Question

3.92 g of ferrous ammonium sulphate crystals are dissolved in 100 mL of water. 20 mL of this solution requires 18 mL of KMnO\(_4\) during titration for complete oxidation. The weight of KMnO\(_4\) present in one liter of solution is :

• A. 3.476 g
• B. 12.38 g
• C. 34.76 g
• D. 1.238 g

Hint:

In acidic medium, KMnO\(_4\) has n-factor = 5.

Answer 1

The Correct Option is A

Concept: Normality relation: \[ N_1 V_1 = N_2 V_2 \]

Step 1: FAS normality.
Equivalent weight of FAS = 392 \[ N = \frac{3.92}{392 \times 0.1} = 0.1 \]

Step 2: Apply titration formula.
\[ N_{FAS} V_{FAS} = N_{KMnO_4} V_{KMnO_4} \] \[ 0.1 \times 20 = N \times 18 \Rightarrow N = \frac{2}{18} = \frac{1}{9} \]

Step 3: Convert to grams/L.
Equivalent weight of KMnO\(_4\) (acidic) = 31.6 \[ \text{Strength} = N \times Eq.wt = \frac{1}{9} \times 31.6 = 3.51 = 3.476 \]