Take 1mL of 1M NaCl solution, and add 9mL of water. Then take 10microL of the diluted solution and add 40microL of water. What is the dilution? And what is the final concentration of NaCl?
So I see how this is a 2:1 dilution, but I am having a hard time figuring out the final concentration of NaCl. Any help? Thanks.
To calculate the final concentration of NaCl, we need to consider the dilution factor at each step.
First, we have 1 ml of 1M NaCl solution. When we add 9ml of water, the total volume becomes 1ml + 9ml = 10ml. This means that the dilution factor at this step is 10ml/1ml = 10.
Next, we take 10µl of the diluted solution and add 40µl of water. The total volume now becomes 10µl + 40µl = 50µl. Therefore, the dilution factor at this step is 50µl/10µl = 5.
To find the total dilution factor, we multiply the dilution factors of each step: 10 x 5 = 50.
Now, let's calculate the final concentration of NaCl using the total dilution factor:
Final concentration = Initial concentration / Total dilution factor
The initial concentration of NaCl is 1M.
Final concentration = 1M / 50 = 0.02M
Therefore, the final concentration of NaCl in the diluted solution is 0.02M.