Question

A solution is prepared by dissolving 32.5 g of a nonvolatile solute in 200 g of water. The vapor pressure above the solution is 21.85 Torr and the vapor pressure of pure water is 23.76 Torr at this temperature. What is the molecular weight of the solute?

Answer 1

Solution-

Given-

Mass of non volatile solute = 32.5 g

Volume of water = 200g/1 = 200 mL

Molar mass of water = 18 g/mol

vapor pressure above the solution = 21.85 Torr

vapor pressure of pure water = 23.76 Torr

By using Raoult's Law

Vapor pressure of solution = mole of solvent/pure pressure of solvent

Mole fraction of solvent (water) = vapor pressure of solvent/vapor pressure of solute

= 21.85torr/23.76 torr = 0.92 Torr

Mole fraction of solute = 1-0.92 = 0.08

Mole of water = mass /molar mass = 200/18 = 11.11 mole

Mole fraction of solute = mole of solute/(moles of solvent + mole of solute)

Lets assume moles of solute = x

0.08 = x/(11.11 + x)

0.08(11.11 + x) = x

0.88 + 0.08x = x

0.88 = x- 0.08

X = 0.88/0.92

Moles of solute(X) = 0.95 mole

We know the equation

Moles = Mass in g/molar mass

Molar mass = Mass of solute/moles of solute =   32.5g/0.95mole = 34.21 g/mole

Answer- Molecular weight of the solute = 34.21 g/mole