Correct Answer - 0.014

`"Molarity of "H_(2)SO_(4)=0.8 M = 0.8" mol L"^(-1)`

`"Mass of "H_(2)SO_(4)="No. of moles"xx"Moles mass"`

`=(0.8"mol")xx(98.0" g mol"^(-1))=78.4 g`

`"Mass of 1 L solution"=(1000cm^(3))xx(1.06" g cm"^(-3))=1060 g`

`"Mass of water in solution"=(1060-78.4)=981.6 g= 0.9816 kg`

`"Molality of solution (m)"=("No. of moles of "H_(2)SO_(4))/("Mass of water in kg")=((0.8"mol"))/((0.9816" kg"))`

`=0.815" mol kg"^(-1)=0.815 m`

`"Mole fraction of"H_(2)SO_(4)=n_(H_(2)SO_(4))/(n_(H_(2)SO_(4))+n_(H_(2)O))`

`=((0.8mol))/((0.8mol)+((98.6g))/(18" g mol"^(-1)))=0.8/(0.8+54.5)=0.8/55.3=0.014`