Let's solve the problem step by step: ### Step 1: Identify what is given and what is asked - **Given:** - Initial concentration (\(C_1\)) = 0.300 M NaF - Final concentration (\(C_2\)) = 0.0335 M NaF - Final volume (\(V_2\)) = 250.0 mL - **Asked:** - Initial volume (\(V_1\)) of 0.300 M NaF needed **Explanation:** We're asked to find out how much of a more concentrated solution is needed to make a diluted solution of a lower concentration and a specific volume. --- ### Step 2: Use the dilution equation The dilution equation is: \[ C_1 V_1 = C_2 V_2 \] **Explanation:** This equation relates the concentrations and volumes before and after dilution for a solution. --- ### Step 3: Plug in the values - \( C_1 = 0.300 \) M - \( V_1 = ? \) - \( C_2 = 0.0335 \) M - \( V_2 = 250.0 \) mL \[ 0.300 \times V_1 = 0.0335 \times 250.0 \] **Explanation:** We substitute the known values into the dilution equation. --- ### Step 4: Solve for \( V_1 \) \[ V_1 = \frac{0.0335 \times 250.0}{0.300} \] Calculate: \[ V_1 = \frac{8.375}{0.300} \] \[ V_1 = 27.92 \text{ mL} \] **Explanation:** We solved for the unknown volume by isolating \( V_1 \) and performing the calculation. --- ### **Final Answer** **27.9 mL** (rounded to three significant figures) --- **Summary:** To prepare 250.0 mL of 0.0335 M NaF from a 0.300 M NaF stock solution, you would need **27.9 mL** of the stock solution.