A student prepared a stock solution by dissolving 25.00 g of NaOH in enough water to make 150.0 mL solution. The student took 20.0 mL of the stock solution and diluted it with enough water to make 250.0 mL solution. Finally taking 75.0 mL of that solution and dissolving it in water to make 500 mL solution. What is the concentration of NaOH for this final solution? (MW of NaOH:40.00 g/mol).

• A. 0.0500 M
• B. 0.025 M
• C. 0.005 M
• D. 0.500 M
• E. 0.0100 M

Answer: (A)

Verified step by step guidance

1. Calculate the molarity of the stock solution. First, determine the number of moles of NaOH using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here, the mass is 25.00 g and the molar mass of NaOH is 40.00 g/mol.
2. Once you have the moles of NaOH, calculate the molarity of the stock solution using the formula: \( \text{Molarity} = \frac{\text{moles}}{\text{volume in liters}} \). Convert the volume from mL to L by dividing by 1000.
3. Determine the molarity of the first diluted solution. Use the dilution formula: \( M_1V_1 = M_2V_2 \), where \( M_1 \) and \( V_1 \) are the molarity and volume of the stock solution, and \( M_2 \) and \( V_2 \) are the molarity and volume of the diluted solution. Solve for \( M_2 \).
4. Calculate the molarity of the second diluted solution using the same dilution formula: \( M_1V_1 = M_2V_2 \). Here, \( M_1 \) and \( V_1 \) are the molarity and volume of the first diluted solution, and \( M_2 \) and \( V_2 \) are the molarity and volume of the second diluted solution. Solve for \( M_2 \).
5. The molarity obtained from the last step is the concentration of NaOH in the final solution. Compare this value with the given options to identify the correct answer.