If you dilute 18.0 mL of the stock solution from part a) to a final volume of 3.00 L, what will be the concentration of the diluted solution?
3000/18 times the volume, so 18/3000 times the concentration
Well, let's see. If we take 18 mL of the stock solution and dilute it to a final volume of 3.00 L, we'll have a lot more liquid to work with. It's like ordering a large soda instead of a small one - you get way more liquid, but the taste is still the same. In this case, the concentration of the diluted solution would be much lower. So, to answer your question, the concentration of the diluted solution would be diluted... just like my sense of humor when I'm trying to be serious.
To find the concentration of the diluted solution, we need to use the equation:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = concentration of the diluted solution
V2 = volume of the diluted solution
From the information given:
C1 = concentration of the stock solution = ?? (not provided in the question)
V1 = volume of the stock solution = 18.0 mL = 0.018 L
C2 = concentration of the diluted solution = ?? (what we need to find)
V2 = volume of the diluted solution = 3.00 L
Using the equation, we can rewrite it as:
C1 * V1 = C2 * V2
Substituting the given values:
C1 * 0.018 L = C2 * 3.00 L
Simplifying the equation:
C1 = (C2 * 3.00 L) / 0.018 L
Finally, we can solve for C2 (concentration of the diluted solution) by rearranging the equation:
C2 = (C1 * 0.018 L) / 3.00 L
However, since the concentration of the stock solution (C1) is not provided in the question, it is not possible to calculate the concentration of the diluted solution (C2) without that information.
To find the concentration of the diluted solution, we can use the formula:
C1V1 = C2V2
where:
C1 = initial concentration of the stock solution
V1 = initial volume of the stock solution
C2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
From the information given:
C1 = concentration of the stock solution from part a)
V1 = initial volume of the stock solution = 18.0 mL = 0.018 L
V2 = final volume of the diluted solution = 3.00 L
Now, let's substitute the known values into the formula:
C1V1 = C2V2
C1 * 0.018 L = C2 * 3.00 L
We can isolate C2 by dividing both sides of the equation by V2:
C2 = (C1 * V1) / V2
Substituting the known values:
C2 = (C1 * 0.018 L) / 3.00 L
Now, you just need to know the value of C1 (concentration of the stock solution from part a) to calculate C2 using the formula above.