Derive the relationship between relative lowering of vapour pressure and molar mass of nonvolatile solute.
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.
`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 x2 is given by
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 A2 is p_2^0 The relative lowering of vapour pressure is given by,
Combining equations (1) and (2)
For dilute solutions n1 >> n2. Hence n2 may be neglected in comparison with n1 in equation (1) and thus equation (3) becomes
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.
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- Colligative Properties and Determination of Molar Mass - Relative Lowering of Vapour Pressure