0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol-1, the lowering in freezing point of the solution is
- 0.56 K
- 1.12 K
- -0.56 K
- -1.12 K
Answer
(2) 1.12 K
Solution
HX H+ + X-
Degree of ionisation = 0.20, number of ions n = 2
⇒ i = 1 + (n-1) = 1 + 0.20 = 1.20
Tf = i Kf m
= 1.20 1.86 0.5
= 1.116 K 1.12 K
∴ The lowering (a positive quantity) is 1.12 K.
Answer: (2)