JEE MAIN - Chemistry (2022 - 29th June Morning Shift - No. 12)
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]
Answer
5
Explanation
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}$
Moles of solute $($ acetic acid $)=\frac{1.2 \times 1.02}{60}$
As moles of solute are very less.
So, take molarity and molality the same.
$0.0198=\mathrm{i} \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}$
$\mathrm{i}=1.05$
$\alpha=\frac{i-1}{n-1}=\frac{0.05}{1}= 0.05 = 5\text{%}$
Moles of solute $($ acetic acid $)=\frac{1.2 \times 1.02}{60}$
As moles of solute are very less.
So, take molarity and molality the same.
$0.0198=\mathrm{i} \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}$
$\mathrm{i}=1.05$
$\alpha=\frac{i-1}{n-1}=\frac{0.05}{1}= 0.05 = 5\text{%}$
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