Question
Make 50.0 mL of a 0.500 M solution of acetic acid (CH3COOH) from the 1.00 M CH3COOH using a 50.0 mL graduated cylinder.
1. Show your calculations and explain how you made the acetic acid solution.
1. Show your calculations and explain how you made the acetic acid solution.
Answers
DrBob222
mL1 x M1 = mL2 x M2
50 x 0.5 = mL2 x 1.00
mL2 = 25 mL.
Add 25 mL of the 1.00 CH3COOH to the empty 50 mL graduated cylinder and add distilled water to the 50.0 mL mark. The result is 50.0 mL of 0.50M CH3COOH.
50 x 0.5 = mL2 x 1.00
mL2 = 25 mL.
Add 25 mL of the 1.00 CH3COOH to the empty 50 mL graduated cylinder and add distilled water to the 50.0 mL mark. The result is 50.0 mL of 0.50M CH3COOH.