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Pure benzene freezes at `5.45^(@)C` at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at `3.55^(@)C`. The `K_(f)` for benzene isA. `5.08 K kg mol^(-1)`B. `508 K kg mol^(-1)`C. `0.508 K kg mol^(-1)`D. `50.8^(@)C kg mol^(-1)`

Answer» Correct Answer - A
`DeltaT_(f) = (5.45 - 3.55) = K_(f) xx m`
or `1.90 = k_(f) xx 0.374`
or `k_(f) = (1.90)/(0.374) = 5.08`


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