Q. Calculate the molal depression constant of a solvent which has freezing point $16.6℃$ and latent heat of fusion 180.75 $Jg^{- 1}$.
A) 2.68
B) 3.86
C) 4.68
D) 2.86
Explanation
(b)
Given,
R=8.314 J$K^{- 1}\ \ mol^{- 1}$
$T_{f} = 273 + 16.6 = 289.6\ K$
$L_{f}\ = 180.75\ \ Jg^{- 1}$
$\ \ \ \ k_{f} = ?$
$k_{f} = \frac{R.T_{f}^{2}}{1000 \times L_{f}}$
$= \frac{8.314 \times (289.6)^{2}}{1000 \times 180.75}$
$k_{f} = 3.86$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.