How many milliliters of 0.0500 M AgNO_3 solution are required to prepare 15.0 mL of a 0.0100 M AgNO_3 solution?

• A. 1.50 mL
• B. 3.00 mL
• C. 0.30 mL
• D. 7.50 mL

Answer: (B)

Verified step by step guidance

1. Identify the dilution problem where a more concentrated solution is diluted to a lower concentration. The key relationship is given by the dilution equation: \(C_1 V_1 = C_2 V_2\), where \(C_1\) and \(V_1\) are the concentration and volume of the initial (more concentrated) solution, and \(C_2\) and \(V_2\) are the concentration and volume of the final (diluted) solution.
2. Write down the known values: \(C_1 = 0.0500\, M\), \(C_2 = 0.0100\, M\), and \(V_2 = 15.0\, mL\). The unknown is \(V_1\), the volume of the concentrated 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: \(V_1 = \frac{0.0100\, M \times 15.0\, mL}{0.0500\, M}\)
5. Calculate \(V_1\) to find the volume of the 0.0500 M AgNO\(_3\) solution required to prepare 15.0 mL of 0.0100 M AgNO\(_3\).