Q. 0.1 molal aqueous solution of $NaBr$ freezes at $- 0.335℃$ at atmospheric pressure $k_{f}$ for water is $1.86℃$. The percentage of dissociation of the salt in solution is

A) 90
B) 80
C) 58
D) 98
Explanation

(b)
$\mathrm{\Delta}T_{f} = ik_{f}m$
$0.335 = i \times 1.86 \times 0.1$
$i = 1.80$
For $NaBr,\ i = 1 + \alpha$
$\alpha = 0.80 = 80\%$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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