NEETChemistrySolutionsMCQ+4 / −1
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86oC/m, the freezing point of the solution will be
- A0.18oC
- B0.54oC
- C0.36oC
- D0.24oC
View written solutionFree
Correct answer: D
We know that Tf
= i × Kf × m
Here i is van’t Hoff’s factor.
i for weak acid is 1 + .
Here is degree of dissociation i.e., 30/100 = 0.3
i = 1 + = 1 + 0.3 = 1.3
Tf
= i × Kf
× m = 1.3 × 1.86 × 0.1 = 0.24
Freezing point = – 0.24
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