`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