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The freezing point of 0.02 mol fraction ...

The freezing point of `0.02 mol` fraction solution of acetic acid (A) in benzene (B) is `277.4K`. Acetic acid exists partly as a dimer `2A = A_(2)`. Calculate equilibrium constant for the dimerisation. Freezing point of benzene is `278.4K` and its heat of fusion `DeltaH_(f)` is `10.042 kJ mol^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
`3.225`
`k_(f) = (8.314 xx(278.4)^(2) xx 78)/(1000 xx 10042) = 5`
`m = (0.02 xx 1000)/(0.98 xx 78) = 0.2614`
`DeltaT_(f) = i xx k_(f) xx m`
`i=1 -(alpha)/(2) = 0.7613`
`alpha = 0.4713`
`{:(2A,hArr,A_(2)),(C,,),(C-Calpha,,Calpha//2):}`
`k = (0.4713)/(2 xx 0.2614(1-0.4713)^(2)) = 3.225`
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