`NaHC_(2)O_(4)` is neutralised by NaOH. It can also be oxidised by `KMnO_(4)` (in acidic medium). Equivalent weight is related to molecular weight (M) of `NaHC_(2)O_(4)` in these two reaction as :

Explanation

To solve the problem, we need to determine the relationship between the equivalent weight and molecular weight of `NaHC₂O₄` in two different reactions: when it is neutralized by `NaOH` and when it is oxidized by `KMnO₄` in acidic medium.

### Step 1: Determine the Molecular Weight of `NaHC₂O₄`
First, we calculate the molecular weight (M) of `NaHC₂O₄` (sodium hydrogen oxalate).
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol (2 Carbon atoms = 2 × 12 = 24 g/mol)
- Oxygen (O): 16 g/mol (4 Oxygen atoms = 4 × 16 = 64 g/mol)

Now, we add these together:
\[ M = 23 + 1 + 24 + 64 = 112 \text{ g/mol} \]

### Step 2: Reaction with `NaOH`
When `NaHC₂O₄` is neutralized by `NaOH`, the reaction can be written as:
\[ \text{NaHC}_2\text{O}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{O} \]

In this reaction, one hydrogen ion (H⁺) from `NaHC₂O₄` is replaced by sodium ion (Na⁺).

### Step 3: Determine the Valence Factor (N) for Neutralization
The valence factor (N) is the number of moles of H⁺ that are replaced or neutralized. Here, only one H⁺ is replaced:
\[ N = 1 \]

### Step 4: Calculate Equivalent Weight for Neutralization
The equivalent weight (EW) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{N} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{112 \text{ g/mol}}{1} = 112 \text{ g/equiv} \]

### Step 5: Reaction with `KMnO₄`
In the second reaction, `NaHC₂O₄` is oxidized by `KMnO₄` in acidic medium. The reaction can be simplified to focus on the carbon component:
\[ \text{NaHC}_2\text{O}_4 \rightarrow \text{CO}_2 + \text{other products} \]

### Step 6: Determine the Change in Oxidation State
- In `NaHC₂O₄`, the oxidation state of carbon is +3.
- In `CO₂`, the oxidation state of carbon is +4.

The change in oxidation state for each carbon atom is:
\[ \Delta = +4 - (+3) = +1 \]
Since there are two carbon atoms, the total change is:
\[ \Delta = 2 \times 1 = 2 \]

### Step 7: Determine the Valence Factor (N) for Oxidation
Thus, the valence factor (N) for the oxidation reaction is:
\[ N = 2 \]

### Step 8: Calculate Equivalent Weight for Oxidation
Using the formula for equivalent weight:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{N} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{112 \text{ g/mol}}{2} = 56 \text{ g/equiv} \]

### Summary of Results
1. For the neutralization reaction with `NaOH`:
- Equivalent Weight = Molecular Weight = 112 g/equiv
2. For the oxidation reaction with `KMnO₄`:
- Equivalent Weight = \(\frac{1}{2}\) Molecular Weight = 56 g/equiv

### Final Answer
- In the neutralization reaction, the equivalent weight is equal to the molecular weight.
- In the oxidation reaction, the equivalent weight is half of the molecular weight.