An aqueous solution that is 30.0%NaOH by mass has a density of $1.33 gm{L}^{\mathit{-}\mathit{1}}$ What is the molarity of NaOH in this solution?

Here, volume of solution is the ratio of mass of the solution to the density of the solution. Thus,

${\mathrm{vol}}^m{\text{ of Sol}}^n= \frac{\mathrm{mass} \mathrm{of} {\mathrm{Sol}}^n}{\& (\mathrm{density} \mathrm{of} {\mathrm{Sol}}^n}$

$\Rightarrow \frac{100\mathrm{g}}{1.330}\mathrm{mL}$

Molarity = $=\frac{.75}{\frac{100}{1.300}}\times 1000$

$\begin{array}{r}\Rightarrow \frac{750\times 1.333}{100}\\ \Rightarrow 9.975=9.98\end{array}$