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If a chemist wishes to dilute a 1.000 x 103 mg/L stock solution to prepare 5.000...

If a chemist wishes to dilute a 1.000 x 103 mg/L stock solution to prepare 5.000 x 102 mL of a working standard that has a concentration of 6.250 mg/L, what volume of the 1.000 x 103 mg/L standard solution is needed?

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Answer #1

concentration of stock solution C1 = 1.000 x 10^3 mg/L

volume = V1 = ??

concentration of working solution C2 = 6.250 mg/L

volume V2 = 5.000 x 10^2 mL

C1 V1 = C2 V2

1.000 x 10^3 x V1 = 6.250 x 5.000 x 10^2

V1 = 3.125 mL

volume of standard solution = 3.125 mL

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