# Question 9a862

Mar 19, 2015

You would need b) 5.3 mL of your stock solution.

The first thing you need to do is figure out how much acrylamide you have in the target 20-mL, 8% solution.

$\text{%v/v" = "volume of acrylamide"/"volume of solution} \cdot 100$

V_("acrylamide") = (V_("solution") * "%v/v")/100 = "20 mL" * 8/100 = "1.6 mL"

Now you need to figure out what volume of the 30% stock solution contains that much acrylamine. Use the same equation as before, only this time solve for the volume of the solution

V_("solution") = (V_("acrylamine") * 100)/"%v/v" = "1.6 mL" * 100/30 = "5.3 mL"#

Watch out for sig figs as well, since you should give the answer with only 1 sig fig, the number of sig figs given for the 20 mL volume.