A coordination isomer of composition CrCl3·6H2O is dissolved in water to make a 0.100 mol kg^-1 solution that freezes 0.558 K below the freezing…

The van't Hoff factor follows from the freezing point depression: $i=\Delta T_f/(K_fm)=0.558/(1.86\times0.100)=3.00$. Among the hydrate isomers of $\mathrm{CrCl_3\cdot6H_2O}$, the species $[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}$ would give $i=4$, $[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2\cdot H_2O}$ gives $i=3$ (one complex cation plus two chloride ions), $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl\cdot2H_2O}$ gives $i=2$, and $[\mathrm{Cr(H_2O)_3Cl_3}]\cdot3\mathrm{H_2O}$ gives $i=1$. Hence the isomer is $[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2\cdot H_2O}$, with two ionisable chlorides per formula unit; the coordinated chloride is not precipitated. Moles of complex in $200\ \mathrm{mL}$ $=0.200\times0.100=0.0200$, so moles of free $\mathrm{Cl^-}=0.0400$ and the mass of $\mathrm{AgCl}$ $=0.0400\times143.5=5.74\ \mathrm{g}$. Had one wrongly taken all three chlorides as ionisable, $8.61\ \mathrm{g}$ would result, which is why the freezing point data are needed to fix the structure.