When 9.45 text{ g} of ClCH_2COOH is added to | vedclass

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(C) Molar mass of $ClCH_2COOH = 94.5 	ext{ g mol}^{-1}$.Molality $(m) = \frac{9.45 	ext{ g}}{94.5 	ext{ g mol}^{-1} 	imes 0.5 	ext{ kg}} = 0.2

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) Molar mass of $ClCH_2COOH = 94.5 	ext{ g mol}^{-1}$.Molality $(m) = \frac{9.45 	ext{ g}}{94.5 	ext{ g mol}^{-1} 	imes 0.5 	ext{ kg}} = 0.2 	ext{ m}$.Using the formula $\Delta T_f = i \cdot K_f \cdot m$:$0.5 = i 	imes 1.86 	imes 0.2$.$i = \frac{0.5}{0.372} \approx 1.344$.For the dissociation $ClCH_2COOH ightleftharpoons ClCH_2COO^- + H^+$,the van't Hoff factor is $i = 1 + \alpha$.$\alpha = i - 1 = 0.344$.The dissociation constant $K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.2 	imes (0.344)^2}{1 - 0.344} = \frac{0.2 	imes 0.118336}{0.656} \approx 0.0361 = 36.1 	imes 10^{-3}$.Thus,$x \approx 36$.

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