Question

Dissolution of $1.5\, g$ of non-volatile solute (Molecular weight = 60) in $250\, g$ of a solvent reduces its freezing point by $0.01^{o}C$ . Find the molal depression constant of the solvent.

• A. 0.01
• B. 0.001
• C. 0.0001
• D. 0.1

Answer 1

Answer: (D)

Depression in freezing point,

$ΔT_{f}=k_{f}\times m$

Where,

$m=Molality=\frac{Weight \, of \, solute \times 1000}{Molecualr \, weight \, of \, solute \, \times Weight \, of \, solvent}$

$=\frac{1.5 \times 1000}{60 \times 250}$

$ΔT_{f}$ $=0.1^{o}C \, molal^{- 1}$

$\Rightarrow ΔT_{f}=k_{f}\times 0.1$

$0.01=k_{f}\times 0.1$

$\therefore \, \, k_{f}=\frac{0.01}{0.1}$

$=0.1$