#### Question

Derive the relationship between relative lowering of vapour pressure and molar mass of nonvolatile solute.

#### Solution

Let W_{2} g of solute of molar mass M_{2} be dissolved in W_{1} g of solvent of molar mass M_{1} . Hence number of

mole of solvent n_{1} and number of mole of solute n_{2} in solution.

`n_1=W_1/M_1 ` and `n_2=W_2/M_2` `(because "Number of moles (n)"="mass of the substance"/"molar mass of the substance")`

The mole fraction of solute x_{2} is given by

`x_2=n_2/(n_1+n_2)`

`x_2=(W_2/M_2)/(W_1/M_1+W_2/M_2)` .....(1)

For a solution of two components A1 and A2 with mole fraction x1 and x2 respectively, if the vapour pressure of pure component A1 is

`P_1^0 ` and that of component A_{2} is p_2^0 The relative lowering of vapour pressure is given by,

`(Deltap)/p_0^1=(p_1^0-p)/p_1^0`

`(Deltap)/p_0^1=(p_1^0x_2)/p_1^0`

`(Deltap)/p_0^1=x_2` .....(2)

Combining equations (1) and (2)

`(Deltap)/p_0^1=(p_1^0-p)/p_1^0=x_2=(W_2/M_2)/(W_1/M_1+W_2/M_2)`

For dilute solutions n1 >> n2. Hence n2 may be neglected in comparison with n1 in equation (1) and thus equation (3) becomes

`(Deltap)/p_0^1=n_2/n_1=(W_2/M_2)/(W_1/M_1)=(W_2M_1)/(W_1M_2)`

Knowing the masses of non-volatile solute and the solvent in dilute solutions and by determining experimentally vapour pressure of pure solvent and the solution, it is possible to determine molar mass of a non-volatile solute.