Q. In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionisation is 0.3 Taking $k_{f}$ for water as 1.85, the freezing point of the solution will be nearest to

A) $- 0.480℃$
B) $- 0.360℃$
C) $- 0.260℃$
D) $+ 0.480℃$
Explanation

(a)
Given, m=0.2
$k_{f} = 1.85\ \ \ \$
$\alpha = 0.3\ \$
$\therefore$ i=1+$\alpha$=1.3
$\mathrm{\Delta}T_{f} = molality\ \times k_{f} \times i$
=0.2$\times 1.85 \times 1.3$
=0.481${^\circ}$
$\therefore$ freezing point =-0.481$℃$

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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