JEE Main 2021 — Solutions Question with Solution
From: JEE Main 2021 (Online) 24th February Morning Shift
Question
When 9.45 g of CICH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH2COOH is x 10-3.
The value of x is ________. (Rounded off to the nearest integer)
[Kf(H20) = 1.86 K kg mol-1]
The value of x is ________. (Rounded off to the nearest integer)
[Kf(H20) = 1.86 K kg mol-1]
Enter your answer
Show full solutionCorrect answer: 34.4
Correct answer
34.4
Step-by-step explanation
Here,
i = 1 + + i = 1 +
Formula used for freezing point; Tf = i Kfm
Here, Kf = freezing constant of H2O
Kf(H2O) = 1.86 K kg mol1
m = molarity
Tf = i Kfm
0.5 = (1 + ) (1.86)
= 1 + 1 =
Percentage of dissociation = 34.4%
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