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The freezing point (in `.^(@)C)` of a solution containing `0.1g` of `K_(3)[Fe(CN)_(6)]` (Mol. wt. `329`) in `100 g` of water `(K_(f) = 1.86 K kg mol^(-1))` is :A. `-2.3 xx 10^(-2)`B. `-5.7 xx 10^(-2)`C. `-5.7 xx 10^(-3)`D. `-1.2 xx 10^(-2)` |
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Answer» Correct Answer - A `K_(3)[Fe(CN)_(6)]rarr3K^(+)+[Fe(CN)_(6)]^(3-)` `i=4` `DeltaT=ixxK_(f)xx(w_(B)xx1000)/(m_(B)xxw_(A))=4xx1.86xx(0.1xx1000)/(329xx100)` `=2.3xx10^(-2)` `:.` Freezing point of solution will be `-2.3 xx 10^(-2) .^(@)C`] |
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