1. Moles of the given compound

Given mass of compound $= 2.675\,\text{g}$

Molar mass $= 267.5\,\text{g mol}^{-1}$

So, moles of complex taken:

$n = \frac{2.675}{267.5} = 0.01\,\text{mol}$

2. Moles of $AgCl$ formed

Mass of $AgCl = 4.78\,\text{g}$

Molar mass of $AgCl = 143.5\,\text{g mol}^{-1}$

$n(AgCl) = \frac{4.78}{143.5} \approx 0.0333\,\text{mol}$

Since each mole of $AgCl$ corresponds to one mole of $Cl^-$,

$n(Cl^-) = 0.0333\,\text{mol}$

3. Chloride ions per mole of complex

From $0.01$ mol of complex, chloride ions obtained $= 0.0333$ mol.

Therefore, chloride ions per mole of complex:

$\frac{0.0333}{0.01} \approx 3.33$

This is very close to $3$ (the small mismatch is due to rounding/experimental approximation), so the compound gives 3 ionisable chloride ions in solution after passing through cation exchanger.

4. Interpretation using coordination compounds

A cation exchanger retains the complex cation and releases the counter anions ($Cl^-$) into solution. Thus, the number of moles of $AgCl$ formed tells us the number of outer-sphere chloride ions.

Now check options:

• A: $[Co(NH_3)_6]Cl_3$
  Outer-sphere $Cl^- = 3$
• B: $[CoCl_2(NH_3)_4]Cl$
  Outer-sphere $Cl^- = 1$
• C: $[CoCl_3(NH_3)_3]$
  Outer-sphere $Cl^- = 0$
• D: $[CoCl(NH_3)_5]Cl_2$
  Outer-sphere $Cl^- = 2$

Only option A gives $3$ chloride ions outside the coordination sphere.

5. Final answer

$\boxed{[Co(NH_3)_6]Cl_3}$

So, the correct option is A.