Question

When $9.45\text{g}$ of $ClC{H}_{2}COOH$ is added to $500\text{mL}$ 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 )$[K_{f\left(H_{2}O\right)}=1.86\text{Kg}mo{l}^{-1}]$

• A. 35.0
• B. 35
• C. 35.00

Answer 1

Answer: (A), (B), (C)

Moles of $ClC{H}_{2}COOH=\frac{9.45}{94.5}=0.1\text{moles}$

$ClC{H}_{2}COOH\rightleftharpoons ClC{H}_{2}CO{O}^{-}+H^{+}$

$Ateqb.C(1-\alpha )C\alpha C\alpha$

$\text{vant't Hoff factor}\left(i\right)=1+\alpha$

$C=\frac{\text{moles}}{\text{volume in L}}=\frac{0.1}{0.5}=0.2\text{M}$

Assuming molarity = molality

$\Delta T_f=i\times K_f\times m=(1+\alpha )\times 1.86\times 0.2$
$0.5=(1+\alpha )\times 1.86\times 0.2$


$\Rightarrow (1+\alpha )=\frac{0.5}{0.2\times 1.86}1+\alpha =1.34$

$\Rightarrow \alpha =0.34$

$K_{a}of\left(ClC{H}_{2}COOH\right)=\frac{C\alpha \cdot C\alpha }{C(1-\alpha )}$

$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.2\times {\left(0.34\right)}^2}{1-0.34}$

$K_a=35\times {10}^{-3}x=35$