Q. What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised? ($k_{f}\ for\ \ water = 1.86\ K\ mol^{- 1}$)
A) 0.85$℃$
B) -3.53$℃$
C) 0$℃$
D) -0.35$℃\$
Explanation
(b)
$\mathrm{\Delta}T_{f} = i \times k_{f} \times m$
i for HBr=1 +$\alpha$
where, $\alpha$ =degree of dissociation
i=1+0.9=1.9
$\therefore\ \ \ \ \ \ \ \mathrm{\Delta}T_{f} =$ 1.9 $\times$1.86$\times \frac{8.1 \times 1000}{100 \times 81}$
=3.534$℃$
Freezing point =-3.534$℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.