JEE Main 2019 — Solutions Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
K2Hgl4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:
Choose an option
Show full solutionCorrect option: D
Correct answer
D1.8
Step-by-step explanation
K2Hgl4 is 40% ionised.
= = 0.4
K2[Hgl4] 2K+ + [Hgl4]2+
N = = 3
i = 1 + (N - 1)
= 1 + (3 - 1)0.4
= 1 + 20.4
= 1.8
= = 0.4
K2[Hgl4] 2K+ + [Hgl4]2+
N = = 3
i = 1 + (N - 1)
= 1 + (3 - 1)0.4
= 1 + 20.4
= 1.8
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