In which case depression freezing point is equal to cryoscopic constant for water:

Question

In which case depression freezing point is equal to cryoscopic constant for water: (A) 6 % by mass of urea in aqueous solution (B) 100 g of sucrose in 100 mL solution (C) 9 g of urea in 59 g aqueous solution (D) 1 M KCl solution

Tardigrade Admin · Accepted Answer

∵ Δ T=Kf × molality 59 g solution has 9 g glucose and 50 g water ∴ Δ T=Kf ×(9/(180×50)1000)=Kf

Q. In which case depression freezing point is equal to cryoscopic constant for water:

6 % by mass of urea in aqueous solution

100 g of sucrose in 100 mL solution

9 g of urea in 59 g aqueous solution

1 M KCl solution

$∵ Δ T = K_{f} \times$ molality
59 g solution has 9 g glucose and 50 g water
$∴ Δ T = K_{f} \times \frac{9}{\frac{180 \times 50}{1000}} = K_{f}$

Q. In which case depression freezing point is equal to cryoscopic constant for water:

6 % by mass of urea in aqueous solution

100 g of sucrose in 100 mL solution

9 g of urea in 59 g aqueous solution

1 M KCl solution

∵ΔT=Kf​× molality 59 g solution has 9 g glucose and 50 g water ∴ΔT=Kf​×1000180×50​9​=Kf​

$∵ Δ T = K_{f} \times$ molality
59 g solution has 9 g glucose and 50 g water
$∴ Δ T = K_{f} \times \frac{9}{\frac{180 \times 50}{1000}} = K_{f}$