19.5 g of fluoro acetic acid (molar mass $=78 \mathrm{~g} \mathrm{~mol}^{-1}$) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was $1^{\circ} \mathrm{C}$. What is $\mathrm{K}_{\mathrm{a}}$ of fluoro acetic acid ? (For water, $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$). Assume molarity and molality to have same values.

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Solutions

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) $3 \times 1 0^{- 3}$ ✓ Correct

Answer: (D)

Solution

Moles of acid

= \frac{19.5}{78} = 0.25

, so the molality is

\frac{0.25}{0.5} = 0.5 m

. From

Δ T_{f} = i K_{f} m

1 = i \times 1.86 \times 0.5

, giving

i = 1.075

. For a weak monobasic acid

i = 1 + α

, so the degree of dissociation is

α = 0.075

. Then

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

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option

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