Correct Answer - 13.61 mL
Step I. Calculation of mass of `H_(2)SO_(4)`
`"Molarity of acid (M)"=0.2 M=0.2"mol L"^(-1), "Volume os solution"=1 L`
Molar mass of acid = 98 g `mol^(-1)`
`"Molarity"=("Mass/Molar mass")/("Volume of solution in Litres")`
`(0.2" mol L"^(-1))=W/((98" g mol"^(-1))xx(1L))`
`1=(0.2" mol L"^(-1))xx(98" g mol"^(-1))(1L)=19.6 g`.
step II. Calculation of volume os 80% `H_(2)SO_(2)`
`"Volume of pure"H_(2)SO_(4)=("Mass")/("density")=((19.6g))/((1.8" g mol"^(-1)))=10.89 mL`
`"Volume of 80% "H_(2)SO_(4)=(10.89mL)xx100/80=13.61 mL`
