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Question: Answered & Verified by Expert
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198 °C. The percentage of dissociation of the acid is (Nearest integer) [Given : Density of acetic acid is 1.02 g mL-1 Molar mass of acetic acid is 60 gmol-1
KfH2O=1.85 K kg mol-1]
ChemistrySolutionsJEE MainJEE Main 2022 (29 Jun Shift 1)
Solution:
1323 Upvotes Verified Answer
The correct answer is: 5

M=d×V=1.02×1.2=1.224gm

Moles of acetic acid =0.0204 moles in 2 L

So molality =0.0102 mol/kg

Now ΔTf=i×Kf×m

i=1+α for acetic acid

0.0198=1+α×1.85×0.0102

α=0.04928

5%

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