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
`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 A2 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.
18. What is the relation between osmotic pressure (π) and relative lowering of vapour pressure? Show Derivation.
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