Question

Question #ab266

Answer 1

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

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(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_"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.

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))`

More specifically, you'd only need

`V_"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 `"174-mL"` sample of the stock solution, then add enough water to get its total volume of `"250.0 mL"`.

This will cause the concentration of the solution to decrease from `"0.500 M"` to `"0.348 M"`, effectively diluting the solution by a factor of

`"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