# Question 17e71

Nov 17, 2015

$\text{80.0 mL}$

#### Explanation:

The new concentration is actually given to you, $\text{1.50 M}$. 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 $\text{6.00 M}$ and that you take a volume of $\text{20.0 mL}$ of this solution. The first thing to do here is determine how many moles of hydrochloric acid you have in this sample.

$c = \frac{n}{V} \implies n = c \cdot V$

$n = 6.00 \text{moles"/color(red)(cancel(color(black)("L"))) * 20.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.120 moles HCl}$

Now the question is what volume of the target solution would contain the same number of moles of $\text{HCl}$.

$c = \frac{n}{V} \implies V = \frac{n}{c}$

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 $\text{20.0-mL}$ sample of the $\text{6.00-M}$ solution and add enough water to make the total volume of the resulting solution equal to $\text{80.0 mL}$.