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Calculate the freezing point of a solution conataining 54 g glucose `(C_(6)H_(12)O_(6)` in 250 gof water will freeze `K_(f)` for waer is 1.86 `Km^(-1))`

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Correct Answer - 272.93K
`W_(B)=0.52g, M_(B)=180" g mol"^(-1), W_(A)=80 g=0.08 kg`
`DeltaTK_(f)=1.86" K m"^(-1)=1.86" K kg mol"^(-1)`.
`DeltaT_(f)=(W_(B)xxK_(f))/(M_(B)xxW_(A))=((0.52g)xx(1.86" K kg mol"^(-1)))/((180" g mol"^(-1))xx(0.08" kg"))=0.067 K`
`"Freezing point of solution" = 273 K-0.067 K=272.93 K`

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