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When `0.6g` of urea dissolved in `100g` of water, the water will boil at `(K_(b)` for water `= 0.52kJ. mol^(-1)` and normal boiling point of water `=100^(@)C)`:A. `373.052 K`B. `273.52K`C. `372.48K`D. `273.052K` |
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Answer» Correct Answer - A `DeltaT_(b) = ((0.6)/(60))/(100) xx 1000 xx 0.52` `= (1)/(1000) xx 1000 xx 0.52 = 0.52` `T - 373 = 0.52 T = 373.52` |
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