32g of a sample of `FeSO_(4) .7H_(2)O` were dissolved in dilute sulphuric aid and water and its volue was made up to 1litre. 25mL of this solution required 20mL of `0.02 M KMnO_(4)` solution for complete oxidation. Calculate the mass% of `FeSO_(4) .7 H_(2)O` in the sample.

Options

A. 34.75
B. 69.5
C. 89.5
D. none of these

Answer

To solve the problem, we need to calculate the mass percentage of `FeSO4.7H2O` in the sample based on the information provided. Here’s a step-by-step solution:

### Step 1: Calculate the moles of KMnO4 used
We know that:
- Volume of KMnO4 solution used = 20 mL = 0.020 L
- Molarity of KMnO4 solution = 0.02 M

Using the formula:
\[ \text{Moles of KMnO4} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of KMnO4} = 0.02 \, \text{mol/L} \times 0.020 \, \text{L} = 0.0004 \, \text{mol} \]

### Step 2: Determine the moles of Fe2+ oxidized
The balanced reaction between KMnO4 and Fe2+ in acidic medium is:
\[ \text{MnO4}^- + 5 \text{Fe}^{2+} + 8 \text{H}^+ \rightarrow \text{Mn}^{2+} + 5 \text{Fe}^{3+} + 4 \text{H}_2\text{O} \]
From the reaction, 1 mole of KMnO4 reacts with 5 moles of Fe2+. Therefore, the moles of Fe2+ oxidized can be calculated as:
\[ \text{Moles of Fe}^{2+} = 5 \times \text{Moles of KMnO4} = 5 \times 0.0004 \, \text{mol} = 0.002 \, \text{mol} \]

### Step 3: Calculate the mass of Fe2+ in the sample
The molar mass of Fe (Iron) is approximately 55.85 g/mol. Therefore, the mass of Fe2+ can be calculated as:
\[ \text{Mass of Fe}^{2+} = \text{Moles of Fe}^{2+} \times \text{Molar mass of Fe} = 0.002 \, \text{mol} \times 55.85 \, \text{g/mol} = 0.1117 \, \text{g} \]

### Step 4: Calculate the mass of FeSO4.7H2O corresponding to the Fe2+
The molar mass of `FeSO4.7H2O` can be calculated as follows:
- Molar mass of `FeSO4` = 55.85 (Fe) + 32.07 (S) + 4 × 16.00 (O) = 55.85 + 32.07 + 64.00 = 151.92 g/mol
- Molar mass of `7H2O` = 7 × (2 × 1.01 + 16.00) = 7 × 18.02 = 126.14 g/mol
- Total molar mass of `FeSO4.7H2O` = 151.92 + 126.14 = 278.06 g/mol

Now, we can find the mass of `FeSO4.7H2O`:
\[ \text{Mass of FeSO4.7H2O} = \text{Mass of Fe}^{2+} \times \frac{\text{Molar mass of FeSO4.7H2O}}{\text{Molar mass of Fe}} = 0.1117 \, \text{g} \times \frac{278.06 \, \text{g/mol}}{55.85 \, \text{g/mol}} \approx 0.555 \, \text{g} \]

### Step 5: Calculate the mass percentage of `FeSO4.7H2O` in the sample
The total mass of the sample is 32 g. Therefore, the mass percentage can be calculated as:
\[ \text{Mass \% of FeSO4.7H2O} = \left( \frac{\text{Mass of FeSO4.7H2O}}{\text{Total mass of sample}} \right) \times 100 = \left( \frac{0.555 \, \text{g}}{32 \, \text{g}} \right) \times 100 \approx 1.73\% \]

### Final Answer
The mass percentage of `FeSO4.7H2O` in the sample is approximately **1.73%**.

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