`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

Elevation in boiling point may be calculated from the relation,
`Delta T_(b) = (K_(b) xx w_(b) xx 1000)/(M_(B) xx w_(A))`
`w_(B) = 0.052 g, w_(A) = 80.2 g, K_(b) = 5.2 K m^(-1) , M_(B) = 180`
`therefore " " Delta T_(b) = (5.2 xx 0.052 xx 1000)/(180 xx 80.2) = 0.0187`
Boiling point of water = 373 K .
Boiling point of solution = `373+0.0187 = 373.0187`
`= 373.02 K`
Now , depression in freezing point,
`Delta T_(f) = (K_(f) xx w_(B) xx 1000)/(M_(B) xx w_(A))`
`w_(B) = 0.052 g , w_(A) = 80.2 g , K_(f) = 1.86 K m^(-1) , M_(B) = 180`
` therefore " " Delta T_(f) = (1.86 xx 0.052 xx 1000)/(180 xx 80.2) = 0.067`
Freezing point of water = 273k .
Freezing point of solution = 273-0.067 = 272.933 K.