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# Dilutions

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#### Solution Preview

7-1. 1:2, first, then 1:3, the 1:8, for a total dilution factor of 1:2*3*8 or 1:48

7-2. 4:10, or 1:2.5 at first. An additional 1:10 will make the final dilution factor 1:25

7-3. 1:6 initial dilution for tube 1. Five more 1:2 dilutions are made, making the final dilution factor 1:6*2*2*2*2*2, or 1:192!

7-4. This is 6 1:3 serial dilutions, 1:3^6, for a final dilution of 1:729

7-5. Add 3 mL of water to each tube, and then make the initial dilution by adding 1 mL of stock. Take 1mL of the first dilution and add to the second tube, repeating until you reach the final tube, assuming we want a 1:4 dilution with each tube.

7-6. Add 0.5 mL of water to each tube. Perform the same procedure as in 7-5, except add 0.5 mL of the previous or stock solution

7-7. This is 1:10, so 10mg/dL

7-8. 1:6, then 1:4. Total dilution=1:24, so 1000/24= 41.7mg/dL

7-9. 2.0g/dL is 2000mg/dL, so we need a total 1:40 dilution factor. If we wanted 1L, then we would take ...

#### Solution Summary

15 dilution problems are included in this solution as well as 19 other problems related to concentrations. This is a valuable solution set which specifically deals with these concepts:

1) Calculating the dilution factor of serial dilutions
2) How to perform and calculate serial dilutions
3) Conversions of metric units (ng to mg, for example)
4) How to prepare solutions of varying molarity, normality, and %
5) The difference between %w/w and %v/v are explored briefly
6) How to change molarity to normality.

\$2.19

## Dilution Factors & Serial Dilution

Seven test ubes each containing 4.5 ml of water are lined up.

0.5 ml of yeast culture is pipetted in to test tube #1. Then 0.5 ml of the solution form test tube #1 is pipetted into test tube #2. Then 0.5 ml of the solution for test tube #2 is pipetted into test tube # 3 and so on until 0.5 ml of the solution from test tube #6 is pipetted into test tube #7.

Question #1 Calculate the dilution factor for all 7 test test tubes. Show all work/calculations.

0.1 ml of the solution from the serial dilution 10^-4, 10^-5 and 10^-6 was pipetted onto three separate agar plates.

For the 10^-4 dilution because 0.1 ml is 1/10 (10^-1) of a milliliter the dilution represented on the plate is a 10^-5 dilution, not a 10^-4 dilution, the same goes for the 10^-5 and 10^-6 dilution.

The plates were left to incubate for a week. On the plate with the 10^-4 dilution sixty-eight colonies formed. On the plate with the 10^-5 dilution eighteen colonies formed. one the plate with the 10^-6 dilution one colony formed.

Question # 2 Based on the numbers of colonies formed calculate the cells per ml in serial dilution 10^-4, 10^-5 and 10^-6,. Show all work/calculations.

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