΔP = P°. Xm

ΔP states the difference between vapor pressure of pure solvent and vapor pressure from solution.

P° is the vapor pressure of pure solvent.

Xm means mole fraction (mol of solute / mol of solute + mol of solvent)

Let's determine the mol of solute

35 g / 60 g/mol = 0.583 mol of urea

Let's determine the mol of solvent

75 g / 18 g/mol = 4.167 mol of water

Total moles = 0.583 + 4.167 = 4.75

Mole fraction of solute → 0.583 / 4.75 = 0.123

Let's replace the data in the formula:

23.8Torr - Vapor pressure of solution = 23.8 Torr. 0.123

Vapor pressure of solution = - (23.8 Torr. 0.123 - 23.8 Torr)

Vapor pressure of solution = 20.8 Torr