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 ${Cl = 35.5 \, g \, mol^{-1}}$)

• A. $0.2\, M$
• B. $0.3\, M$
• C. $0.6\, M$
• D. $1.8\, M$

Answer 1

Answer: (A)

$0.2\, M$

Answer 2

3NaOH + FeCl $\rightarrow$ Fe (CH) + NaCl
100 ml 2.14 gm
m = ?
Moles of Fe(CH) $=\frac{2.14}{107}=2\times10^{-2}$ mol
moles FeCl $=2\times10^{-2}$ mol
$M=\frac{2\times10^{-2}}{100}\times1000=0.2$ M