Question

When the solutions, $50$ mL of $0.12MFe{\left(N{O}_3\right)}_3$, $100$ mL of $0.1MFe{Cl}_3$ and $100$ mL of $0.26MMg{\left(N{O}_3\right)}_2$ are mixed, the final concentrations of $C{l}^{-}$ ions is (write it as multiples of $100$ (only two digits))

Answer 1

$\because mMofFe{\left(N{O}_3\right)}_3=50\times 0.12=6$
$\therefore mMof{Fe}^{3+}=6$
and mM of $N{O}_3^{-}=6\times 3=18$
Now mM of $Fe{Cl}_3=100\times 0.1=10$
$\therefore mMof{Fe}^{3+}=10$
mM of ${Cl}^{-}=10\times 3=30$
Also, mM of $Mg{N{O}_3}_2=100\times 0.26=26$
$\therefore mMof{Mg}^{2+}=26$
mM of $N{O}_3^{-}=26\times 2=52$
Thus, mM of ${Fe}^{3+}=6+10=16$
mM of $N{O}_3^{-}=18+52=70$
mM of ${Cl}^{-}=30$
mM of ${Mg}^{2+}=26$
Also, totall volume of solution becomes$50+100+100=250mL$
$\left[concentrationofspecies\right]=mM/volume$ in mL
$\left[{Fe}^{3+}\right]=16/250=0.064M$
$\left[N{O}_3^{-}\right]=70/250=0.28M$
$\left[{Cl}^{-}\right]=30/250=0.12M$
$\left[{Mg}^{2+}\right]=26/250=0.104M$
So, answer is $12$.