State Raoult’s law.(b) Derive the relationship to determine molar mass of non volatile solute by depression of freezing point.

Answer:

(a) Raoult’s law: the partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution, \(P_1 = P_1^0 x_1\). For a solution of a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: \(\dfrac{P_1^0 - P}{P_1^0} = x_2\).

(b) The depression of freezing point is proportional to the molality \(m\) of the solution: \(\Delta T_f = K_f\, m\), where \(K_f\) is the cryoscopic constant.

If \(W_2\) g of solute of molar mass \(M_2\) is dissolved in \(W_1\) g of solvent, moles of solute \(= W_2/M_2\) and

\(m = \dfrac{W_2/M_2}{W_1/1000} = \dfrac{1000\,W_2}{M_2 W_1}\)

\(\therefore \Delta T_f = \dfrac{1000\,K_f W_2}{M_2 W_1}\), so \(M_2 = \dfrac{1000\,K_f W_2}{\Delta T_f\, W_1}\) (g mol^–1).

(With masses in kg: \(M_2 = \dfrac{K_f W_2}{\Delta T_f W_1}\).)