The volume of water to be added to 400 ml of N/8 HCl to make it exactly N/12 is -

Options

A. 400 ml
B. 300 ml
C. 200 ml
D. 100 ml

Explanation

To solve the problem of how much water to add to 400 ml of N/8 HCl to dilute it to N/12, we can use the concept of dilution which is based on the relationship between normality and volume.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume (V1) = 400 ml
- Initial normality (N1) = N/8
- Final normality (N2) = N/12

2. **Use the Dilution Formula:**
The formula for dilution is given by:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where:
- \(N_1\) = initial normality
- \(V_1\) = initial volume
- \(N_2\) = final normality
- \(V_2\) = final volume

3. **Substitute the Known Values:**
- Substitute \(N_1 = \frac{1}{8}\), \(V_1 = 400 \, \text{ml}\), and \(N_2 = \frac{1}{12}\) into the equation:
\[ \left(\frac{1}{8}\right) \times 400 = \left(\frac{1}{12}\right) \times V_2 \]

4. **Calculate the Left Side:**
- Calculate \( \frac{1}{8} \times 400 \):
\[ \frac{400}{8} = 50 \]

5. **Set Up the Equation:**
- Now we have:
\[ 50 = \left(\frac{1}{12}\right) \times V_2 \]

6. **Solve for \(V_2\):**
- Rearranging gives:
\[ V_2 = 50 \times 12 = 600 \, \text{ml} \]

7. **Calculate the Volume of Water to Add:**
- The volume of water to be added is:
\[ \text{Volume of water} = V_2 - V_1 = 600 \, \text{ml} - 400 \, \text{ml} = 200 \, \text{ml} \]

### Final Answer:
The volume of water to be added is **200 ml**.

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