Asked by Anonymous

Prepare 45.00ml of 0.125M gold III chloride solution, given 500.00ml of a 0.500M gold III chloride solution. Explain how you will prepare the solution

Answers

Answered by DrBob222
c1v1 = c2v2
0.500*x = 0.125*45
Solve for x = volume of the 0.500 M stuff you have. Pipet that amount into a 45 mL volumetric flask and make to the mark with DI water.
Yes, I know 45 mL volumetric flasks are not common. Neither are 11.25 mL pipets.
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