Correct Answer - B
One mole of `CuSO_(4)` gives `1 "mole of" Cu^(2+)` and `1 "mole of" SO_(4)^(2-)`. Also `1 "mole of" Al_(2)(SO_(4))_(3)` gives `2 "mole of" Al^(3+)` and `3"mole of" SO_(4)^(2-)`. Thus total moles of all the ions present in solution having `0.1 M of CuSO_(4)` and `0.1M of Al_(2)(SO_(4))_(3)` is
`0.1+0.1+0.2+0.3=0.7`