Question

An excess of NaOH was added to $100$ ml of a ferric chloride solution. This caused the precipitation of $1.425$ g of $Fe(OH{)}_3$. The normality of the ferric chloride solution is :

• A. 0.20 N
• B. 0.50 N
• C. 0.25 N
• D. 0.40 N

Answer 1

The correct option is D 0.40 N

The balanced equation governing this reaction is:
$3NaOH+FeC{l}_3\to 3NaCl+Fe(OH{)}_3$

Molecular wt. of $Fe(OH{)}_3$ is $56+(16+1)\times 3=107g/mol$

Thus no. of moles of $Fe(OH{)}_3$ precipitated$=1.425/107=0.01332moles$

No. of moles of $FeC{l}_3$ initially present (a/c to balanced eqn.) $=0.01332moles$

No. of equivalents of $FeC{l}_3$ initailly present $=no.ofmoles\times valency$$=0.01332\times 3=0.0399$

Normality of $FeC{l}_3$ solution$=\frac{No.ofequivalents}{Vol.ofsol.inlitres}$

$=\frac{0.0399}{0.1}$

$=0.399N\cong 0.40N$

Thus the correct option is (d).