Derive relationship between relative lowering of vapour pressure and molar mass of non volatile solute.(ii) Write statement of second law of thermodynamics.

Answer: (i) \(\dfrac{P_1^\circ - P_1}{P_1^\circ} = \dfrac{W_2 M_1}{M_2 W_1}\), so \(M_2 = \dfrac{W_2 M_1 P_1^\circ}{W_1 (P_1^\circ - P_1)}\). (ii) Statement below.

(i) By Raoult’s law, for a solution of a non-volatile solute the vapour pressure is due to the solvent only: \(P_1 = x_1 P_1^\circ\). The lowering of vapour pressure is \(\Delta P = P_1^\circ - P_1 = P_1^\circ(1 - x_1) = P_1^\circ x_2\). Hence the relative lowering of vapour pressure equals the mole fraction of the solute:

For a dilute solution \(n_2 \ll n_1\), so \(x_2 \approx \dfrac{n_2}{n_1}\). With \(n_1 = W_1/M_1\) (solvent) and \(n_2 = W_2/M_2\) (solute),

So the molar mass of the solute can be found by measuring the relative lowering of vapour pressure.

(ii) Second law of thermodynamics: The total entropy of a system and its surroundings increases in a spontaneous process: \(\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0\). (Equivalent statement: heat cannot flow by itself from a colder body to a hotter body.)