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?

Answer

269.27 K

Explanation

`DeltaT_(f)=K_(f)m therefore K_(f)=(DeltaT_(f))/(m)=(273.15-271)/(0.1539)="13.97K/m, "DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2))=(1000xx13.97xx5)/(100xx180)=3.88`
`therefore" Freezing point of the solution "=273.15-3.88K=269.27K.`