`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

Here, we are given `w_(2)=45g, w_(1)=600g, K_(f)="1.86 K kg mol"^(-1)`
`M_(2)"for solute "C_(2)H_(6)O_(2)=24+6+32="62 g mol"^(-1)`
Substituting these value in the formula, `DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2)),` we get
`DeltaT_(f)=("1000 g kg"^(-1)xx"1.86 K kg mol"^(-1)xx45g)/(600 gxx"62 g mol"^(-1))="2.25 K"`
Freezing point of pure water = 273.15 K
`therefore" "` Freezing point of the solution `=T_(f)^(@)-DeltaT_(f)=273.15-2.25K=270.9K`