What volume of a 2.00 M KCl solution is required to prepare 500. mL of a 0.100 M KCl solution?

• A. 250.0 mL
• B. 100.0 mL
• C. 50.0 mL
• D. 25.0 mL

Answer: (D)

Verified step by step guidance

1. Identify the known values: the concentration of the stock solution \(C_1 = 2.00\,M\), the desired concentration of the diluted solution \(C_2 = 0.100\,M\), and the volume of the diluted solution \(V_2 = 500.\,mL\).
2. Recall the dilution equation which relates the concentrations and volumes before and after dilution: \(C_1 \times V_1 = C_2 \times V_2\), where \(V_1\) is the volume of the stock solution needed.
3. Rearrange the dilution equation to solve for \(V_1\): \(V_1 = \frac{C_2 \times V_2}{C_1}\).
4. Substitute the known values into the equation: \(V_1 = \frac{0.100\,M \times 500.\,mL}{2.00\,M}\).
5. Calculate \(V_1\) to find the volume of the 2.00 M KCl stock solution required to prepare 500 mL of 0.100 M KCl solution.