JEE MAIN - Chemistry (2019 - 11th January Evening Slot - No. 19)
K2Hgl4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:
1.6
2.2
2.0
1.8
Explanation
K2Hgl4 is 40% ionised.
$$ \therefore $$ $$\alpha $$ = $${{40} \over {100}}$$ = 0.4
K2[Hgl4] $$ \to $$ 2K+ + [Hgl4]2+
N = $${{2 + 1} \over 1}$$ = 3
i = 1 + (N - 1)$$\alpha $$
= 1 + (3 - 1)0.4
= 1 + 2$$ \times $$0.4
= 1.8
$$ \therefore $$ $$\alpha $$ = $${{40} \over {100}}$$ = 0.4
K2[Hgl4] $$ \to $$ 2K+ + [Hgl4]2+
N = $${{2 + 1} \over 1}$$ = 3
i = 1 + (N - 1)$$\alpha $$
= 1 + (3 - 1)0.4
= 1 + 2$$ \times $$0.4
= 1.8
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