An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 `"g mol"^(-1)`) in 600 g of water. Calculate the freezing point of the solution. (`K_(f)` for water `=1.86" K kg mol"^(-1)`)

Explanation

Given : `w_(1)=600" g,"w_(2)=31" g,"M_(2)=62" g mol"^(-1)`
Molality of the solution `=(w_(2)//M_(2))/(w_(1)//1000)=(31//62)/(600//1000)=(31)/(62)xx(1000)/(600)=5/6m`
Apply the relation :
`DeltaT_(f)=K_(f)m`
`=1.86" K kg mol"^(-1)xx5/6" mol kg"^(-1)`
`=1.86xx5/6=1.55`
Freezing point of the solution `=-1.55^(@)C` or `271.45K`