The volume of water which must be added to `0.4 dm^(3)` of 0.25 N oxalic acid in order to make it exactly decinormal is :

Options

A. `0.2 dm^(3)`
B. `0.4 dm^(3)`
C. `0.6 dm^(3)`
D. `0.8 dm^(3)`

Answer

To solve the problem of determining the volume of water that must be added to 0.4 dm³ of 0.25 N oxalic acid to make it exactly decinormal (0.1 N), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Normality of oxalic acid (N₁) = 0.25 N
- Volume of oxalic acid solution (V₁) = 0.4 dm³
- Desired normality (N₂) = 0.1 N

2. **Use the normality equation:**
The equation relating the normalities and volumes of two solutions is:
\[ N_1 \times V_1 = N_2 \times V_2 \]
Where:
- \(N_1\) = initial normality
- \(V_1\) = initial volume
- \(N_2\) = final normality
- \(V_2\) = final volume after adding water

3. **Substitute the known values into the equation:**
\[ 0.25 \, \text{N} \times 0.4 \, \text{dm}^3 = 0.1 \, \text{N} \times V_2 \]

4. **Calculate the left side:**
\[ 0.25 \times 0.4 = 0.1 \, \text{N} \times V_2 \]
\[ 0.1 = 0.1 \times V_2 \]

5. **Solve for \(V_2\):**
\[ V_2 = \frac{0.1}{0.1} = 1 \, \text{dm}^3 \]

6. **Calculate the volume of water to be added:**
The volume of water to be added is the difference between the final volume \(V_2\) and the initial volume \(V_1\):
\[ \text{Volume of water} = V_2 - V_1 = 1 \, \text{dm}^3 - 0.4 \, \text{dm}^3 = 0.6 \, \text{dm}^3 \]

### Final Answer:
The volume of water that must be added is **0.6 dm³**.

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