Correct Answer - 3
Moles of the complexes `= (2.675)/(267.5) = 0.01`
moles of AgCl precipitated `= (4,78)/(143.5) = 0.033`
Thus, 1 mole of the complex will precipitate `AgCl(0.033)/(0.01) = 3` moles. That means 1 molecule of the complex contains 3 ionizable Cl.
Hence, the formula is `[Co(NH_(3))_(6)]Cl_(3)`