Question

If latent heat of fusion of ice is $80\text{cal}$ per mole at $0{}^{\circ }C$, calculate molal depression constant for water.

• A. $0.1863{\text{K kg mol}}^{-1}$
• B. $1.863{\text{K kg mol}}^{-1}$
• C. $186.3{\text{K kg mol}}^{-1}$
• D. $18.63{\text{K kg mol}}^{-1}$

Answer 1

The correct option is B $1.863{\text{K kg mol}}^{-1}$

We know that,
$K_f=\frac{R{T}_f^2}{1000\times L_f}$
Here
$K_f=\text{molal depression constant}$
$\text{Gas constant}\left(R\right)=2{\text{Cal K}}^{-1}{\text{mol}}^{-1}$
$\text{Freezing point of water},T_f=0{}^{\circ }C=273\text{K}$
$\text{Latent heat of fusion},L_f=80{\text{cal mol}}^{-1}$
$\therefore K_f=\frac{2\times 273\times 273}{1000\times 80}=1.863{\text{K kg mol}}^{-1}$