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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 0.0198^oC. 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 $60 g {mol}^{- 1}$
$K_{f} (H_{2} O) = 1.85 {Kkgmol}^{- 1}$ ]

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answer is 5.

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Detailed Solution

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 $∆$ T_f = i × K_f × M
i = 1 + α for acetic acid
0.0198 = (1 + α) × 1.85 × 0.0102
α = 0.04928
$≅ 5 %$

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