Respuesta :
Answer:
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.
Explanation:
Convert the given original molarity to molar as follows.
[tex]500mM = 500mM \times (\frac{1M}{1000M})= 0.5M[/tex]
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:
[tex]M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M[/tex]
10 mL of this solution is diluted to 250 ml
[tex]M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M[/tex]
Convert μM :
[tex]2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M[/tex]
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:
[tex]M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M[/tex]
10 mL of this solution is diluted to 1000 ml
[tex]M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M[/tex]
Convert μM :
[tex]2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M[/tex]
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:
[tex]M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M[/tex]
5 mL of this solution is diluted to 1000 ml
[tex]M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M[/tex]
Convert μM :
[tex]0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M[/tex]
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:
[tex]M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M[/tex]
10 mL of this solution is diluted to 500 ml
[tex]M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M[/tex]
Convert μM :
[tex]2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M[/tex]
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:
[tex]M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M[/tex]
10 mL of this solution is diluted to 1000 ml
[tex]M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M[/tex]
Convert μM :
[tex]2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M[/tex]
Therefore, The dilution scheme will result in a 200μM solution.
