A 100.0 mL sample is prepared from a 250.0 mL stock solution by pipetting 5.00 mL of the stock solution and diluting it until it had a molarity of 2.5 * 10^-4 M. What is the molarity of the original stock solution?

• A. 0.0125 M
• B. 0.00625 M
• C. 0.0250 M
• D. 0.0025 M

Answer: (A)

Verified step by step guidance

1. Identify the relationship between the concentrations and volumes of the stock and diluted solutions using the dilution equation: \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( V_1 \) are the concentration and volume of the stock solution, and \( C_2 \) and \( V_2 \) are the concentration and volume of the diluted solution.
2. Substitute the known values into the dilution equation. Here, \( V_1 = 5.00 \) mL, \( C_2 = 2.5 \times 10^{-4} \) M, and \( V_2 = 100.0 \) mL.
3. Rearrange the equation to solve for \( C_1 \), the concentration of the original stock solution: \( C_1 = \frac{C_2V_2}{V_1} \).
4. Perform the calculation by substituting the values: \( C_1 = \frac{(2.5 \times 10^{-4} \text{ M})(100.0 \text{ mL})}{5.00 \text{ mL}} \).
5. Simplify the expression to find the molarity of the original stock solution.