Join thousands of students who trust us to help them ace their exams!Watch the first video
Multiple Choice
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
Verified step by step guidance
1
Calculate the molarity of the stock solution: First, determine the number of moles of NaOH in the stock solution by using the formula: moles = mass (g) / molar mass (g/mol). Here, the mass of NaOH is 25.00 g and the molar mass is 40.00 g/mol.
Calculate the molarity of the stock solution: Use the formula for molarity, M = moles of solute / liters of solution. Convert the volume of the stock solution from mL to L (150.0 mL = 0.150 L) and use the moles calculated in the previous step to find the molarity.
Determine the concentration after the first dilution: Use the dilution formula M1V1 = M2V2, where M1 is the molarity of the stock solution, V1 is the volume of the stock solution taken (20.0 mL = 0.020 L), M2 is the molarity after dilution, and V2 is the final volume after dilution (250.0 mL = 0.250 L). Solve for M2.
Determine the concentration after the second dilution: Again, use the dilution formula M1V1 = M2V2. Here, M1 is the molarity after the first dilution, V1 is the volume taken from the first dilution (75.0 mL = 0.075 L), M2 is the molarity after the second dilution, and V2 is the final volume after the second dilution (500.0 mL = 0.500 L). Solve for M2.
The molarity calculated in the previous step is the concentration of NaOH in the final solution. Compare this value with the given options to identify the correct answer.