19.5 of $\mathrm{CH}_2 \mathrm{FCOOH}$ is dissolved in $500 \mathrm{~g}$ of water. The depression in the freezing point of water observed is $1 \cdot 0^{\circ} \mathrm{C}$. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Question

19.5 of $\mathrm{CH}_2 \mathrm{FCOOH}$ is dissolved in $500 \mathrm{~g}$ of water. The depression in the freezing point of water observed is $1 \cdot 0^{\circ} \mathrm{C}$. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.,

MARKS App · Accepted Answer

Using the relation, $=\frac{1000 \times 1.86 \times 19.5}{500 \times 1}=72.54 \mathrm{~g} \mathrm{~mol}^{-1}$ $M_2($ Calculated $)=12+2+19+12+2 \times 16+1$ $=78 \mathrm{~g} \mathrm{~mol}^{-1}$ $\therefore i=\frac{M_2 \text { (calculated) }}{M_2 \text { (observed) }}=\frac{78}{72.54}=1.0753$. Calculation of dissociation constant : Let $\alpha$ be the degree of dissociation and $\mathrm{C}$ be the initial concentration of $\mathrm{CH}_2 \mathrm{FCOOH}$ $\begin{aligned} &i=\frac{C(1+\alpha)}{C}=1+\alpha \ &\alpha=i-1=1.0753-1=0.0753 \ &K_a=\frac{C \alpha C \alpha}{C(1-\alpha)}=\frac{C \alpha^2}{1-\alpha} \end{aligned}$ Taking the volume of solution as $500 \mathrm{~mL}$ $\begin{aligned} &\mathrm{C}=\frac{19.5}{\frac{78}{500} \times 1000} \mathrm{M}=0.5 \mathrm{M} \ &\therefore \quad K_a=\frac{0.5 \times(0.0753)^2}{(1-0.0753)}=3.07 \times 10^{-3} \end{aligned}$

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