An excess of NaOH was added to 100 mL of a FeCl 3 solution which gives 2.14 of Fe ( OH )3. Calculate the normality of FeCl 3 solution

Question

An excess of NaOH was added to 100 mL of a FeCl 3 solution which gives 2.14 of Fe ( OH )3. Calculate the normality of FeCl 3 solution (A) 0.2 N (B) 0.3 N (C) 0.6 N (D) 1.8 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/606f2bff3a99a8692ff2f89ae995a40d-.png" /> FaCl 3 ≡ Fe ( OH )3 M × 100 ≡ (2.14/107) × 1000 M FeCl 3=0.2 N FeCl 3=0.2 × 3( valency factor =3) =0.6 N

Q. An excess of $N a O H$ was added to $100 m L$ of a $F e C l_{3}$ solution which gives $2.14$ of $F e (O H)_{3}$ . Calculate the normality of $F e C l_{3}$ solution

$0.2 N$

$0.3 N$

$0.6 N$

$1.8 N$

$F a C l_{3} \equiv F e (O H)_{3}$
$M \times 100 \equiv \frac{2.14}{107} \times 1000$
$M_{F e C l_{3}} = 0.2$
$N_{F e C l_{3}} = 0.2 \times 3 ($ valency factor $= 3)$
$= 0.6 N$

Q. An excess of NaOH was added to 100mL of a FeCl3​ solution which gives 2.14 of Fe(OH)3​. Calculate the normality of FeCl3​ solution

0.2N

0.3N

0.6N

1.8N