`45 g` of ethylene glycol `C_(2)H_(6)O_(2)` is mixed with `600 g` of water. Calculate (a) the freezing point depression and (b) the freezing point of solution.
Given`K_(f)=1.86 K kg mol^(-1)`.

Explanation

Depression in freezing point is related to the molality, therefore, the molality
of the solution with respect to ethylene glycol `=("moles of ethylene glycol")/("mass of water in kilogram")`
Moles of ethylene glycol `=(45g)/(1000g" "kg^(-1))=0.6kg`
Hence molality of ethylene glycol `=(0.73" "mol)/(0.60" "kg)=1.2" mol "kg^(-1)`
Therefore freezing point depression
`overset(..)(A)T_(f)=1.86" K kg "mol^(-1)xx1.2" mol "kg^(-1)=2.2K`
Freezing point of the aqueous solution `=273.15K-2.2K=270.95K`