# Question 1f9b2

May 1, 2017

${\text{30 mg L}}^{- 1}$

#### Explanation:

The first thing you need to do here is to figure out the number of milligrams of solute present in the $\text{20-mL}$ aliquot of the ${\text{1000 mg L}}^{- 1}$ standard solution.

As you know, solutions are homogeneous mixtures, which means that they have the same composition throughout. You can thus use the initial concentration of the solution to figure out the mass of the solute in the first aliquot

20 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "1000 mg"/(1 color(red)(cancel(color(black)("L")))) = "20 mg"

Now, you dilute this aliquot to a final volume of $\text{100 mL}$. The trick here is to realize that the mass of solute remains unchanged after the dilution.

This means that the standard $\text{A}$ solution contains $\text{20 mg}$ of solute in $\text{100 mL}$ of solution.

Consequently, the $\text{15-mL}$ aliquout of Standard $\text{A}$ solution will contain

15 color(red)(cancel(color(black)("mL"))) * "20 mg"/(100color(red)(cancel(color(black)("mL")))) = "3 mg"

of solute. You dilute this solution to a final volume of $\text{100 mL}$, which means that the Standard $\text{B}$ solution will now contain $\text{3 mg}$ of solute in $\text{100 mL}$ of solution.

Now, use the known composition of the Standard $\text{B}$ solution to figure out how many milligrams of solute would be present in $\text{1 L}$ of solution

1 color(red)(cancel(color(black)("L"))) * (10^3 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "3 mg"/(100 color(red)(cancel(color(black)("mL")))) = "30 mg"#

Therefore, you can say that the concentration of the Standard $\text{B}$ solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{Standard B" = "30 mg L}}^{- 1}}}}$