Q. The molecular weight of $NaCl$ determined by studying freezing point depression of its 0.5% aqueous solution is 30. The apparent degree of dissociation of NaCl is
A) 0.60
B) 0.50
C) 0.30
D) 0.95
Explanation
(d)
$i = 1 + \alpha$
$= \frac{cal.mol.weight}{exp.\ mol.\ wt} = \frac{58.5}{30} = 1.95$
$\therefore 1 + \alpha = 1.95$
$\alpha = 0.95$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.