Chemistry - Liquid Solution Question with Solution | TestHub

The formula of dichlorotetraqua chromium (III) chloride is $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$. On ionisation it generates only one ${\mathrm{Cl}}^{-}$ion. $$\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}⟶{\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]}^{+}+{\mathrm{Cl}}^{-}$$ Initial $100\times 0.01$ 0 $\mathrm{mmol}=1\mathrm{mmol}$ After ionisation 0 $1\mathrm{mmol}1\mathrm{mmol}$ One mole of ${\mathrm{Cl}}^{-}$ions react with only 1 mole of ${\mathrm{AgNO}}_3$ molecule to produce 1 mole of $\mathrm{AgCl}$. $\therefore 1$ mmol or $1\times 10^{-3}$ mole reacts with ${\mathrm{AgNO}}_3$ to give $\mathrm{AgCl}$ $$=\frac{1\times 1\times 10^{-3}}{1}=10^{-3}\text{ or }0.001\mathrm{mol}\mathrm{AgCl}$$