Question #31802

Your first goal here will be to determine how many grams of amoxicillin you need in order to have in your target solution.

A solution's parts per million concentration, ppm, expresses the number of grams of solute present in #10^6# grams of solvent. Simply put, a #"1 ppm"# solution will have #"1 g"# of solute in #10^6"g"# of solvent.

If you take water's density to be equal to #"1.0 g mL"^(-1)#, you can say that the target solution contains

#5 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "5 g"#

of water. Notice that because your solution contains a very, very small amount o amoxicillin, you can safely assume that the mass of the solution is equal to the mass of the water.

Now, a #"20 ppm"# amoxicillin solution will contain #"20 g"# of amoxicillin for every #10^6"g"# of water. You can thus say that your sample will contain

#5 color(red)(cancel(color(black)("g water"))) * "20 g amoxicillin"/(10^6color(red)(cancel(color(black)("g water")))) = 1 * 10^(-4)"g amoxicillin"#

At this point, you should focus on determining the volume of amoxicillin solution that contains #1 * 10^(-4)"g"# of solute. Your stock solution is #1%"m/v"# amoxicillin, which basically means that it contains #"1 g"# of solute for every #"100 mL"# of solution.

You can thus say that you need

#1 * 10^(-4) color(red)(cancel(color(black)("g amoxicillin"))) * "100 mL stock"/(1color(red)(cancel(color(black)("g amoxicillin")))) = "0.01 mL stock"#

Therefore, you can prepare your solution by taking #"0.01 mL"# of your #"1% m/v"# amoxicillin stock solution and adding enough water to get the final volume to #"5 mL"#.

This will get you #"5 mL"# of #"20 ppm"# amoxicillin solution.