Question

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?

Answer 1

Here, the vapor pressure of the solution at normal boiling point (p_s) = 1.004 bar

Vapour pressure of pure water at normal boiling point (p°) = 1 atm = 1.013 bar

Let mass of solution = 100 g

Mass of solute (w₂) = 2 g

Mass of solvent (water), (w₁) = 100 − 2 = 98 g

Molar mass of solvent (water), (M₁) = 18 g mol⁻¹

According to Raoult’s law,

`(p^circ - p_s)/p^circ = n_2/(n_1 + n_2) = n_2/n_1`

= `(w_2//M_2)/(w_1//M_1) = w_2/M_2 xx M_1/w_1`

`=> (1.013 - 1.004)/1.013 = (2 xx 18)/(M_2 xx 98)`

`=> 0.009/1.013 = (2 xx 18)/(M_2 xx 98)`

`=> M_2 = (1.013 xx 2 xx 18)/(0.009 xx 98)`

= 41.35 g mol⁻¹

Hence, the molar mass of the solute is 41.35 g mol⁻¹.