To solve the problem step by step, we will follow these calculations:
### Step 1: Calculate the Molecular Weight of Ferrous Ammonium Sulphate
The formula for ferrous ammonium sulphate hexahydrate is \( \text{FeSO}_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O \).
- Atomic weights:
- Fe = 56 g/mol
- S = 32 g/mol
- O = 16 g/mol
- N = 14 g/mol
- H = 1 g/mol
Calculating the molecular weight:
\[
\text{Molecular Weight} = 56 + (32) + (4 \times 16) + (2 \times 14) + (8 + 6 \times 1 + 6 \times 16) = 392 \text{ g/mol}
\]
### Step 2: Calculate the Molarity of the Ferrous Ammonium Sulphate Solution
Given that \( 3.92 \, \text{g} \) of ferrous ammonium sulphate is dissolved in \( 100 \, \text{ml} \) of water:
1. Calculate the number of moles of ferrous ammonium sulphate:
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{3.92 \, \text{g}}{392 \, \text{g/mol}} = 0.01 \, \text{mol}
\]
2. Convert the volume from ml to liters:
\[
\text{Volume in liters} = 100 \, \text{ml} = 0.1 \, \text{L}
\]
3. Calculate the molarity:
\[
\text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{0.01 \, \text{mol}}{0.1 \, \text{L}} = 0.1 \, \text{M}
\]
### Step 3: Determine the Reaction and n-Factor
The reaction between \( KMnO_4 \) and \( Fe^{2+} \) is:
\[
\text{MnO}_4^- + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5 \text{Fe}^{3+}
\]
- The n-factor for \( KMnO_4 \) is 5 (change in oxidation state from +7 to +2).
- The n-factor for \( Fe^{2+} \) is 1 (change from +2 to +3).
### Step 4: Calculate the Number of Equivalents
From the titration, \( 20 \, \text{ml} \) of the ferrous ammonium sulphate solution requires \( 18 \, \text{ml} \) of \( KMnO_4 \).
1. Calculate the equivalents of ferrous ammonium sulphate in \( 20 \, \text{ml} \):
\[
\text{Equivalents of } Fe^{2+} = \text{Molarity} \times \text{Volume in L} \times \text{n-factor} = 0.1 \, \text{mol/L} \times 0.02 \, \text{L} \times 1 = 0.002 \, \text{equivalents}
\]
2. The equivalents of \( KMnO_4 \) used:
\[
\text{Equivalents of } KMnO_4 = \text{Equivalents of } Fe^{2+} = 0.002 \, \text{equivalents}
\]
### Step 5: Calculate the Normality of KMnO4
Using the volume of \( KMnO_4 \):
\[
\text{Normality of } KMnO_4 = \frac{\text{Equivalents}}{\text{Volume in L}} = \frac{0.002 \, \text{equivalents}}{0.018 \, \text{L}} = 0.1111 \, \text{N}
\]
### Step 6: Calculate the Weight of KMnO4 in 1 Litre
1. The molecular weight of \( KMnO_4 \):
\[
\text{Molecular Weight} = 39 + 55 + (4 \times 16) = 158 \, \text{g/mol}
\]
2. Calculate the weight of \( KMnO_4 \) in 1 Litre:
\[
\text{Weight} = \text{Normality} \times \text{Molecular Weight} \times \text{Volume in L} = 0.1111 \, \text{N} \times 158 \, \text{g/mol} \times 1 \, \text{L} = 17.6 \, \text{g}
\]
### Final Answer
The weight of \( KMnO_4 \) present in one litre of the solution is approximately \( 3.476 \, \text{g} \).
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