Question

A diluted solution is prepared by adding water to 375 mL of a 4.83 M sodium hydroxide solution. If the resultant solution is 3.67 M, how much water was added to the original solution? Report your answer to two decimal points in mL.

a) 185.00 mL
b) 150.00 mL
c) 225.00 mL
d) 200.00 mL

Answer 1

After performing calculations using the dilution equation M1V1 = M2V2, it was determined that approximately 120.00 mL of water was added to the original solution, which differs from the provided answer options.

Step-by-step explanation:

To find out how much water was added to the original solution, we can use the concept of dilution, which states that the number of moles of solute in the solution remains constant during the dilution process. The equation for dilution is M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume after dilution.

We have a 4.83 M (M1) sodium hydroxide solution with an initial volume of 375 mL (V1). The resultant solution has a molarity of 3.67 M (M2). We need to find the final volume (V2).

Using the equation M1V1 = M2V2, we have 4.83 M * 375 mL = 3.67 M * V2. Solving for V2 yields V2 = (4.83 M * 375 mL) / 3.67 M ≈ 495.00 mL. The amount of water added (Vwater) is the difference between the final and initial volumes, which is Vwater = V2 - V1 = 495.00 mL - 375 mL ≈ 120.00 mL. None of the provided answer options (a, b, c, d) are correct.