What volume of 5.00 M KI stock solution is required to prepare 250.0 mL of a 1.50 M KI solution?

• A. 83.3 mL
• B. 125.0 mL
• C. 37.5 mL
• D. 75.0 mL

Answer: (D)

Verified step by step guidance

1. Identify the known values: the concentration of the stock solution \(C_1 = 5.00\, M\), the desired concentration of the diluted solution \(C_2 = 1.50\, M\), and the volume of the diluted solution \(V_2 = 250.0\, mL\).
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: \(V_1 = \frac{1.50\, M \times 250.0\, mL}{5.00\, M}\).
5. Calculate \(V_1\) to find the volume of the 5.00 M KI stock solution required to prepare 250.0 mL of 1.50 M KI solution.