The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in
0.3 of [Co(NH3)6]Cl3 is .
To react completely with one mole of [ML6]Cl3 , 3 moles of AgNO3 is required.
0.3 g[ML6]Cl3 means 0.3267.46 moles of [ML6]Cl3
So, moles of AgNO3 required will 0.3×3267.46 moles
To find the volume,
V(L) = 0.02692
V(mL) = 26.92