JEE Main 2021 — Solutions Question with Solution
From: JEE Main 2021 (Online) 18th March Morning Shift
Question
2 molal solution of a weak acid HA has a freezing point of 3.885C. The degree of dissociation of this acid is ___________ 103. (Round off to the Nearest Integer).
[Given : Molal depression constant of water = 1.85 K kg mol1 Freezing point of pure water = 0 C]
[Given : Molal depression constant of water = 1.85 K kg mol1 Freezing point of pure water = 0 C]
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Show full solutionCorrect answer: 50
Correct answer
50
Step-by-step explanation
Tf = Kf (im)
3.885 = i 1.85 2
i = 1.05
Also, we know,
i = 1 + (n 1)
here n = number of particle obtained upon the dissociation of one particle.
here from one particle HA we get two particle H+ and A.
n = 2
So, i = 1 + (2 1)
1.05 = 1 +
= 0.05 = 50 103
3.885 = i 1.85 2
i = 1.05
Also, we know,
i = 1 + (n 1)
here n = number of particle obtained upon the dissociation of one particle.
here from one particle HA we get two particle H+ and A.
n = 2
So, i = 1 + (2 1)
1.05 = 1 +
= 0.05 = 50 103
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