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QJEEM21Chbb7e: JEE Main 2021 Chemistry - Class 12 JEEM21Chbb7e Solution: When 9.45 g of CICH 2 COOH is added to 500 mL of water, its…

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]

Step-by-Step Verified Solution

JEE Main 2021 (Online) 24th February Morning Shift Chemistry - Solutions Question 95 English Explanation
Here, i=Final molesInitial molesi = {{Final\,moles} \over {Initial\,moles}}

i = 1 −- α\alpha + α\alpha + α\alpha ⇒\Rightarrow i = 1 + α\alpha

Formula used for freezing point; Δ\Delta Tf = i Kfm

Here, Kf = freezing constant of H2O

Kf(H2O) = 1.86 K kg mol −- 1

m = molarity

Δ\Delta Tf = i Kfm

0.5 = (1 + α\alpha ) (1.86) (9.45/94.5)(500/1000){{(9.45/94.5)} \over {(500/1000)}}

53.72{5 \over {3.72}} = 1 + α\alpha ⇒\Rightarrow 53.72{5 \over {3.72}} −- 1 = α\alpha

⇒\Rightarrow α\alpha =1.283.72=3293=0.344= {{1.28} \over {3.72}} = {{32} \over {93}} = 0.344

Percentage of dissociation = 34.4%

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