Verified Solution • 2025–26 Session moderate

QJEEM22Ch14ab: JEE Main 2022 Chemistry - Class 12 JEEM22Ch14ab Solution: 1.2 mL of acetic acid is dissolved in water to make 2.0 L of…

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 ∘^\circ 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 g mol −- 1, Kf(H2O) = 1.85 K kg mol −- 1]

Step-by-Step Verified Solution

ΔTb=i×Kb×m\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}

Moles of solute (( acetic acid )=1.2×1.0260)=\frac{1.2 \times 1.02}{60}

As moles of solute are very less.

So, take molarity and molality the same.

0.0198=i×1.85×1.2×1.0260×20.0198=\mathrm{i} \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}

i=1.05\mathrm{i}=1.05

\alpha=\frac{i-1}{n-1}=\frac{0.05}{1}= 0.05 = 5\text{%}

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