1.

In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. An immiscible mixture of water and quinoline is prepared and distillation is done. If p_("quinoline")^(@) = 7.9 torr and p_("water")^(@) = 732.04 torr at the boiling point of 98.9^(@)C. Find out the relative masses of water and quinoline in the distillate. If it is given that no of moles of any component in the vapour phase prop its vapour pressure. (Molar mass of water = 18 g/mol and molar mass of quinoline= 129g//mol)

Answer»

`W_("water"): W_("quinoline") : : 13:1`
`W_("water"): W_("quinoline"):: 1:1`
`W_("water") : W_("quinoline") ::1:14`
`W_("Water"):W_("quinoline") :: 7.1:1`

SOLUTION :`(W_("water"))/(W_("quinoline"))=(n_("water"))/(n_("quinoline"))XX(M_("water"))/(M_("quinoline"))`
and `(n_("water"))/(n_("quinoline"))=(p_("water"))/(p_("quinoline"))=(732.04)/(7.9)`
`(W_("water"))/(W_("quinoline"))=(73204)/(790)xx(18)/(129)~=(13)/(1)`.


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