Subjective Type

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 $$ClCH_{2} COOH$$ is $$x \times 10^{-3}$$. The value of $$x$$ is _________. (Rounded off to the nearest integer) $$[K_{f(H_{2}O)} = 1.86\ \text{kg mol}^{-1}] $$

Solution

$$ClCH_{2}COOH \rightleftharpoons ClCH_{2} COO^{\ominus} + H^{+} $$
$$i = 1 + (2 - 1) \alpha $$
$$ i = 1 + \alpha $$
$$\Delta T_{f} = iK_{f} m $$

$$0.5 = (1 + \alpha) (1.86) \left (\dfrac{\left (\dfrac{9.45}{94.5} \right )}{\left (\dfrac{500}{1000} \right )} \right ) $$

$$ \dfrac{5}{3.72} = 1 + \alpha \Rightarrow \alpha = \dfrac{1.28}{3.72} $$

$$ \alpha = \dfrac{32}{93} $$

$$ClCH_{2} COOH \rightleftharpoons ClCH_{2} COO^{\ominus} + H^{+} $$
$$C - C\alpha$$ $$C \alpha $$ $$C \alpha $$

$$K_{a} = \dfrac{(C \alpha)^{2}}{C - C \alpha} = \dfrac{C \alpha^{2}}{1 - \alpha} ; C = \dfrac{0.1}{500/1000} = 0.2 $$

$$K_{a} = \dfrac{0.2 (32/93)^{2}}{(1 - 32 / 93)} = \dfrac{0.2 \times (32)^{2}}{93 \times 61} = 0.036$$

$$K_{a} = 36 \times 10^{-3} $$

Answer is $$36$$

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