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Question

The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in
0.3 of [Co(NH3)6]Cl3 is  .
M[Co(NH3)6]Cl3=267.46 g/mol
MAgNO3=169.87 g/mol


Solution

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,
0.3×3267.46=0.125×V(L)
V(L) = 0.02692
V(mL)  = 26.92

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