Question

An excess of $NaOH$ was added to $100$ mL of a $FeC{l}_3$ solution which gives $2.14$ grams of $Fe(OH{)}_3$. Calculate the normality of $FeC{l}_3$ solution.

• A. 0.2 N
• B. 0.3 N
• C. 0.6 N
• D. 1.8 N

Answer 1

Answer: (C)

0.6 N

$FeC{l}_3+3NaOH\to Fe(OH{)}_3+3NaCl$
Thus, $1molFeC{l}_3\equiv 1molFe(OH{)}_3$
The molar masses of $FeC{l}_3$ and $Fe(OH{)}_3$ are 162.2 g/mol and 106.9 g/mol respectively.
2.14 g of $Fe(OH{)}_3$ corresponds to $\frac{2.14}{106.9}=0.02$ moles.
This is also equal to 0.02 moles of $FeC{l}_3$. This is present in 100 mL of solution.
Hence, the molarity of $FeC{l}_3$ solution is $\frac{0.02mol}{\frac{100mL}{1000mL/L}}=0.2M$.
The normality of $FeC{l}_3$ solution is $3\times 0.2=0.6N$.