To prepare 600 mL of 2 N solution of `NH_(4)OH`, volume of 10 N `NH_(4)OH` required is

Options

A. 60 mL
B. 120 mL
C. 300 mL
D. 600 mL

Explanation

To solve the problem of how much volume of 10 N NH₄OH is required to prepare 600 mL of a 2 N solution of NH₄OH, we can use the dilution formula:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Normality of the concentrated solution (N₁) = 10 N
- Normality of the diluted solution (N₂) = 2 N
- Volume of the diluted solution (V₂) = 600 mL

2. **Use the Dilution Formula:**
The dilution formula is given by:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where:
- \(N_1\) = Normality of the concentrated solution
- \(V_1\) = Volume of the concentrated solution required
- \(N_2\) = Normality of the diluted solution
- \(V_2\) = Volume of the diluted solution

3. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ 10 \, \text{N} \times V_1 = 2 \, \text{N} \times 600 \, \text{mL} \]

4. **Calculate the Right Side:**
Calculate the right side of the equation:
\[ 2 \, \text{N} \times 600 \, \text{mL} = 1200 \, \text{N mL} \]

5. **Set Up the Equation:**
Now the equation looks like this:
\[ 10 \, \text{N} \times V_1 = 1200 \, \text{N mL} \]

6. **Solve for \(V_1\):**
To find \(V_1\), divide both sides by 10 N:
\[ V_1 = \frac{1200 \, \text{N mL}}{10 \, \text{N}} = 120 \, \text{mL} \]

7. **Conclusion:**
Therefore, the volume of 10 N NH₄OH required is **120 mL**.