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1.0 g of a non-electolyte solute (molar mass 250 g `mol^(-1)`) was dissolved in 51.2 g of benzene. If the `K_(f)` for benzene is 5.12 K kg `mol^(-1)`, the freezing point of benzene will be lowerd by :A. 0.2 KB. 0.4 KC. 0.3 KD. 0.5 K. |
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Answer» Correct Answer - b `DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))` `=((5.12 kg mol^(-1))xx(1.0g))/((250gmol^(-1))xx(0.0512kg))=0.4 K` |
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