An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Given:
2% non-volatile solute in an aqueous solution means that if the solution is of mass 100 g then the mass of solute in the solution is,
Mass of solute = [2 × 100]/100
= 2 g
∴ Mass of Solvent = 100-2
= 98 g
Vapour pressure of solution at normal boiling point, P1 = 1.004 bar
Vapour pressure of pure water at normal boiling point, P1° = 1 atm = 1.013 bar
To find: Molar mass of solute
Formula:
From Raoult’s Law, we have,
………….. [1]
Now we know that number of moles of solute is given by the following relationship,
⇒ Number of moles 
Using the above relationship the equation [1] can be modified as follows:
Where
W1 = mass of solvent
W2 = mass of solute
M1 = molecular mass of solvent
M2 = molecular mass of solute
Solution:
M2 = 
M2 = 41.3469 g
∴ M2 ≈ 41.35 g
Therefore the molar mass of the solute is 41.35 g.