Explanation:
Mole fraction of solute \({{\rm{X}}_{\rm{2}}}{\rm{ = 0}}{\rm{.2}}\).
Therefore, mole fraction of solvent \({{\rm{X}}_{\rm{1}}}{\rm{ = 0}}{\rm{.8}}\)
\({\rm{Or}}{\mkern 1mu} \,\,{\mkern 1mu} \frac{{{{\rm{n}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}}}{\rm{ = }}\,\,{\rm{0}}{\rm{.2}}\,\,{\rm{and}}\,\,{\mkern 1mu} {\mkern 1mu} \frac{{{{\rm{n}}_{\rm{1}}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}}}{\rm{ = }}\,\,{\rm{0}}{\rm{.8}}\)
\({\rm{Or}}\,\,\frac{{{{\rm{n}}_{\rm{2}}}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.8}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\)
Now, if \({{\rm{n}}_{\rm{1}}}\) (solvent moles) = 1000/78 = 12.8 moles
\({{\rm{n}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{12}}{\rm{.8}}}}{{\rm{4}}}\) =3.2moles. Therefore, 3.2 moles of the compound are present in one Kg of solvent benzene and so molality = 3.2.