Question #a673e
1 Answer
Here's what I got.
Explanation:
For starters, I think that there's a typo in your question. Instead of
I say this because a
Let's assume that you are indeed working with a
#color(blue)(ul(color(black)("1:20 dilution " implies V_"diluted" = 20 * V_"stock")))#
In your case, a
To make the calculations easier, pick a
#V_"diluted" = 20 * "1 L" = "20 L"#
So if
#1 color(red)(cancel(color(black)("L diluted solution"))) * "5 nmoles solute"/(20color(red)(cancel(color(black)("L diluted solution")))) = "0.25 nmoles solute"#
So this solution contains
You can play around with this value a bit to convince yourself that none of the possible answers match.
Now, look what happens with a
#1 color(red)(cancel(color(black)("L diluted solution"))) * "5 mmoles solute"/(20color(red)(cancel(color(black)("L diluted solution")))) = "0.25 mmoles solute"#
The diluted solution would thus have a concentration of
#(0.25 color(red)(cancel(color(black)("mmoles"))))/(1color(red)(cancel(color(black)("L")))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(white)(.)"mL") * (1color(red)(cancel(color(black)("mole"))))/(10^3color(red)(cancel(color(black)("mmoles")))) * (10^9color(white)(.)"nmoles")/(1color(red)(cancel(color(black)("mole")))) = "250 nmoles/mL"#
In this case, the answer would be (b)