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Calculate the temperature at which a solution containing 54g of glucose, `(C_(6)H_(12)O_(6))` in 250g of water will freeze. (`K_(f)` for water = 1.86 K `mol^(-1)` kg) |
Answer» Molecular mass of glucose, `M_(B) = 72 + 12 + 96 = 180` `DeltaT_(f)=(K_(f)xxW_(B)xx1000)/(M_(B)xxW_(A)) = (1.86 xx 54 xx 1000)/(180xx250)=(100440)/(45000)=2.23` Freezing point of solution = 0 - 2.23 = - 2.23 `.^(@)C` |
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