Question

Derive the relation between relative lowarimg of vapour pressure and molar mass of non volaite solute

Answer 1

Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1 . Hence number of
mole of solvent n1 and number of mole of solute n2 in solution.

n1=W1M1n1=W1M1 and  n2=W2M2n2=W2M2 (∵Number of moles (n)=mass of the substancemolar mass of the substance)(∵Number of moles (n)=mass of the substancemolar mass of the substance)

The mole fraction of solute x2 is given by

x2=n2n1+n2x2=n2n1+n2

x2=W2M2W1M1+W2M2x2=W2M2W1M1+W2M2 .....(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

P01P10 and that of component A2 is p_2^0 The relative lowering of vapour pressure is given by,

Δpp10=p01−pp01Δpp01=p10-pp10

Δpp10=p01x2p01Δpp01=p10x2p10

Δpp10=x2Δpp01=x2 .....(2)

Combining equations (1) and (2)

Δpp10=p01−pp01=x2=W2M2W1M1+W2M2Δpp01=p10-pp10=x2=W2M2W1M1+W2M2

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

Δpp10=n2n1=W2M2W1M1=W2M1W1M2Δpp01=n2n1=W2M2W1M1=W2M1W1M2

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.