`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 realed to the molality, therefore the molality of the solution with respect to ethylene glycol,
`DeltaT_(f)=K_(f)m`
Mole of ethylene glycol = `45 gxx(1 mol)/(62g) =0.73 mol`
Mass of water in kg =`600 xx (1 kg)/(1000 g)=0.60 kg`
Hence, molality of ethylene glycol
`=(0.73 mol)/(0.60 kg)=1.20 mol kg`
Therefore,freezing point depression
`(DeltaT_(f)=1.86 K kg mol^(-1))xx(1.2 mol kg^(-1) )=2.2 K`
Freezing point of the aqueous solution
`T_(f) =T_(f)^(@) - Delta_(f)T`
`273.15 K -2.2 K =270.95 K`