Identify the compound and chloride ions released

The compound is: $[Co(NH_3)_6]Cl_3$

This is a coordination compound in which the complex ion is $[Co(NH_3)_6]^{3+}$ and there are 3 chloride ions outside the coordination sphere.

Hence, 1 mole of $[Co(NH_3)_6]Cl_3$ gives 3 moles of $Cl^-$.

Calculate moles of the compound

Given mass: $m = 0.3\text{ g}$

Molar mass: $M = 267.46\text{ g mol}^{-1}$

So, $n\big([Co(NH_3)_6]Cl_3\big)=\frac{0.3}{267.46}$

$n\big([Co(NH_3)_6]Cl_3\big) \approx 1.12166\times 10^{-3}\text{ mol}$

Calculate moles of chloride ions

Since each mole gives 3 moles of $Cl^-$, $n(Cl^-)=3\times 1.12166\times 10^{-3}$

$n(Cl^-)\approx 3.36498\times 10^{-3}\text{ mol}$

Reaction with silver nitrate

The precipitation reaction is: $Ag^+ + Cl^- \rightarrow AgCl(s)$

Thus, $1$ mole of $AgNO_3$ provides $1$ mole of $Ag^+$ and precipitates $1$ mole of $Cl^-$.

Therefore, $n(AgNO_3)=n(Cl^-)=3.36498\times 10^{-3}\text{ mol}$

Use molarity to find volume

Given: $M = 0.125\text{ mol L}^{-1}$

Using $V=\frac{n}{M}$

$V=\frac{3.36498\times 10^{-3}}{0.125}$

$V=2.69198\times 10^{-2}\text{ L}$

Convert to mL: $V=26.92\text{ mL}$

Final answer

The required volume of $0.125\,M$ $AgNO_3$ is: $\boxed{27\text{ mL}}$ (more precisely, $26.92\text{ mL}$)

Comparison with stored correct answer

Stored correct answer: $26.80\text{ to }27.00$

Our calculated value is $26.92\text{ mL}$, which lies in this range. So the answer agrees.