The van't Hoff factor for ${\mathrm{BaCl}}_2$ at $0.01\mathrm{M}$ concentration is $1.98$. The percentage of dissociation of ${\mathrm{BaCl}}_2$ at this concentration is

$$\begin{gathered} {\mathrm{BaCl}}_2 \rightleftharpoons {\mathrm{Ba}}^{2+} + 2{\mathrm{Cl}}^{-} \\ \mathrm{at} \mathrm{t}=0 0.01 \mathrm{M} \\ \mathrm{at} \mathrm{time} \mathrm{t}, \left(0.01-\mathrm{x}\right)\mathrm{M} \mathrm{xM} 2\mathrm{xM} \\ \therefore \mathrm{i}=\frac{\mathrm{no}. \mathrm{of} \mathrm{particles} \mathrm{after} \mathrm{dissociation}}{\mathrm{no}. \mathrm{of} \mathrm{particles} \mathrm{before} \mathrm{dissociation}} \\ =\frac{\left(0.01-\mathrm{x}\right) + \mathrm{x} + 2\mathrm{x}}{0.01} \\ =\frac{0.01+2\mathrm{x}}{0.01} \\ \mathrm{i}\times 0.01=0.01 + 2\mathrm{x} \\ \Rightarrow 1.98\times 0.01=0.01 + 2\mathrm{x} \\ \Rightarrow 2\mathrm{x}=0.0198-0.01 \\ \Rightarrow \mathrm{x}=\frac{0.0098}{2}=0.0049 \\ \therefore \mathrm{dissociation} \mathrm{percentage}\left(\mathrm{\alpha }\right)=\frac{\mathrm{x}}{0.01}\times 100 \\ =\frac{0.0049}{0.01}\times 100 \\ =49 \% \end{gathered}$$