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Depression in Freezing Point

1.2 mL of ace...

Question

$1.2 m L$ 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 ${0.0198}^{o} C$ . The percentage of dissociation of the acid is (Nearest integer)
[Given: Density of acetic acid is $1.02 g m L^{- 1}$ , Molar mass of acetic acid is $60 g {mol}^{- 1}$ , $K_{f} (H_{2} O) = 1.85 K k g {mol}^{- 1}$ ]

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Solution

$Δ T_{f} = i \times K_{f} \times m$
Moles of solute (acetic acid) $= \frac{1.2 \times 1.02}{60}$
As moles of solute are very less.
So, take molarity and molality same
$0.0198 = 1 \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}$
$i = 1.05$
$α = \frac{i - 1}{n - 1} = \frac{0.05}{1} = 0.05$

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