Ncert Exercises & Solutions

2.32 Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. Ka = 1.4 × 10^–3, K_f = 1.86 K kg mol^–1.

NEETprep Answer:

Molar mass of ${CH}_{3} {CH}_{2} CHClCOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1 = 122.5 g {mol}^{- 1}$

$∴$ No. of moles present in 10 g of ${CH}_{3} {CH}_{2} CHClCOOH = \frac{10 g}{122.5 g {mol}^{- 1}}$ ${CH}_{3} {CH}_{2} CHCICOOH = \frac{10 g}{122.5 g {mol}^{- 1}}$

$= 0.0816 mol$

It is given that 10 g of is added to 250 g of water.

Molality of the solution,

Let α be the degree of dissociation of CH₃CH₂CHClCOOH.

CH₃CH₂CHClCOOH undergoes dissociation according to the following equation:

${CH}_{3} {CH}_{2} CHCICOOH \leftrightarrow {CH}_{3} {CH}_{2} {CHCICOOH}^{-} + H^{+}$
$Initial conc . C mol L^{- 1} 0 0$
$At equlibrium C (1 - a) Ca Ca$
$∴ K_{a} = \frac{Ca . Ca}{C (1 - a)}$
$= \frac{{Ca}^{2}}{1 - a}$

Since α is very small with respect to 1, 1 − α ≈ 1

Now, $K_{a} = \frac{{Ca}^{2}}{1}$

$\Rightarrow K_{a} = {Cα}^{2}$
$\Rightarrow a = \frac{K_{a}}{C}$
$= \frac{1.4 \times 10^{- 3}}{0.3264} (∴ K_{a} = 1.4 \times 10^{- 3})$
$= 0.0655$

Again,

${CH}_{3} {CH}_{2} CHClCOOH \leftrightarrow {CH}_{3} {CH}_{2} {CHClCOOH}^{-} + H^{+}$
$Initial conc . 1 0 0$
$At equilibrium 1 - α α α$

Total moles of equilibrium = 1 − α + α + α = 1 + α

$∴ i = \frac{1 + α}{1}$
$= 1 + α$
$= 1 + 0.0655$
$= 1.0655$

Hence, the depression in the freezing point of water is given as:

$∆ T_{f} = i . K_{f} m$
$= 1.0665 \times 1.86 K kg {mol}^{- 1} \times 0.3264 mol {kg}^{- 1}$
$= 0.65 K$

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