Find out % of oxalate ion ina given sample of an alkali metal oxalate salt, 0.30g of it is dissolve in 100mL water and its required 90mL OF Centimolar `KMnO_(4)` solution in aicdic medium:

Options

A. 66%
B. 55%
C. 44%
D. 6.6%

Answer

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

### Step 1: Understand the Reaction
The reaction between oxalate ions (C₂O₄²⁻) and potassium permanganate (KMnO₄) in acidic medium can be represented as:
\[ \text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Mn}^{2+} + \text{CO}_2 \]

### Step 2: Determine the n-factor for KMnO₄
In this reaction, the oxidation state of manganese changes from +7 in MnO₄⁻ to +2 in Mn²⁺. Therefore, the change in oxidation state is:
\[ 7 - 2 = 5 \]
Thus, the n-factor for KMnO₄ is 5.

### Step 3: Determine the n-factor for oxalate ion
The oxidation state of carbon in oxalate (C₂O₄²⁻) changes from +3 to +4 when it is oxidized to CO₂. For two carbon atoms, the change in oxidation state is:
\[ 4 - 3 = 1 \text{ (for one carbon)} \]
Thus, for two carbon atoms, the total change is:
\[ 2 \]
So, the n-factor for C₂O₄²⁻ is 2.

### Step 4: Calculate the number of equivalents of KMnO₄ used
We know the volume of KMnO₄ solution used is 90 mL and it is a centimolar (0.01 M) solution. The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \]

Converting the volume from mL to L:
\[ 90 \text{ mL} = 0.090 \text{ L} \]

Now, substituting the values:
\[ \text{Number of equivalents of KMnO}_4 = 0.01 \, \text{mol/L} \times 0.090 \, \text{L} \times 5 = 0.0045 \, \text{equivalents} \]

### Step 5: Calculate the number of equivalents of oxalate ion
Since the number of equivalents of KMnO₄ is equal to the number of equivalents of oxalate ions:
\[ \text{Number of equivalents of C}_2\text{O}_4^{2-} = 0.0045 \, \text{equivalents} \]

### Step 6: Calculate the number of moles of oxalate ion
Using the n-factor for oxalate (which is 2):
\[ \text{Number of moles of C}_2\text{O}_4^{2-} = \frac{\text{Number of equivalents}}{\text{n-factor}} = \frac{0.0045}{2} = 0.00225 \, \text{moles} \]

### Step 7: Calculate the weight of oxalate ion
The molecular weight of oxalate (C₂O₄²⁻) is 88 g/mol. Therefore, the weight of the oxalate ion can be calculated as:
\[ \text{Weight of C}_2\text{O}_4^{2-} = \text{Number of moles} \times \text{Molecular weight} \]
\[ = 0.00225 \, \text{moles} \times 88 \, \text{g/mol} = 0.198 \, \text{g} \]

### Step 8: Calculate the percentage of oxalate ion in the sample
The total weight of the sample is 0.30 g. The percentage of oxalate ion in the sample is given by:
\[ \text{Percentage of C}_2\text{O}_4^{2-} = \left( \frac{\text{Weight of C}_2\text{O}_4^{2-}}{\text{Total sample weight}} \right) \times 100 \]
\[ = \left( \frac{0.198 \, \text{g}}{0.30 \, \text{g}} \right) \times 100 = 66\% \]

### Final Answer
The percentage of oxalate ion in the given sample of alkali metal oxalate salt is **66%**.

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