1.

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

Answer»


SOLUTION :`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.`


Discussion

No Comment Found

Related InterviewSolutions