Question

Dissolution of $1.5g$ of non-volatile solute (Molecular weight = 60) in $250g$ 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,

$\Delta T_f=k_f\times m$

Where,

$m=Molality=\frac{Weightofsolute\times 1000}{Molecualrweightofsolute\times Weightofsolvent}$

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

$\Delta T_f$ $=0.1^{o}Cmola{l}^{-1}$

$\Rightarrow \Delta T_f=k_f\times 0.1$

$0.01=k_f\times 0.1$

$\therefore k_f=\frac{0.01}{0.1}$

$=0.1$