Correct Answer - C
We have
`C_(H^(+))= 10^(-pH)mol L^(-1)`
`"Initial pH = 1, thus, initial" C_(H^(+))=10^(-1)mol L^(-1)`
`"Final pH= 2, thus, final" C_(H^(+))=10^(2)mol L^(-1)`
In case of dilution, number of moles of solute do not change. Thus,
`M_(i)V_(i)=M_(f)V_(f)`
`(0.1)(1)=(0.01)(V_(f))`
`V_(f)=10L`
Hence, 9L of `H_(2)O` must be added.