Solve the following

Solution:

It is given that:

$w_{1}=500 \mathrm{~g}$

$w_{2}=19.5 \mathrm{~g}$

$K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

$\Delta T_{f}=1 \mathrm{~K}$

We know that:

$M_{2}=\frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$

$=\frac{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 19.5 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{500 \mathrm{~g} \times 1 \mathrm{~K}}$

$=72.54 \mathrm{~g} \mathrm{~mol}^{-1}$

Therefore, observed molar mass of $\mathrm{CH}_{2} \mathrm{FCOOH},\left(M_{2}\right)_{\text {obs }}=72.54 \mathrm{~g} \mathrm{~mol}$

The calculated molar mass of $\mathrm{CH}_{2} \mathrm{FCOOH}$ is:

$\left(M_{2}\right)_{\text {cal }}=14+19+12+16+16+1$

$=78 \mathrm{~g} \mathrm{~mol}^{-1}$

Therefore, van't Hoff factor, $i=\frac{\left(M_{2}\right)_{\mathrm{cal}}}{\left(M_{2}\right)_{\mathrm{obs}}}$

$=\frac{78 \mathrm{~g} \mathrm{~mol}^{-1}}{72.54 \mathrm{~g} \mathrm{~mol}^{-1}}$

$=1.0753$

Let α be the degree of dissociation of 

$\therefore i=\frac{C(1+\alpha)}{C}$

$\Rightarrow i=1+\alpha$

$\Rightarrow \alpha=i-1$

$=1.0753-1$

$=0.0753$

Now, the value of Ka is given as:

$K_{a}=\frac{\left[\mathrm{CH}_{2} \mathrm{FCOO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{2} \mathrm{FCOOH}\right]}$

$=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}$

$=\frac{C \alpha^{2}}{1-\alpha}$

Taking the volume of the solution as 500 mL, we have the concentration:

$C=\frac{\frac{19.5}{78}}{500} \times 1000 \mathrm{M}$

$=0.5 \mathrm{M}$

Therefore, $K_{a}=\frac{C \alpha^{2}}{1-\alpha}$

$=\frac{0.5 \times(0.0753)^{2}}{1-0.0753}$

$=\frac{0.5 \times 0.00567}{0.9247}$

$=0.00307$ (approximately)

$=3.07 \times 10^{-3}$