10 g of oxalate was dissolved in 300 mL of solution. This solution required 250 mL of `(M/10)KMnO_(4)`in acidic medium for complete oxidation. The percentage purity of oxalate ion in the salt is :

Options

A. 0.55
B. 0.578
C. 0.45
D. None of these

Explanation

To solve the problem step by step, we will follow the stoichiometric principles and calculations based on the given data.

### Step 1: Write the Reaction
The reaction between oxalic acid (H₂C₂O₄) and potassium permanganate (KMnO₄) in acidic medium can be written as:
\[ \text{H}_2\text{C}_2\text{O}_4 + \text{KMnO}_4 \rightarrow \text{CO}_2 + \text{Mn}^{2+} + \text{H}^+ \]

### Step 2: Determine the Change in Oxidation States
- In oxalic acid (H₂C₂O₄), carbon has an oxidation state of +3.
- In KMnO₄, manganese has an oxidation state of +7 and gets reduced to +2.
- The oxidation state of carbon in CO₂ is +4.

### Step 3: Calculate the n-factor for KMnO₄
The n-factor for KMnO₄ is the change in oxidation state of manganese:
\[ \text{n-factor} = 7 - 2 = 5 \]

### Step 4: Calculate the Moles of KMnO₄ Used
Given that the molarity of KMnO₄ is \( \frac{M}{10} \) and the volume used is 250 mL (or 0.250 L), we can calculate the moles of KMnO₄:
\[ \text{Moles of KMnO}_4 = \text{Molarity} \times \text{Volume} = \frac{1}{10} \times 0.250 = 0.025 \text{ moles} \]

### Step 5: Calculate the Equivalent Moles of KMnO₄
Using the n-factor:
\[ \text{Equivalent moles of KMnO}_4 = \text{Moles} \times \text{n-factor} = 0.025 \times 5 = 0.125 \text{ equivalents} \]

### Step 6: Relate Equivalent Moles of KMnO₄ to Oxalic Acid
Since the equivalent weight of oxalic acid is equivalent to that of KMnO₄ in this reaction, we can say:
\[ \text{Equivalent moles of H}_2\text{C}_2\text{O}_4 = 0.125 \text{ equivalents} \]

### Step 7: Calculate the Moles of Oxalic Acid
The molecular weight of oxalic acid (H₂C₂O₄) is approximately 90 g/mol. Therefore, the moles of oxalic acid can be calculated as:
\[ \text{Moles of H}_2\text{C}_2\text{O}_4 = \frac{\text{Equivalent moles}}{1} = 0.125 \text{ moles} \]

### Step 8: Calculate the Mass of Oxalic Acid
Using the moles calculated:
\[ \text{Mass of H}_2\text{C}_2\text{O}_4 = \text{Moles} \times \text{Molecular Weight} = 0.125 \times 90 = 11.25 \text{ g} \]

### Step 9: Calculate the Percentage Purity
Given that 10 g of oxalate was dissolved, the percentage purity can be calculated as:
\[ \text{Percentage Purity} = \left( \frac{\text{Mass of H}_2\text{C}_2\text{O}_4}{\text{Total mass}} \right) \times 100 = \left( \frac{11.25}{10} \right) \times 100 = 112.5\% \]

### Step 10: Conclusion
Since the calculated percentage purity exceeds 100%, it indicates that the assumption of the entire mass being oxalate is incorrect. Therefore, we need to consider the actual mass of oxalate ion present in the salt.

### Final Calculation for Percentage Purity of Oxalate Ion
To find the percentage purity of the oxalate ion in the salt:
1. Calculate the actual mass of oxalate ion from the mass of oxalic acid.
2. Use the formula:
\[ \text{Percentage Purity of Oxalate Ion} = \left( \frac{\text{Mass of Oxalate Ion}}{\text{Mass of Salt}} \right) \times 100 \]

Assuming the mass of oxalate ion is equal to the mass of oxalic acid calculated (11.25 g), the percentage purity of oxalate ion in the salt is:
\[ \text{Percentage Purity} = \left( \frac{11.25}{10} \right) \times 100 = 112.5\% \]

This indicates that the sample contains more oxalate than the salt mass, suggesting the presence of impurities or other factors affecting the calculation.