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Q.

1.2 mLof 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 0.0198oC. The percentage of dissociation of the acid is (Nearest integer)

[Given: Density of acetic acid is 1.02gmL1
Molar mass of acetic acid is 60gmol1
KfH2O=1.85Kkgmol1]

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answer is 5.

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Detailed Solution

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M = d × V = 1.02 × 1.2 = 1.224 gm
Moles of acetic acid = 0.0204 moles in 2L
So molality = 0.0102 mol/kg
Now Tf = 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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