JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 18)
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 $$ \times $$ 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]
Answer
34.4
Explanation
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Here, $$i = {{Final\,moles} \over {Initial\,moles}}$$
i = 1 $$-$$ $$\alpha$$ + $$\alpha$$ + $$\alpha$$ $$\Rightarrow$$ i = 1 + $$\alpha$$
Formula used for freezing point; $$\Delta$$Tf = i Kfm
Here, Kf = freezing constant of H2O
Kf(H2O) = 1.86 K kg mol$$-$$1
m = molarity
$$\Delta$$Tf = i Kfm
0.5 = (1 + $$\alpha$$) (1.86)$${{(9.45/94.5)} \over {(500/1000)}}$$
$${5 \over {3.72}}$$ = 1 + $$\alpha$$ $$\Rightarrow$$ $${5 \over {3.72}}$$ $$-$$ 1 = $$\alpha$$
$$\Rightarrow$$ $$\alpha$$ $$ = {{1.28} \over {3.72}} = {{32} \over {93}} = 0.344$$
Percentage of dissociation = 34.4%
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