Consider the ionisation of `H_(2) SO_(4)` as follow"
`H_(2) SO_(4) + 2H_(2)P rarr 2H_(3) O^(o+) SO_(4)^(2-)`
The total number of ions furnised by `100 mL` of `0.1 M` `H_(2)SO_(4)` will be

Options

A. `1.2 xx 10^(23)`
B. `0.12 xx 10^(23)`
C. `0.18 xx 10^(23)`
D. `1.8 xx 10^(23)`

Answer

C

Explanation

1 mol `H_(2) SO_(4) -= 2 "mol" H_(3)O^(O+) -= 1 "mol" SO_(4)^(2-)`
`((100)/(1000) xx 0.1 = 0.01) -= 2 xx 0.01 + 0.01 = 0.03 "mo"`
`= 0.18069 xx 10^(23)`