Question

$9.45g$ of ${\mathrm{CH}}_2\mathrm{ClCOOH}$ is dissolved in $500mL$ of ${\mathrm{H}}_2\mathrm{O}$ solution and the depression in freezing point of the solution is $0.5°C$. Find the percentage dissociation.

Answer 1

Step 1: Calculate the molality of solution

Molar mass of ${\mathrm{CH}}_2\mathrm{ClCOOH}=12+2+35.5+12+16+16+1=94.5\mathrm{g}/\mathrm{mol}$

Molality of solution, $m=\frac{{\displaystyle \frac{9.45}{94.5}}}{{\displaystyle \frac{500}{1000}}}=0.2molal$

Step 2: Calculate van't hoff factor

$∆T_f=i\times K_f\times m$

$$\begin{gathered} ∆T_f=i\times K_f\times m \\ 0.5=i\times 1.86\times 0.2 \\ \Rightarrow i=1.344 \end{gathered}$$

Step 3: Calculate percentage dissociation

$$\begin{gathered} i=1+\alpha \\ 1.344=1+\alpha \\ \Rightarrow \alpha =0.344 \end{gathered}$$

Thus, percentage dissociation$=0.344\times 100=34.4\%$

Hence, the percentage dissociation of ${\mathrm{CH}}_2\mathrm{ClCOOH}=34.4\%$