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A `5%` solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.

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Correct Answer - `269.07 K`
`W_(2)=5g,W_(solution)=100g, Mw_(2)=342g`
`(C_(12)H_(22)O_(11))`
First case
`W_(1)=100-5=95g`
`DeltaT_(f)=273.15-271=2.15K`
`DeltaT_(f)=K_(f)xx(W_(2)xx1000)/(Mw_(2)xxW_(1))`
`2.15=K_(f)xx(5xx1000)/(342xx95)`
Solve for`K_(f)impliesK_(f)=13.97`
Second case
`W_(2)=5g, W_(1)=95g, ` ltbr. `Mw_(2)` of glucose `(C_(6)H_(12)O_(6))=180g`
`DeltaT_(g)=13.97xx(5xx1000)/(180xx95)=4.085`
`273.15-T_(2)=4.085`
`T_(2)=273.15-4.085=269.07K`

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