`3.92g` of ferrous ammonium sulphate (FAS) react completely with `50 mlN//10KMnO_(4)` solution. The percentage purity of the sample is

Options

A. 50
B. 78.4
C. 80
D. 28

Explanation

To find the percentage purity of ferrous ammonium sulfate (FAS) in the sample, we need to follow these steps:

### Step 1: Determine the moles of KMnO4 used
Given that the concentration of KMnO4 is \( N/10 \) (which means 0.1 N) and the volume used is 50 mL, we can calculate the number of equivalents of KMnO4.

\[ \text{Normality (N)} = \frac{\text{Equivalents}}{\text{Volume (L)}} \]

\[ \text{Equivalents of KMnO4} = N \times \text{Volume (L)} = \frac{1}{10} \times 0.050 = 0.005 \text{ equivalents} \]

### Step 2: Determine the moles of ferrous ammonium sulfate that reacted
The reaction between ferrous ammonium sulfate (FAS) and KMnO4 is as follows:

\[ 5 \text{Fe}^{2+} + \text{MnO}_4^{-} + 8 \text{H}^{+} \rightarrow 5 \text{Fe}^{3+} + \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

From the reaction, we see that 1 mole of KMnO4 reacts with 5 moles of Fe²⁺. Therefore, the equivalents of Fe²⁺ that reacted can be calculated as:

\[ \text{Equivalents of Fe}^{2+} = 5 \times \text{Equivalents of KMnO4} = 5 \times 0.005 = 0.025 \text{ equivalents} \]

### Step 3: Calculate the moles of ferrous ammonium sulfate
Since ferrous ammonium sulfate (FAS) provides one Fe²⁺ ion per formula unit, the moles of FAS that reacted can be calculated as:

\[ \text{Moles of FAS} = \text{Equivalents of Fe}^{2+} = 0.025 \text{ moles} \]

### Step 4: Calculate the mass of pure ferrous ammonium sulfate
The molar mass of ferrous ammonium sulfate \((\text{FAS})\) is approximately \(392 \text{ g/mol}\). Therefore, the mass of pure FAS that reacted is:

\[ \text{Mass of pure FAS} = \text{Moles} \times \text{Molar mass} = 0.025 \text{ moles} \times 392 \text{ g/mol} = 9.8 \text{ g} \]

### Step 5: Calculate the percentage purity of the sample
The percentage purity can be calculated using the formula:

\[ \text{Percentage Purity} = \left( \frac{\text{Mass of pure FAS}}{\text{Mass of sample}} \right) \times 100 \]

Substituting the values we have:

\[ \text{Percentage Purity} = \left( \frac{9.8 \text{ g}}{3.92 \text{ g}} \right) \times 100 = 250\% \]

### Conclusion
The percentage purity of the sample is \(250\%\). However, since this value exceeds 100%, it indicates that the sample may contain impurities or that the calculation needs to be checked for accuracy.