Question

The volume (in mL) of 0.125 M $AgNO_3$ required to quantitatively precipitate chloride ions in 0.3 g of $[Co(NH_3)]Cl_3$ is__________.
$^M[Co(NH_3)_6]Cl_3 = 267.46$ g/mol $M_{AgNO_3} = 169.87$ g/mol
Given -

Answer 1

Number of moles of $Cl^-$ precipitated in $[Co(NH_3)_6]Cl_3$ is equal to number of moles of $AgNO_3$ used.
$\frac{0.3}{267.46}\times3=\frac{0.125\times V}{1000}$
where $V$ is volume of $AgNO_3$ (in $mL$)
$V = 26.92 \,mL$