Question

An excess of NaOH was added to $100$ ml of a $FeC{l}_3$ solution which gives $2.14gm$ 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

The correct option is C 0.6 N

$\underset{\left(excess\right)}{3NaOH}+\underset{\underset{1mmol\equiv 1mmol}{100mL}}{FeC{l}_3}\to Fe(OH{)}_3+3NaCl$
$\text{milli equivalents of}FeC{l}_3\equiv \text{milli equivalents of}Fe(OH{)}_3$
$M\times 100\equiv \frac{2.14}{107}\times 1000$
$M_{FeC{l}_3}=0.2$
$N_{FeC{l}_3}=0.2\times 3$ (valency factor =3)
$=0.6N$