When 9 . 45g of ClCH _ 2 COOH is added to 500mL of water, its...

${ClCH}_{2}COOH ⇌ {ClCH}_{2}{COO}^{⊖} + {H}^{+}$

$i = 1 + ( 2 - 1 ) \alpha$

$i = 1 + \alpha$

${\Delta T}_{f} = i . {K}_{f} . m$

$0 . 5 = ( 1 + \alpha ) ( 1 . 86 ) \left(\dfrac{\left(\dfrac{9 . 45}{94 . 5}\right)}{\left(\dfrac{500}{1000}\right)}\right)$

$\dfrac{5}{3 . 72} = 1 + \alpha ⇒ \alpha = \dfrac{1 . 28}{3 . 72}$

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

${K}_{a} = \dfrac{{\left(C\alpha\right)}^{2}}{C - C\alpha} = \dfrac{C{\alpha}^{2}}{1 - \alpha}C = \dfrac{0 .1}{500 / 1000} = 0 .2$

${K}_{a} = \dfrac{0 .2 {\left(32 / 93\right)}^{2}}{\left(1 - 32 / 93\right)} = \dfrac{0 .2 \times {\left(32\right)}^{2}}{93 \times 61}$

$= 0 .036$

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