The mole fraction of a given sample of ${\mathrm{I}}_2\mathrm{in}{\mathrm{C}}_6{\mathrm{H}}_6$ is 0.2. The molality of ${\mathrm{I}}_2\mathrm{in}{\mathrm{C}}_6{\mathrm{H}}_6$ is

$\begin{array}{l}\mathrm{Mole}\mathrm{fraction}\mathrm{of}{\mathrm{I}}_2={\mathrm{X}}_{\mathrm{A}}; \mathrm{Mole}\mathrm{fraction}\mathrm{of} {\mathrm{C}}_6{\mathrm{H}}_6={\mathrm{X}}_{\mathrm{B}}\\ {\mathrm{X}}_{\mathrm{A}}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}\text{;\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{X}}_{\mathrm{B}}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}\\ {\mathrm{X}}_{\mathrm{A}}+{\mathrm{X}}_{\mathrm{B}}=1;{\mathrm{X}}_{\mathrm{A}}=0.2\\ {\mathrm{X}}_{\mathrm{B}}=0.8;\frac{{\mathrm{X}}_{\mathrm{A}}}{{\mathrm{X}}_{\mathrm{B}}}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}\\ {\mathrm{n}}_1:{\mathrm{n}}_2=1:4\end{array}$

$\begin{array}{c}\mathrm{mass} \mathrm{of} \mathrm{solvent}\left({\mathrm{C}}_6{\mathrm{H}}_6\right), \mathrm{w}={\mathrm{n}}_2\times \mathrm{M}=4\times 78\left(\mathrm{g}\right)\\ \because {\mathrm{n}}_2=\frac{\mathrm{w}}{\mathrm{M}}; \mathrm{M}=\mathrm{molecular} \mathrm{wt}. \mathrm{of} \mathrm{solvent}=78\mathrm{g}\end{array}$

$\begin{array}{l}\mathrm{Molality}=\frac{{\mathrm{n}}_1}{\mathrm{mass}\mathrm{of}\mathrm{solvent}\left(\mathrm{in}\mathrm{kg}\right)}\\ =\frac{1}{4\times 78}\times 1000=3.21\mathrm{m}\end{array}$