Preparation of aqueous solution 
In Chemistry class, Samantha finds that she can make a certain solution by mixing $.04\,\rm L$ of chemical $\mathrm{A}$ with $.02\,\rm L$ of water (this gives her $.06\,\rm L$ of the solution). She wants to make a total of $.48\,\rm L$ of this new solution. To do so, how many liters of water will she use?

How can I set this up? I'm having a hard time understanding this.
 A: That is the ratio of chemical $\mathrm{A}$ in the solution:
$$
\frac{.04}{.06} = \frac{2}{3}
$$
That means $\frac{2}{3}$ of the desired amount is chemical $\mathrm{A}$.
$$
\frac{2}{3} \times .48\,\mathrm{L} = .32\,\rm L
$$
A: Since it's been demonstrated that volumes are additive ($.04\,\rm L + .02\,\rm L = .06\,\rm L$, but this is not always the case in chemistry), it's very convenient to use volume concentration ($\phi_i$):
$$
\phi_i = \frac{V_i}{V} \tag{1}
$$
where $V_i$ – volume of chemical $\mathrm{A}$; $V$ – volume of $\mathrm{A}$ and volume of water $\mathrm{H_2O}$ together. Samantha wants to preserve the same concentration, hence $\phi_i(\mathrm{A}) = \mathrm{const}$:
$$
\frac{V_1(\mathrm{A})}{V_1(\mathrm{A}) + V_1(\mathrm{H_2O})} = \frac{V_2(\mathrm{A})}{V_2(\mathrm{A}) + V_2(\mathrm{H_2O})} \tag{2}
$$
Now, let's plug in the known values and solve the proportion for $V_2(\mathrm{A})$:
$$
\frac{.04\,\rm L}{.06\,\rm L} = \frac{V_2(\mathrm{A})}{.48\,\rm L} \quad \Rightarrow \quad V_2(\mathrm{A}) = .32\,\rm L \tag{3}
$$
To find the volume of water required, one need to subtract the volume of chemical $\mathrm{A}$ from the resulting volume of the solution $V_2$ (as it's been already established, volumes are additive):
$$
V(\mathrm{H_2O}) = V_2 - V(\mathrm{A}) = .48\,\mathrm{L} - .32\,\mathrm{L} = .16\,\rm L \tag{4}
$$
A: Firstly, we find that $\frac{.02}{.06}=\frac{1}{3}$ is the fraction of water in the solution. Thus, since Samantha wants $.48$ total liters of the solution, we have that she must use $.48 \times \frac{1}{3}=\boxed{0.16}$ liters of water.
