Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurately prepared in a single step due to the small amount of mass that would need to be measured. Therefore, you decided to conduct this as a two step process. You will create a high concentration stock solution in one 10 mL flask. You then transfer 0.1 mL into a second 25 mL volumetric flask that will be the concentration of 6 ng/μL. What is the mass of chemical you need to add to the stock flask to achieve the desired concentration after dilution?

Question

Alexzander Lamb

Chemistry

22 April, 18:01

Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurately prepared in a single step due to the small amount of mass that would need to be measured. Therefore, you decided to conduct this as a two step process. You will create a high concentration stock solution in one 10 mL flask. You then transfer 0.1 mL into a second 25 mL volumetric flask that will be the concentration of 6 ng/μL. What is the mass of chemical you need to add to the stock flask to achieve the desired concentration after dilution?

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Answers (1)

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Rachel Faulkner · Answer 1

Mass of chemical = 1.5 mg

Explanation:

Step 1: First calculate the concentration of the stock solution required to make the final solution.

Using C1V1 = C2V2

C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution

C1 = C2V2/V1

C1 = (6 * 25) / 0.1

C1 = 1500 ng/μL = 1.5 μg/μL

Step 2: Mass of chemical added:

Mass of sample = concentration * volume

Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL

Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg

Therefore, mass of sample = 1.5 mg