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An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:A. `0.0585 K`B. `0.0512 K`C. `1.864 K`D. `0.0265K` |
Answer» Correct Answer - B `0.186 = 1.86 xm` `m = 0.1` `DeltaT_(b) = 0.1 xx 0.512 = 0.0512` |
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