One hundred gram of a 5 m urea solution are cooled to ` - 6^(@)C` .What amount of urea will seprate out ? `(K_(f) = 1.86 K m^(-1))` ?

Explanation

Molality of solution at ` - 6^(@)C`
`Delta T_(f) = k_(f) m`
or `" " m = (Delta T_(f))/(K_(f)) = (6)/(1.86) = 3.23 mol kg ^(-1)`
5 m solution of urea means that 5 moles or `5 xx 60` g of urea are present in 1000 g of solvent or `(1000+5xx6)`g of solution , so that .
`(100+300)` g of solution contain urea = 300 g .
100g of solution contain urea ` = (300)/(1300) xx 100`
` =23.08 g`
Mass of water in the solution `= 100- 23.08`
` = 76.92 g`
When urea separates out , mass of water does not change. At ` - 6^(@)C` , the molality is `3.23 mol kg^(-1)` . Let us calculate the amount of urea to make the molality `3.23 mol kg^(-1)` .
1000 g of water contain urea ` = 60 xx 3.23 g`
`76.92` g of water contain urea ` = (60 xx 3.23 xx 76.92)/(1000)`
` = 14.91 g`
Mass of urea separated ` = 23.08 - 14.91`
` = 8.17g`