You are given 500 mL of 2N HCl and 500 mL of 5N HCl. What will be the maximum volume of 3 M HCl that you can make from these two solutions ?

Options

A. 250mL
B. 750mL
C. 500mL
D. 1000mL

Explanation

To solve the problem of determining the maximum volume of 3 M HCl that can be made from the given solutions, we will use the concept of normality and the dilution equation.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- You have 500 mL of 2N HCl (Normality N1 = 2N, Volume V1 = 500 mL).
- You have 500 mL of 5N HCl (Normality N2 = 5N, Volume V2 = 500 mL).
- You want to prepare a solution with a normality of 3N (N = 3N).

2. **Use the Dilution Equation:**
The dilution equation can be expressed as:
\[ N_1V_1 + N_2V_2 = N_fV_f \]
where:
- \(N_1\) = Normality of the first solution (2N)
- \(V_1\) = Volume of the first solution (to be determined)
- \(N_2\) = Normality of the second solution (5N)
- \(V_2\) = Volume of the second solution (to be determined)
- \(N_f\) = Final normality (3N)
- \(V_f\) = Final volume of the mixture (to be determined)

3. **Set Up the Equation:**
Let \(V_1\) be the volume of the 2N solution used and \(V_2\) be the volume of the 5N solution used. The total volume \(V_f\) will be \(V_1 + V_2\).

The equation becomes:
\[ 2V_1 + 5V_2 = 3(V_1 + V_2) \]

4. **Rearrange the Equation:**
Expanding the right-hand side gives:
\[ 2V_1 + 5V_2 = 3V_1 + 3V_2 \]
Rearranging this gives:
\[ 5V_2 - 3V_2 = 3V_1 - 2V_1 \]
Simplifying leads to:
\[ 2V_2 = V_1 \]

5. **Volume Constraints:**
Since you have 500 mL of each solution, we know:
- \(V_1 \leq 500\) mL
- \(V_2 \leq 500\) mL

From \(V_1 = 2V_2\), substituting gives:
\[ 2V_2 \leq 500 \implies V_2 \leq 250 \text{ mL} \]

6. **Calculate the Maximum Volume:**
If \(V_2 = 250\) mL, then:
\[ V_1 = 2 \times 250 = 500 \text{ mL} \]
Therefore, the total volume \(V_f\) is:
\[ V_f = V_1 + V_2 = 500 + 250 = 750 \text{ mL} \]

### Final Answer:
The maximum volume of 3 M HCl that can be made from the two solutions is **750 mL**.