Density of a sodium hydroxide solution The specific gravity of a $50\%$ aqueous solution $\ce{NaOH}$ is $\pu{1.5298 g cm-3}.$
To my understanding, the $50\%$ means $50$ mass fraction $w,$ i.e. $\pu{500 g}$ $\ce{NaOH}$ in $\pu{500 g}$ $\ce{H2O}:$
$$w(\ce{NaOH}) = \frac{m(\ce{NaOH})}{m(\ce{NaOH}) + m(\ce{H2O})} = \frac{\pu{500 g}}{\pu{500 g} + \pu{500 g}} = 0.5.\tag{1}$$
However, when I try to calculate the density of that solution I end up with
$$
\begin{align}
\rho(\ce{NaOH}) &amp;= \pu{2.13 g cm-3}; \\
\rho(\ce{H2O}) &amp;= \pu{1.00 g cm-3}.
\end{align}
$$
$$V = \frac{m(\ce{NaOH})}{\rho(\ce{NaOH})} + \frac{m(\ce{H2O})}{\rho(\ce{H2O})} = \pu{234.74 cm^3} + \pu{500 cm^3} = \pu{734.74 cm^3},\tag{2}$$
$$\rho_\text{sln} = \frac{\pu{1000 g}}{\pu{734.74 cm^3}} = \pu{1.361 g cm-3}.\tag{3}$$
What am I missing here?
What's your first miss is the definition of $50\%(w/w)$ $\ce{NaOH}$ solution (although it doesn't matter here). Actual definition is $\pu{50 g}$ of $\ce{NaOH}$ in $\pu{100 g}$ of solution. Since water is the solvent, your interpretation is correct.
Your second mistake is the lethal one. You have considered water and solid $\ce{NaOH}$ are additive. Assuming $V_\text{total} = V_\ce{NaOH} + V_\ce{H2O}$ is a mistake.
Using a given density of $\ce{NaOH}$ ($\pu{1.5298 g}$),we can calculate the volume of $\pu{1000 g}$ of solution: $\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$. Therefore, dissolving $\pu{500 g}$ of $\ce{NaOH}$ in $\pu{500 mL}$ of water (assuming temperature is $\pu{25 ^\circ C}$, hence density of water is $\pu{1.00 g/mL}$), the volume increased by only $\pu{153.68 mL}$. Do you still think volumes are additive?