Question

Excess of $NaOH (aq)$ was added to $100\, mL$ of $FeCl _{3} (aq)$ resulting into $2.14 \,g$ of $Fe ( OH )_{3}$. The molarity of $FeCl _{3} (aq)$ is
(Given molar mass of $Fe =56 \,g \,mol ^{-1}$ and molar mass of $Cl =35.5 \,g \,mol ^{-1}$ )

Answer 1

Moles of $Fe ( OH )_{3}=\frac{2.14}{107}$
$=2 \times 10^{-2} mol$
moles of $FeCl _{3}$
$=2 \times 10^{-2} mol$
$M=\frac{2 \times 10^{-2}}{100} \times 1000=0.2 \,M$