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A 0.1539 molal aqueus solution of cane sugar `("mol mass" = 342 g mol^(-1))` has a freezing point of 271 K while freezing point ofpure water is 273.15 K. What will be the freezing point of an aqueus solution containing 5 g of glucose `("mol. Mass" =180 g mol^(-1))` per 100 g of water?

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`Delta T_(f)=K_(f)xxm`
`DeltaT_(f)=(273.1-271)=2.15 K, M=0.1539 molal`
`(2.15K)=K_(f)xx0.1539 M`
`m=((5g)/(180 g mol^(-1)))/((0.1 kg))=0.278m`
`DeltaT_(f)=K_(f)xx0.278m`
Divding eqn (ii) by eqn (i)
`(DeltaT_(f))/((2.15K))=(K_(f)xx(0.278m))/(K_(f)xx(0.1539m))`
`DeltaT_(f)=((0.278m)xx(2.15K))/((0.139m))=3.88 K`
Freezing point temperature of the solution=(273.15-3.88)=269.27 K

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