Question

the vapor pressure of a solution containing 0.024 kg of a solute dissolved in 0.200 kg of water at 300 k is 3.50×10³ Pa.calculate the molar mass of the solute if the vapoure pressure of water at 300 K is 3.70×10³ Pa

Answer 1

Let molar mass of the solute be = M grams
Moles of solute in the solution = 24 gm/M gms = 24/M

solvent = water ,  molar mass = 18 gm
weight of solvent = 0.200 kg 
moles of solvent = 200/18 = 100/9
Total number of moles in the solution = [100/9 + 24/M] = (100M+ 216) / 9M
Mole Fraction of solvent = 100 M /(100M + 216)

Apply Raoult's law.

Vapor pressure of solution = 100 M / (100M+216) * 3.70 * 10³  Pa
So     3.5 * 10³ = 100 M /(100M+ 216) * 3.70 * 10³
          1 + 216/(100M) = 37/35
          100M/216 = 35/2
          200 M = 35*216        
 =>  M = 37.8  gms/mole