Question

When 9.45 g of $ClC{H}_{2}COOH$ is added to 500 mL of water , its freezing point drops by ${0.5}^{0}C$ . The dissociation constant of $ClC{H}_{2}COOH$ is $x\times {10}^{-3}$ . The value of x is ( Rounded off to the nearest integer ) $[K_{f\left(H_{2}O\right)}=1.86KKgmo{l}^{-1}]$
(JEE MAIN 2021)

Answer 1

Moles of $ClC{H}_{2}COOH=\frac{9.45}{94.5}=0.1moles$ $ClC{H}_{2}COOH\rightleftharpoons ClC{H}_{2}CO{O}^{-}+H^{+}$ $Ateqb.C(1-\alpha )C\alpha C\alpha$ $\text{vant't Hoff factor}\left(i\right)=1+\alpha$ $C=\frac{\text{moles}}{\text{volume in L}}=\frac{0.1}{0.5}=0.2M$ Assuming molarity = molality $\Delta T_f=i\times K_f\times m=(1+\alpha )\times 1.86\times 0.2$ $0.5=(1+\alpha )\times 1.86\times 0.2$ $\Rightarrow (1+\alpha )=\frac{0.5}{0.2\times 1.86}1+\alpha =1.34$ $\Rightarrow \alpha =0.34$ $K_{a}of\left(ClC{H}_{2}COOH\right)=\frac{C\alpha \cdot C\alpha }{C(1-\alpha )}$ $K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.2\times {\left(0.34\right)}^2}{1-0.34}$ $K_a=35\times {10}^{-3}x=35$