`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

Moles of ethylene glycol `=(45g)/(62g " mol"^(-))=0.73 " mol"`.
Moles of water in kg `=(600gm)/(1000g kg^(-1))=0.6kg`
Molality,
`m=("Moles of ethylene glycol")/("Mass of water in kg" )=(0.73 mol )/(0.60kg) =1.2"mol" kg ^(-1)`
(i) Therefore, freezing point depression,
`Delta T_p=k_(F)m`
`=1.86Kkg mol^(-1)xx1.2molkg^(-1)=2.2K`
(ii) Freezing point of the aqueous solution.
`=T_(F)^(@) -DeltaT_(F) =273.15K-2.2K=270.95K`