# 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 **dilution** implies that you take **part** of stock solution and add **parts** of water to get a total volume of diluted solution equal to **times** that of the stock solution.

#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 *diluted solution* will contain

#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)**