# How is a 5*%v/v solution of methanol in water prepared?

Take $45 \cdot m L$ of water and add $5 \cdot m L$ of methanol. The volumes should be additive and you are left with 5% methanol, $\frac{v}{v}$.
Of course that method works to a first approximation, and here the dilution is $\text{volume of solute"/"volume of solution}$. If you want $\text{mass of solute"/"mass of solution}$ you add appropriate masses. The actual concentrations will not be that different.