Question

In lab (write this down in your lab protocol), you will be given a stock solution that has a glucose concentration of 60 mg/dL. You need to make 1 mL of each of the following glucose concentrations: 0.94 mg/dL, 1.88 mg/dL, 3.75 mg/dL, 7.5 mg/dL, 15 mg/dL, and 30 mg/dL. Distilled water will be the solvent in these dilutions. Enter your numbers only. 1. What is the dilution factor for this serial dilution? 2. What is the V2 for this serial dilution in mL? 3. What is the V1 for this serial dilution in mL?

Answer 1

Answer:

1. The dilution factor for the serial dilution = 2

2. V2 = 1 mL

3. V1 = 0.5 mL

Explanation:

1. Dilution factor is the ratio of the initial concentration to the final concentration.

Dilution factor = initial concentration / final concentration

First dilution: initial concentration = 60 mg/dL

final concentration = 30 mg/dL

Dilution factor = 60 mg/dL / 30 mg/dL = 2

Second dilution: initial concentration = 30 mg/dL

final concentration = 15 mg/dL

Dilution factor = 30 mg/dL / 15 mg/dL = 2

Therefore, the dilution factor for the serial dilution = 2

2. From the dilution formula, C1V1 = C2V2; V2 = final volume to be prepared.

Since 1 mL of the various glucose solutions are to be prepared, the final concentration, V2 = 1 mL

3. From the dilution formula, C1V1 = C2V2; V1 = initial concentration of the solution to be prepared.

C1/C2 = V2/V1

Since the dilution factor, C1/C2 is 2, V2/V1 = 2

V1 = V2/2

V1 = 1 mL / 2

V1 = 0.5 mL