To solve this problem, we need to find the volume of the 0.2 M $H_2SO_4$ solution that should be mixed with 400 mL of the 0.6 M $H_2SO_4$ solution to obtain a mixed solution with a molarity of 0.36 M.

We use the formula for mixing solutions:

$M_1V_1 + M_2V_2 = M_f(V_1 + V_2)$

where:

• $M_1$ is the molarity of the first solution ($0.2$ M),
• $V_1$ is the volume of the first solution (to be found),
• $M_2$ is the molarity of the second solution ($0.6$ M),
• $V_2$ is the volume of the second solution (400 mL),
• $M_f$ is the final molarity of the mixed solution ($0.36$ M),
• $V_1 + V_2$ is the total volume of the mixed solution.

Plugging in the known values:

$0.2 \times V_1 + 0.6 \times 400 = 0.36 \times (V_1 + 400)$

Let's solve for $V_1$.The required volume of the 0.2 M $H_2SO_4$ solution is 600 mL.

Therefore, the correct answer is:

E. 600 mL

Would you like more details or have any questions about this solution?

Here are 5 related questions to consider:

1. What is the final volume of the mixed solution?
2. How would the answer change if the final molarity required was 0.4 M?
3. What is the total number of moles of $H_2SO_4$ in the mixed solution?
4. How would you prepare a solution with a molarity of 0.5 M using different concentrations?
5. Can this method be applied to mixtures of other substances?

Tip: When mixing solutions of different concentrations, always ensure the final volume and concentration meet the desired requirements using the dilution equation.