Equivalent weight of `KHC_(2)O_(4).3NaHC_(2)O_(4)` in reaction with acidic `KMnO_(4)` is `(M=` Molar mass )

Options

A. `(M)/(8)`
B. `(M)/(4)`
C. `M`
D. `(M)/(3)`

Explanation

To find the equivalent weight of `KHC_(2)O_(4).3NaHC_(2)O_(4)` in its reaction with acidic `KMnO_(4)`, we will follow these steps:

### Step 1: Identify the Components and Their Roles
The compound `KHC_(2)O_(4).3NaHC_(2)O_(4)` contains potassium (K), sodium (Na), hydrogen (H), carbon (C), and oxygen (O). In the reaction with acidic `KMnO_(4)`, this compound acts as a reducing agent.

### Step 2: Determine the Oxidation States of Carbon
1. **For `KHC_(2)O_(4)`**:
- The oxidation state of K = +1
- The oxidation state of H = +1
- Let the oxidation state of C be x (for 2 carbon atoms, it is 2x).
- The oxidation state of O = -2 (for 4 oxygen atoms, it is -8).

Setting up the equation:
\[ 1 + 1 + 2x - 8 = 0 \]
Simplifying gives:
\[ 2x - 6 = 0 \implies 2x = 6 \implies x = 3 \]
So, the oxidation state of carbon in `KHC_(2)O_(4)` is +3.

2. **For `3NaHC_(2)O_(4)`**:
- The oxidation state of Na = +1
- The oxidation state of H = +1
- Let the oxidation state of C be x (for 2 carbon atoms, it is 2x).
- The oxidation state of O = -2 (for 4 oxygen atoms, it is -8).

Setting up the equation:
\[ 1 + 1 + 2x - 8 = 0 \]
Simplifying gives:
\[ 2x - 6 = 0 \implies 2x = 6 \implies x = 3 \]
So, the oxidation state of carbon in `3NaHC_(2)O_(4)` is also +3.

### Step 3: Calculate the Change in Oxidation State
In the reaction, carbon is oxidized from +3 to +4. Since there are 4 carbon atoms in total (2 from `KHC_(2)O_(4)` and 2 from `3NaHC_(2)O_(4)`), the total change in oxidation state is:
\[ \Delta \text{oxidation state} = 4 \text{ (from +3 to +4)} \]

### Step 4: Calculate the Equivalent Weight
The formula for equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass (M)}}{n_f} \]
where \( n_f \) is the number of electrons transferred per molecule in the reaction. Here, \( n_f = 4 \) (the total change in oxidation state).

Thus, the equivalent weight of `KHC_(2)O_(4).3NaHC_(2)O_(4)` is:
\[ \text{Equivalent weight} = \frac{M}{4} \]

### Final Answer
The equivalent weight of `KHC_(2)O_(4).3NaHC_(2)O_(4)` in reaction with acidic `KMnO_(4)` is \( \frac{M}{4} \).

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