Aqueous solutions of `Na_(2)C_(2)O_(4)` and `CaCI_(2)` are mixed and precipitate of `CaC_(2)O_(4)` formed is filered and dried. `250 mL` of the saturated solution of `CaC_(2)O_(4)` required `6.0 mL` of `0.001M KMnO_(4)` solution in acidic medium for complete titration.
Equivalent of `KMNO_(4)` required in the titration and equivalent of `C_(2)O_(4)^(2-)` ion present in `CaC_(2)O_(4)`, respectively, are

Options

A. `3xx10^(-5),3xx10^(-5)`
B. `1.8xx10^(-5),3xx10^(-6)`
C. `3xx10^(-6),6xx10^(-6)`
D. `6xx10^(-6),3xx10^(-6)`

Explanation

To solve the problem step by step, we will first determine the equivalents of \( KMnO_4 \) used in the titration and then find the equivalents of \( C_2O_4^{2-} \) ions present in \( CaC_2O_4 \).

### Step 1: Calculate the equivalents of \( KMnO_4 \)

The formula for calculating equivalents is given by:

\[ \text{Equivalents} = n \times M \times V \]

Where:
- \( n \) is the number of electrons transferred (n-factor)
- \( M \) is the molarity of the solution
- \( V \) is the volume in liters

#### Step 1.1: Determine the n-factor for \( KMnO_4 \)

In acidic medium, \( KMnO_4 \) is reduced from \( MnO_4^{-} \) (where Mn has an oxidation state of +7) to \( Mn^{2+} \) (where Mn has an oxidation state of +2).

The change in oxidation state is:

\[ \text{Change in oxidation state} = 7 - 2 = 5 \]

Thus, the n-factor for \( KMnO_4 \) is 5.

#### Step 1.2: Calculate the volume in liters

The volume of \( KMnO_4 \) solution used is given as 6.0 mL. We need to convert this to liters:

\[ V = \frac{6.0 \, \text{mL}}{1000} = 0.006 \, \text{L} \]

#### Step 1.3: Calculate the equivalents of \( KMnO_4 \)

Now substituting the values into the equivalents formula:

\[ \text{Equivalents of } KMnO_4 = n \times M \times V = 5 \times 0.001 \, \text{mol/L} \times 0.006 \, \text{L} \]

Calculating this:

\[ \text{Equivalents of } KMnO_4 = 5 \times 0.001 \times 0.006 = 3 \times 10^{-5} \, \text{equivalents} \]

### Step 2: Calculate the equivalents of \( C_2O_4^{2-} \)

Since the reaction between \( KMnO_4 \) and \( C_2O_4^{2-} \) is a 1:1 reaction (1 equivalent of \( KMnO_4 \) reacts with 1 equivalent of \( C_2O_4^{2-} \)), the equivalents of \( C_2O_4^{2-} \) will be the same as the equivalents of \( KMnO_4 \).

Thus,

\[ \text{Equivalents of } C_2O_4^{2-} = 3 \times 10^{-5} \, \text{equivalents} \]

### Final Answer

- Equivalents of \( KMnO_4 \) required in the titration: \( 3 \times 10^{-5} \, \text{equivalents} \)
- Equivalents of \( C_2O_4^{2-} \) present in \( CaC_2O_4 \): \( 3 \times 10^{-5} \, \text{equivalents} \)