Question

When $9.45g$ of $ClCH_{2}COOH$ is added to $500mL$ 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)
$\left[K_{f \left(H_{2} O\right)} = 1 . 86 K kg \left(mol\right)^{- 1}\right]$

Answer 1

Total no. of moles $=C+Cα=C\left(\right.1+\alpha \left.\right)$
$i=\frac{observed C . P .}{calculated C . P . }=\frac{C \left(\right. 1 + \alpha \left.\right)}{C}=\left(1 + \alpha \right)$
$M.W.=94.5$
$ΔT_{f}=i\times k_{f}\times mΔT_{f}=0.5^\circ C$
$i=1+\alpha $
$0.5=\left(1 + \alpha \right)\times 1.86\times \frac{\frac{9 . 45}{94 . 5}}{\frac{500}{1000}}$
$m=\frac{mole}{k . g \left(\right. Solvent \left.\right)}$
$k_{t}=1.86kkg/mol$
$\left(\right.1+\alpha \left.\right)=\frac{2 . 5}{1 . 86}$
$\alpha =\frac{0 . 64}{1 . 86}=\frac{32}{93}$
$C=\frac{9 . 45}{94 . 5 \times 0 . 5}=0.2M$
$K_{a}=\frac{Cα^{2}}{1 - \alpha }=\frac{0 . 2 \times 1024}{93 \times 93 \times \frac{61}{93}}$
$K_{a}=0.0351=35.1\times 10^{- 3}$