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?

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?

Solution :

 Given values

Concentration is given in percent so that take

mass of the solution = 100 g

mass of non-volatile solute = 2% = 2g mass of the solvent = (100 — 2) = 98 g molecular mass of solvent (water) = 18

We have to find molecular mass of solute

The vapour pressure of pure boiling water = 1atm = 1.013 bar.

Change in vapour pressure  = (1.013 — 1.004)

      = 0.009 bar

Formula of Raoult’s law

Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 < < n1, hence neglecting n2 in the denominator we have

Use this formula we get 

Here w₁ and w₂ are the masses and M₁ and M₂ are the molar masses of the solvent and solute respectively.

Plug the values in above formula we get 

 Cross multiply we get