`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

(i) `Delta T_(b) = (K_(f) xx w_(B) xx 1000)/(w_(A) xx M_(B))`
`w_(A) = 600 g , w_(B) = 45 g, K_(f) = 1.86 Kg mol^(-1)`
`M_(B) = 2 xx 12 + 6 xx 1 + 2 xx 16 = 62 g mol^(-1)`
`therefore " " Delta T_(f) = (1.86 xx 45 xx 1000)/(600 xx 62)`
`" " = 2.25 K`
` therefore " "` Freezing point depression = 2.25 k .
(ii) Freezing point of aqueous solution `= 273.1 - 2.25`
` = 270.90` K