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A `0.1` molal aqueous solution of a weak acid is `30%` ionized. If `K_(f)` for water is `1.86^(@)C//m`, the freezing point of the solution will be.A. `-0.24^(@)C`B. `-0.18^(@)C`C. `-0.54^(@)C`D. `-0.36^(@)C` |
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Answer» Correct Answer - A `i=1-alpha+nalpha` i=1-0.3+2(0.3) i=1.3 `DeltaT_f=iK_fm` `=1.3xx1.86xx0.1` `DeltaT_f=+0.24^@C` Freezing point of solution = `-0.24^@C` |
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