An excess of $\mathrm{AgNO}_3$ is added to $100 \mathrm{~mL}$ of a $0.01 \mathrm{M}$ solution of…
An excess of $\mathrm{AgNO}_3$ is added to $100 \mathrm{~mL}$ of a $0.01 \mathrm{M}$ solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitate would be
0.001
0.002
0.003
0.01
Solution
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.
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{+}+\mathrm{Cl}^{-}$
Initial $100 \times 0.01$
0
$\mathrm{mmol}=1 \mathrm{mmol}$
After ionisation 0
$1 \mathrm{mmol} \quad 1 \mathrm{mmol}$
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 $\mathrm{AgCl}$
$=\frac{1 \times 1 \times 10^{-3}}{1}=10^{-3} \text { or } 0.001 \mathrm{~mol} \mathrm{AgCl}$