Question

Question #17e71

Answer 1

`"80.0 mL"`

Explanation:

The new concentration is actually given to you, `"1.50 M"`. The problem wants you to determine what volume of the stock solution must be diluted in order to prepare that target solution.

You know that you when dilute a solution, you essentially keep the number of moles of solute constant and increase the volume of the solution.

Now, you know that your stock solution has a concentration of `"6.00 M"` and that you take a volume of `"20.0 mL"` of this solution. The first thing to do here is determine how many moles of hydrochloric acid you have in this sample.

`c = n/V implies n = c * V`

`n = 6.00"moles"/color(red)(cancel(color(black)("L"))) * 20.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.120 moles HCl"`

Now the question is what volume of the target solution would contain the same number of moles of `"HCl"`.

`c = n/V implies V = n/c`

`V = (0.120color(red)(cancel(color(black)("moles"))))/(1.50color(red)(cancel(color(black)("moles")))/"L") = "0.080 L"`

Convert this to mililiters to get

`V = 0.080color(red)(cancel(color(black)("L"))) * "1000 mL"/(1color(red)(cancel(color(black)("L")))) = color(green)("80.0 mL")`

So, to prepare this solution, you would take the `"20.0-mL"` sample of the `"6.00-M"` solution and add enough water to make the total volume of the resulting solution equal to `"80.0 mL"`.