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

Step 1: Understanding the Concept:
This is a redox titration problem. Ferrous Ammonium Sulphate (FAS) is a primary standard used to determine the concentration of \( \text{KMnO}_{4} \) through the Law of Equivalents.
: Key Formula or Approach:
1. Normality (\( N \)) \( = \frac{\text{Weight}}{\text{Equivalent Weight}} \times \frac{1000}{\text{Volume (mL)}} \)
2. Law of Equivalents: \( N_{1}V_{1} = N_{2}V_{2} \)
3. Strength (g/L) \( = \text{Normality} \times \text{Equivalent Weight} \)
Step 2: Detailed Explanation:
1. Calculate Normality of FAS solution (\( N_{\text{FAS}} \)):
FAS formula: \( \text{FeSO}_{4}(NH_{4})_{2}\text{SO}_{4} \cdot 6H_{2}O \)
Molecular weight \( = 392 \text{ g/mol} \).
In titration with \( \text{KMnO}_{4} \), \( \text{Fe}^{2+} \to \text{Fe}^{3+} + \text{e}^{-} \). n-factor for FAS \( = 1 \).
Equivalent weight of FAS \( = 392/1 = 392 \).
\[ N_{\text{FAS}} = \frac{3.92}{392} \times \frac{1000}{100} = 0.01 \times 10 = 0.1 N \]
2. Find Normality of \( \text{KMnO}_{4} \) (\( N_{\text{K}} \)):
Using \( N_{1}V_{1} (\text{FAS}) = N_{2}V_{2} (\text{KMnO}_{4}) \):
\[ 0.1 \times 20 = N_{K} \times 18 \]
\[ N_{K} = \frac{2}{18} = \frac{1}{9} N \]
3. Calculate Weight of \( \text{KMnO}_{4} \) per Liter:
In acidic medium, n-factor for \( \text{KMnO}_{4} = 5 \).
Molecular weight of \( \text{KMnO}_{4} = 158 \).
Equivalent weight \( = 158/5 = 31.6 \).
Strength (g/L) \( = \text{Normality} \times \text{Equivalent Weight} \)
\[ \text{Strength} = \frac{1}{9} \times 31.6 \approx 3.51 \text{ g/L} \]
The value closest to 3.51 in the options is 3.476 g.
Step 3: Final Answer:
The weight of \( \text{KMnO}_{4} \) in one liter of solution is 3.476 g.