The volume (in mL) of 0.125 M ${\mathrm{AgNO}}_3$ required to quantitatively precipitate chloride ions in 0.3 g of $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$ is______. 

$\begin{array}{l}{ }^{\mathrm{M}}\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3=267.46\text{\hspace{0.17em}\hspace{0.17em}g/mol}\\ { }^{\mathrm{M}}{\mathrm{AgNO}}_3=169.87\text{\hspace{0.17em}\hspace{0.17em}g/mol}\end{array}$

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3+3{\mathrm{AgNO}}_3\to 3\mathrm{AgCl}+\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\left({\mathrm{NO}}_3\right)}_3$

Moles present in $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$ is:

$\mathrm{number} \mathrm{of} \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}=\frac{0.3 \mathrm{g}}{267.46 \mathrm{g}/\mathrm{mol}}$

One mole of $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$ requires three moles of ${\mathrm{AgNO}}_3$, hence moles of ${\mathrm{AgNO}}_3$ required is:

$\mathrm{number} \mathrm{of} \mathrm{moles}=\frac{0.3 \mathrm{g}}{267.46 \mathrm{g}/\mathrm{mol}}\times 3$

Molarity of ${\mathrm{AgNO}}_3$ is 0.125 M and moles is $\frac{0.3 \mathrm{g}}{267.46 \mathrm{g}/\mathrm{mol}}\times 3$. Volume is calculated as:

$$\begin{gathered} \mathrm{Volume}=\frac{\mathrm{number} \mathrm{of} \mathrm{moles}}{\mathrm{molarity}}=\frac{0.3\times 3}{267.46\times 0.125} \\ \mathrm{Volume}=0.02692 \mathrm{L}\times \frac{1000 \mathrm{mL}}{1 \mathrm{L}}=26.92 \mathrm{mL} \end{gathered}$$