Question #ab266

1 Answer
May 8, 2016

Answer:

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