Question

Calculate the molal depression constant of a solvent which has freezing point of ${27}^{\circ }C$ and latent heat of fusion of $180J{g}^{-1}$

• A. $5.04Kgmo{l}^{-1}$
• B. $4.16Kgmo{l}^{-1}$
• C. $2.56Kgmo{l}^{-1}$
• D. $3.34Kkgmo{l}^{-1}$

Answer 1

The correct option is B $4.16Kgmo{l}^{-1}$

We know,
Freezing point depression constant,
$K_f=\frac{R{T}_f^2}{1000\times L_f}\text{where,}R=8.314J{K}^{-1}mo{l}^{-1}\text{Freezing point of pure solvent,}{T}_f={27}^{\circ }C=273+27=300K,\text{Latent heat of fusion,}{L}_f=180J{g}^{-1}$
Substituting the values in the above equation,
$K_f=\frac{8.314\times (300{)}^2}{1000\times 180}=4.16Kgmo{l}^{-1}$