Question ab266

May 8, 2016

Here's what you'd need to do.

Explanation:

Well, the first thing to do here is calculate how many moles of sucrose you have in that target solution.

To do that, use the molarity and volume of the solution

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

n_"sucrose" = "0.348 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(250.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

${n}_{\text{sucrose" = "0.0870 moles sucrose}}$

Now, notice that the stock solution is more concentrated than the target solution. This means that you'd need a smaller volume of the concentrated solution to get the same number of moles of solute.

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies V_"solution" = n_"solute}} / c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

More specifically, you'd only need

${V}_{\text{stock" = (0.0870 color(red)(cancel(color(black)("moles"))))/(0.500 color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.174 L}}$

Expressed in milliliters, this will be equivalent to

0.174 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "174 mL"

So, you need to instruct your assistance to take a $\text{174-mL}$ sample of the stock solution, then add enough water to get its total volume of $\text{250.0 mL}$.

This will cause the concentration of the solution to decrease from $\text{0.500 M}$ to $\text{0.348 M}$, effectively diluting the solution by a factor of

$\text{D.F." = V_"final"/V_"initial}$

"D.F." = (250.0 color(red)(cancel(color(black)("mL"))))/(174color(red)(cancel(color(black)("mL")))) = 1.44 -># rounded to three sig figs