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The freezing point of 0.02 mole fraction...

The freezing point of `0.02` mole fraction acetic acid in benzene is `277.4 K`. Acetic acid exists partly as dimer. Calculate the equilibrium constant for dimerization. The freezing point of benzene is `278.4 K` and the heat the fusion of benzene is `10.042 kJ mol^(-1)`. Assume molarity and molality same.

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The correct Answer is:
`3.39` ;
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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) .

What is the freezing of 0.4 molal solution of acetic acid in benzene in which it dimerises to the extent of Freezing point of benzene is 278.K and its molar heat of fusion is 10-042KJ mol^(-1)