60 g of urea is dissolved in 1100 g solution. To keep Δ Tf/Kf as 1 mol kg-1 , water separated in the form of ice is

Question

60 g of urea is dissolved in 1100 g solution. To keep Δ Tf/Kf as 1 mol kg-1 , water separated in the form of ice is (A) 40 g (B) 60 g (C) 100 g (D) 200 g

Tardigrade Admin · Accepted Answer

1100 g solution has 60 g urea Water =1040 g Δ Tf=(1000Kfw2/M2w1) (Δ Tf/Kf)=(1000× 60/60× w1)=1 or w1=1000 g The ice formed =1040-1000=40 text g

Q. 60 g of urea is dissolved in 1100 g solution. To keep $Δ T_{f} / K_{f}$ as 1 mol $k g^{- 1}$ , water separated in the form of ice is

40 g

60 g

100 g

200 g

1100 g solution has 60 g urea Water =1040 g $Δ T_{f} = \frac{1000 K _{f} w _{2}}{M _{2} w _{1}}$ $\frac{Δ T _{f}}{K _{f}} = \frac{1000 \times 60}{60 \times w _{1}} = 1$ or $w_{1} = 1000 g$ The ice formed $= 1040 - 1000 = 40 g$

Q. 60 g of urea is dissolved in 1100 g solution. To keep ΔTf​/Kf​ as 1 mol kg−1 , water separated in the form of ice is