When 9.45 g of $\mathrm{ClCH}_{2}\mathrm{COOH}$ is added to 500 mL of water, its freezing point drops by $0.5^{\circ}\mathrm{C}$. The dissociation constant of $\mathrm{ClCH}_{2}\mathrm{COOH}$ is $x \times 10^{-3}$. The value of x is ____. (Rounded off to the nearest integer) \[\left[\mathrm{K}_{\mathrm{f}(\mathrm{H}_{2}\mathrm{O})}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]\]

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Solutions

When 9.45 g of

ClCH_{2} COOH

is added to 500 mL of water, its freezing point drops by

0. 5^{\circ} C

. The dissociation constant of

ClCH_{2} COOH

x \times 1 0^{- 3}

. The value of x is ____. (Rounded off to the nearest integer)

[K_{f (H_{2} O)} = 1.86 K kg mol^{- 1}]

Answer: 36

Chloroacetic acid (

ClCH_{2} COOH

) is a weak acid that partially dissociates:

ClCH_{2} COOH ⇌ ClCH_{2} COO^{-} + H^{+}

Step 1: Calculate molality

Step 2: Calculate van't Hoff factor (i)

Δ T_{f} = i \cdot K_{f} \cdot m

0.5 = i \times 1.86 \times 0.2

i = \frac{0.5}{0.372} = 1.344

Step 3: Relate i to degree of dissociation (α)
For a weak acid:

i = 1 + α

α = i - 1 = 1.344 - 1 = 0.344

Step 4: Calculate dissociation constant

K_{a} = \frac{c α ^{2}}{1 - α}

where c = 0.2 mol/kg

K_{a} = \frac{0.2 \times ( 0.344 ) ^{2}}{1 - 0.344} = \frac{0.2 \times 0.1183}{0.656} = \frac{0.02366}{0.656} = 0.0361

K_{a} \approx 36 \times 1 0^{- 3}

Rounding to nearest integer:

x = 35

Answer: 35

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer