Molecular mass of $78\mathrm{}\mathit{g}∕\mathrm{}\mathit{m}\mathit{o}\mathit{l}$
$\left(\mathit{F}-{\mathit{C}\mathit{H}}_2\mathit{C}\mathit{O}\mathit{O}\mathit{H}\right)$
No. of moles of fluoroacetic acid is $\frac{19.5}{78}=0.25$
Molality is the number of moles of solute in $1\mathrm{}\mathit{k}\mathit{g}$ of solvent
Molality $=\frac{0.25}{\frac{500}{1000}}=0.50\mathrm{}\mathit{m}$
$\mathrm{\Delta }{\mathit{T}}_{\mathit{f}}={\mathit{K}}_{\mathit{f}}\times \mathit{m}=1.86\times 0.50$ $=0.93\mathrm{}\mathit{K}$
$\mathit{i}=\frac{1.0}{0.93}=1.0753$
${\mathit{C}\mathit{H}}_2\mathit{F}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\to {\mathit{C}\mathit{H}}_3{\mathit{F}\mathit{C}\mathit{O}\mathit{O}}^{-}+{\mathit{H}}^{+}$
$\mathit{C}\left(1-\mathrm{\alpha }\right)\mathit{C}\mathrm{\alpha }\mathit{C}\mathrm{\alpha }$
Total number of moles $=\mathit{C}\left(1-\mathrm{\alpha }\right)+\mathit{C}\mathrm{\alpha }+\mathit{C}\mathrm{\alpha }$ $=\mathit{C}\left(1+\mathrm{\alpha }\right)$
$\mathit{i}=\frac{\mathit{C}\left(1+\mathrm{\alpha }\right)}{\mathit{C}}=1+\mathrm{\alpha }=1.0753$
$\mathrm{\alpha }=0.0753$
(ii) $(\mathrm{\alpha }=0.50\times 0.0753=0.03765$
$\mathit{C}\left(1-\mathrm{\alpha }\right)=0.50\left(1-0.0753\right)=0.462$