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`1.0 g` of a non-electrolyte solute( mol. Mass `250.0 g mol^(-1)`) was dissolved in `5.12 g` benzene. If the freezing point depression constant, `K_(f)` of benzene is `51.2 K kg mol^(-1)`, the freezing point of benzene will be lowered by:A. `0.2 K`B. `0.4 K`C. `0.3 K`D. `0.5 K` |
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Answer» Correct Answer - 2 We need to calculate `DeltaT_(f)` (depression of freezing point of solvent) which is given as `DeltaT_(f)=iK_(f)m=iK_(f) W_(solute)/(mm_(solute)W_(solv))` we are given `i=1` as the solute is nonelectrolyte `K_(f)=5.12 K kg mol^(-1)` `W_(solute)=1.00 g` `W_(solvent)=51.2/1000kg` `mm_(solute)=250 g mol^(-1)` Substituting these results, we get `DeltaT_(f)=(1)(5.12 K kg mol^(-1))((1.00 g))/((250 g mol^(-1))(5.12/1000kg))` `=0.4 K` |
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