Find out the `%` of oxalate ion in given sample of oxalate salt of which 0.3 g is present in 100 mL of solution required 90 mL. `N//20KMnO_(4)` for complete oxidation.

Answer

To find the percentage of oxalate ion in the given sample of oxalate salt, we can follow these steps:

### Step 1: Calculate Milliequivalents of KMnO4
We know that the normality (N) of KMnO4 is \( \frac{1}{20} \) N and the volume (V) used is 90 mL.

Using the formula for milliequivalents:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (mL)} \]
\[ \text{Milliequivalents of KMnO4} = \frac{1}{20} \times 90 = 4.5 \text{ meq} \]

### Step 2: Determine Milliequivalents of Oxalate Ion
Since oxalate ion (C2O4^2-) reacts with KMnO4 in a 1:2 ratio (1 mole of KMnO4 reacts with 2 moles of oxalate), we can calculate the milliequivalents of oxalate ion:
\[ \text{Milliequivalents of Oxalate} = \frac{4.5 \text{ meq}}{2} = 2.25 \text{ meq} \]

### Step 3: Calculate the Equivalent Weight of Oxalate
The equivalent weight of oxalate ion (C2O4^2-) can be calculated as follows:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
The molar mass of oxalate (C2O4) is approximately 88 g/mol, and since it has a valency of 2 (n=2):
\[ \text{Equivalent Weight} = \frac{88 \text{ g/mol}}{2} = 44 \text{ g/equiv} \]

### Step 4: Calculate the Weight of Oxalate Ion
Now, we can calculate the weight of oxalate ion that reacted:
\[ \text{Weight of Oxalate} = \text{Milliequivalents} \times \text{Equivalent Weight} \times \frac{1 \text{ g}}{1000 \text{ meq}} \]
\[ \text{Weight of Oxalate} = 2.25 \times 44 \times \frac{1}{1000} = 0.099 \text{ g} \]

### Step 5: Calculate the Percentage of Oxalate Ion in the Sample
Now, we can find the percentage of oxalate ion in the original sample:
\[ \text{Percentage of Oxalate} = \left( \frac{\text{Weight of Oxalate}}{\text{Total Weight of Sample}} \right) \times 100 \]
\[ \text{Percentage of Oxalate} = \left( \frac{0.099 \text{ g}}{0.3 \text{ g}} \right) \times 100 = 33\% \]

### Final Answer
The percentage of oxalate ion in the given sample is **33%**.

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