Question

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your disposal, you have 5-mL and 10-mL transfer pipets and volumetric flasks of sizes 100, 250, 500, and 1000mL. Which of the following serial dilutions will give you the 200.0μM solution?

Answer 1

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Step-by-step explanation:

Convert the given original molarity to molar as follows.

`500mM = 500mM * ((1M)/(1000M))= 0.5M`

Consider the following serial dilutions.

1)

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

`M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5.00mL))/(500 mL)= 5 * 10^(-3)M`

10 mL of this solution is diluted to 250 ml

`M_(final)= (M_(2)V_(2))/(V_(final))= ((5 * 10^(-3)M)(10.0mL))/(250 mL)= 2 * 10^(-4)M`

Convert μM :

`2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M`

Therefore, The dilution scheme will result in a 200μM solution.

2)

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

`M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5.00mL))/(100 mL)= 2.5 * 10^(-2)M`

10 mL of this solution is diluted to 1000 ml

`M_(final)= (M_(2)V_(2))/(V_(final))= ((2.5 * 10^(-2)M)(10.0mL))/(1000 mL)= 2.5 * 10^(-4)M`

Convert μM :

`2.5 * 10^(-4)M = (2.5 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 250 \mu M`

Therefore, The dilution scheme will not result in a 200μM solution.

3)

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

`M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(10mL))/(100 mL)= 0.05M`

5 mL of this solution is diluted to 1000 ml

`M_(final)= (M_(2)V_(2))/(V_(final))= ((0.05M)(5mL))/(1000 mL)= 0.25 * 10^(-4)M`

Convert μM :

`0.25 * 10^(-4)M = (0.25 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 25 \mu M`

Therefore, The dilution scheme will not result in a 200μM solution.

4)

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

`M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(5mL))/(250 mL)= 0.01M`

10 mL of this solution is diluted to 500 ml

`M_(final)= (M_(2)V_(2))/(V_(final))= ((0.01M)(10mL))/(500 mL)= 2 * 10^(-4)M`

Convert μM :

`2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M`

Therefore, The dilution scheme will result in a 200μM solution.

5)

Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

`M_(2)= (M_(1)V_(1))/(V_(2))= ((0.5M)(10mL))/(250 mL)= 0.02M`

10 mL of this solution is diluted to 1000 ml

`M_(final)= (M_(2)V_(2))/(V_(final))= ((0.02M)(10mL))/(1000 mL)= 2 * 10^(-4)M`

Convert μM :

`2 * 10^(-4)M = (2 * 10^(-4)M)((1 \mu M)/(10^(-6)M))= 200 \mu M`

Therefore, The dilution scheme will result in a 200μM solution.