Answer
$c(AlCl_3)=4.52\times 10^{-3}\frac{mol}{l}$
Work Step by Step
Considering stoichiometric coefficients, we can conclude that $n(AlCl_3) : n(AgCl)=1:3$, hence $n(AlCl_3)=\frac{1}{3}n(AgCl)=\frac{1}{3}\frac{m(AgCl)}{M(AgCl)}=\frac{1}{3}\frac{0.215g}{143.5\frac{g}{mol}}=0.0005mol$
The concentration can now be easily obtained:
$c(AlCl_3)=\frac{n(AlCl_3)}{V(AlCl_3)}=\frac{0.0005mol}{0.1105l}=4.52\times 10^{-3}\frac{mol}{l}$
