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Three particles of a solute 'A', associa...

Three particles of a solute 'A', associate in benzene to form species `(A)_(3)`. Calclate the freezing pointof sthe 0.25 molal solution. The degree of association of solute 'A' is found to be 0.80. TheFreezing point of benzene is `5.5^(@)C` and its cryoscopic constanat is 5.12 K `m^(-1)`.

Text Solution

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Calculation of Van't Hoff factor (i)
Since three particles of solute 'A' assoviate to forem `(A)_(3)`
Value of n=3, alpha=0.80
`alpha=(i-1)/(1//n-1) or 0.80=(i-1)/((1//3-1))`
(i-01)=0.8 (-0.667)=(-0.533)
i=1-0.533=0.467
Calculate of freezing point `(DeltaT_(f))`
`DeltaT_(f)=iK_(f)xxm`
`i=0.467, K_(f)=5.12 Km^(-1),m=0.25 molal`
`DeltaT_(f)=(0.467)xx(5.12 Km^(-1))xx(0.25m)= 0.6^(@)C`
Calculation of freezing of solution
Freezing point of solution `(T_(f))=T_(f)^(@)-DeltaT_(f)=5.5^(@)C=4.9^(@)C`
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