Correct Answer - C
`Co^(3+) + 2 Cl^(-) + 2 (en)`
net charge = +1
Thus , complex is
`[Co(en)_(2) Cl_(2)]^(+) Cl^(-) to [Co(en)_(2) Cl_(2)]^(+) + Cl^(-)`
Only one `Cl^(-)` which is precipitated as AgCl
100 mL of 0.024 M complex = 2.4 millimol = 0.0024 mol