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`1.0 g`of a non-electrolyte solute (molar mass`250gmol^(-1)` was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is `5.12K kgmol^(-1)` the lowering in freezing point will beA. 0.4 KB. 0.3 KC. 0.5 KD. 0.2 K |
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Answer» Correct Answer - A Molality of non-electrolyte solute `=(("Weight of solute in (g)")/("molecular weight of solute"))/("weight of solvent in (kg)")=((1)/(250))/(0.0512)` `=(1)/(250xx0.0512)=0.0781 m` `Delta T_(f)=K_(f)xx` molality of solution `=5.12xx0.0781 ~~ 0.4 K` |
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