JEE Main 2026 — Apr 2, Shift 1 — Chemistry solution

JEE Main2026Apr 2, Shift 1Chemistry

$19.5$ g of fluoro acetic acid (molar mass $= 78$ g mol $^{- 1}$ ) is dissolved in $500$ g of water at $298$ K. The depression in the freezing point of water was $1^{\circ}$ C. What is $K_{a}$ of fluoro acetic acid ?

(For water, $K_{f} = 1.86$ K kg mol $^{- 1}$ ). Assume molarity and molality to have same values.

A
$1 0^{- 6}$
B
$4 \times 1 0^{- 4}$
C
$3 \times 1 0^{- 5}$
D
$3 \times 1 0^{- 3}$

Solution

Moles of acid $= 19.5/78 = 0.25$ mol; molality $m = 0.25/0.5 = 0.5$ mol/kg.

From $Δ T_{f} = i K_{f} m$ : $1 = i \times 1.86 \times 0.5 \Rightarrow i = 2/1.86$ .

For a monoprotic weak acid (HA $⇌$ H $^{+}$ + A $^{-}$ , $n = 2$ ):

$i = 1 + α \Rightarrow α = i - 1 = \frac{2}{1.86} - 1 \approx 0.075$

Using $K_{a} = \frac{C α ^{2}}{1 - α}$ with $C = 0.5$ M:

$K_{a} = \frac{0.5 \times ( 0.075 ) ^{2}}{1 - 0.075} = \frac{0.5 \times 0.005625}{0.925} \approx 3 \times 1 0^{- 3}$

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