Question

The volume (in mL) of 0.125 $MAgN{O}_3$ required to quantitatively precipitate chloride ions in
$0.3$ of $\left[Co\right(N{H}_3{)}_6]C{l}_3$ is .
$M\left[Co\right(N{H}_3{)}_6]C{l}_3=267.46\text{g/mol}$
$M_{AgN{O}_3}=169.87\text{g/mol}$

Answer 1

To react completely with one mole of $\left[M{L}_6\right]C{l}_3$ , 3 moles of $AgN{O}_3$ is required.
$0.3g\left[M{L}_6\right]C{l}_3$ means $\frac{0.3}{267.46}$ moles of $\left[M{L}_6\right]C{l}_3$
So, moles of $AgN{O}_3$ required will $\frac{0.3\times 3}{267.46}$ moles
To find the volume,
$\frac{0.3\times 3}{267.46}=0.125\times V\left(L\right)$
V(L) = 0.02692
V(mL) = 26.92