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

$$ \begin{align} \rho(\ce{NaOH}) &= \pu{2.13 g cm-3}; \\ \rho(\ce{H2O}) &= \pu{1.00 g cm-3}. \end{align} $$

$$V = \frac{m(\ce{NaOH})}{\rho(\ce{NaOH})} + \frac{m(\ce{H2O})}{\rho(\ce{H2O})} = \pu{234.74 cm^3} + \pu{500 cm^3} = \pu{734.74 cm^3},\tag{2}$$

$$\rho_\text{sln} = \frac{\pu{1000 g}}{\pu{734.74 cm^3}} = \pu{1.361 g cm-3}.\tag{3}$$

What am I missing here?

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    $\begingroup$ You don't calculate density. $\endgroup$ – Ivan Neretin Aug 31 '20 at 10:21
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    $\begingroup$ Densities are rarely, if ever, directly proportional to mass percentage of a solute. You need to determine the values experimentally. $\endgroup$ – MaxW Aug 31 '20 at 10:22
  • $\begingroup$ Plotting the densities to mass percentage creates an almost linear function (in this case). It is considered proportional in many cases. $ϱ_{sol}=0.0106W_{NaOH} + 1.0034$ $\endgroup$ – Andrew Kovács Aug 31 '20 at 10:26
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    $\begingroup$ Thanks for the input so far. I'm aware that density needs to be measured. I'm thinking of mixing ethanol and water, where the volume of the solution will change. I was suspecting that I get the weight percentage part wrong. My calculations would yield that 1 L of 50% solution contains 680 g NaOH. According to the table I originally linked it would be 764 g / L. So basically my calculation of the solution is correct but I would need to look up the density in a table or use a correlation like @AndrewKovács suggested? $\endgroup$ – idkfa Aug 31 '20 at 10:52
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    $\begingroup$ You have added volumes. This is wrong. Volumes in mixtures are never additive. Masses are additive, but not volumes. $234$ mL pure NaOH +$ 500$ mL water does not give $734$ mL solution. It gives less. $\endgroup$ – Maurice Aug 31 '20 at 11:38

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?


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