An aqueous solution contains `10%` by weight by urea `(60.00)`and `5%` by weight of glucose `(180.00)`. What will be its freezing point ? `K_(f)` for water is `1.86`.

Answer

`-3.03^(@)C`

Explanation

`"Mass of solution = 100 g , Mass of urea = 5 g"`
`"Mass of glucose = 10 "therefore " Mass of water "=100-(5+10)=85h`
`"Molality of urea in the solution "=(5)/(60)xx(1000)/(85)=0.98m`
`"Molality of glucose in the solution "=(10)/(180)xx(1000)/(85)=0.65m`
Total `DeltaT_(f)=K_(f)(m_(1)+m_(2))=1.86(0.98+0.65)=3.03^(@)C`
`therefore" Freezing point of the solution "=-3.03^(@)C.`