Q. The freezing point of the 0.05 molal solution of non-electrolyte in water is
A) $- 0.093℃$
B) $1.86℃$
C) $0.93℃$
D) $0.093℃$
Explanation
(a)
For non-electrolyte
$\mathrm{\Delta}T_{f} = k_{f} \times m$
Given, m =0.05,
$\therefore\ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}T_{f} = 1.86 \times 0.05 = 0.093℃$
Freezing point of solution
$k_{f} = 1.86 = 0 - \mathrm{\Delta}T_{f}$
=0-0.093=-0.093$℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.