What volumes of 2 M and 6 M solutions of HCl have to be mixed to prepare 500 mL of a 3 M solutions? Disgard the change in the volume in mixing .

Explanation

To solve the problem of mixing 2 M and 6 M solutions of HCl to prepare 500 mL of a 3 M solution, we can follow these steps:

### Step 1: Define Variables
Let:
- \( V_1 \) = volume of the 2 M solution (in mL)
- \( V_2 \) = volume of the 6 M solution (in mL)

Since the total volume of the mixture must be 500 mL, we can express \( V_2 \) in terms of \( V_1 \):
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