1.

Three particles of a solute 'A' assoviate in benzene to form species (A)_(3). The degree of association of 'A' is found to be 0.92. Calculate the freezing point of benzene is 5.5^(@)C and its cryscopic constant is 5.12 Km^(-1).

Answer»

Solution :`alpha=(i-1)/(1/n-1)or 0.92=(i-1)/(1/3-1)`
(i-1)=0.92(-0.667)=(-0.61)
i-1-(0.61)=0.39
`DeltaT_(f)=ixxK_(f)xxm=(0.93)xx(5.12 KM^(-1))xx(0.25m)`
=0.5 K or `0.5^(@)C`
Frezing POINT temperaturte of solution.
`T_(f)=T_(f)^(@)-DeltaT_(f)=5.5^(@)C-0.5^(@)C`
`=5.0^(@)C`


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