# Question 31802

Nov 15, 2016

Here's how you can do that.

#### Explanation:

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 $\text{1 ppm}$ solution will have $\text{1 g}$ of solute in ${10}^{6} \text{g}$ of solvent.

If you take water's density to be equal to ${\text{1.0 g mL}}^{- 1}$, 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 $\text{20 ppm}$ amoxicillin solution will contain $\text{20 g}$ of amoxicillin for every ${10}^{6} \text{g}$ 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 $1 \cdot {10}^{- 4} \text{g}$ of solute. Your stock solution is 1%"m/v" amoxicillin, which basically means that it contains $\text{1 g}$ of solute for every $\text{100 mL}$ 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 $\text{0.01 mL}$ of your $\text{1% m/v}$ amoxicillin stock solution and adding enough water to get the final volume to $\text{5 mL}$.

This will get you $\text{5 mL}$ of $\text{20 ppm}$ amoxicillin solution.