Considering acetic acid dissociates in water, its dissociation constant is \(6.25 \times 10^{-5}\). If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at \(-\mathrm{x} \times 10^{-2}{ }^{\circ} \mathrm{C}\), provided pure water freezes at \(0^{\circ} \mathrm{C}\). \(\mathrm{x}=\) ______. (Nearest integer) Given: \((\mathrm{K}_{\mathrm{f}})_{\text{water}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). density of acetic acid is \(1.2 \mathrm{~g} \mathrm{~mol}^{-1}\) molar mass of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\). molar mass of acetic acid \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\). density of water \(=1 \mathrm{~g} \mathrm{~cm}^{-3}\) Acetic acid dissociates as \[\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{\Theta}+\mathrm{H}^{\oplus}\]

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Solutions

Considering acetic acid dissociates in water, its dissociation constant is

6.25 \times 1 0^{- 5}

. If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at

- x \times 1 0^{- 2}^{\circ} C

, provided pure water freezes at

0^{\circ} C

x =

______. (Nearest integer)
Given:

(K_{f})_{water} = 1.86 K kg mol^{- 1}

.
density of acetic acid is

1.2 g mol^{- 1}

molar mass of water

= 18 g mol^{- 1}

.
molar mass of acetic acid

= 60 g mol^{- 1}

.
density of water

= 1 g cm^{- 3}

Acetic acid dissociates as

CH_{3} COOH ⇌ CH_{3} COO^{Θ} + H^{\oplus}

Answer: 19

Mass of acetic acid

= 5 mL \times 1.2 g/mL = 6

g, so

n = \frac{6}{60} = 0.1

mol in 1 kg of water (molality

= 0.1

).
Degree of dissociation from

K_{a} = C α^{2}

α = \frac{K _{a}}{C} = \frac{6.25 \times 1 0 ^{- 5}}{0.1} = 6.25 \times 1 0^{- 4} = 0.025

Van't Hoff factor for dissociation into 2 ions:

i = 1 + α = 1.025

Δ T_{f} = i K_{f} m = 1.025 \times 1.86 \times 0.1 \approx 0.19^{\circ} C = 19 \times 1 0^{- 2}^{\circ} C

x = 19

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Integer answer