When a solution containing w g of urea in 1 kg of water is cooled to -0.372° C , 200 g of ice is separated. If Kf for water is 1.86 K ⋅ kg / mol , w is

Question

When a solution containing w g of urea in 1 kg of water is cooled to -0.372° C , 200 g of ice is separated. If Kf for water is 1.86 K ⋅ kg / mol , w is (A) 4.8 g (B) 12.0 g (C) 9.6 g (D) 6.0 g

Tardigrade Admin · Accepted Answer

Mass of water present in the solution at the freezing pt. =1000 -200=800 g Δ Tf =0.372=(1000 × kf w/60 × 800) =(1000 × 1.86 × w/60 × 800) or w =9.6 g .

Q. When a solution containing $w g$ of urea in $1 k g$ of water is cooled to $- 0.37 2^{\circ} C, 200 g$ of ice is separated. If $K_{f}$ for water is $1.86 K \cdot k g / m o l, w$ is

4.8 g

12.0 g

9.6 g

6.0 g

Mass of water present in the solution at the freezing pt. $= 1000$

$- 200 = 800 g$

$Δ T_{f} = 0.372 = \frac{1000 \times k _{f} w}{60 \times 800}$

$= \frac{1000 \times 1.86 \times w}{60 \times 800}$

or $w = 9.6 g .$

Q. When a solution containing wg of urea in 1kg of water is cooled to −0.372∘C,200g of ice is separated. If Kf​ for water is 1.86K⋅kg/mol,w is