While preparing w/w 50% NaOH solution I noticed a decrease in solution volume (5-10mL) in a 2L volumetric flask, so I am wondering if there is an increase in water volume during the water evaporation or concentration of NaOH? Is it the same amount and is it measurable?
QuantumYitian question regarding the decrease of volume during the dissolution of NaOH
There is no water volume in solution. $\ce{Na+}$ and $\ce{OH-}$ ions and $\ce{H2O}$ molecules are packed together in solution in average tighter than is the average packing tightness for $\ce{NaOH(s)}$ and $\ce{H2O(l)}$ alone.
At water evaporation, the mass concentration of $\ce{NaOH}$ obviously increases until solution saturation. After that, extra $\ce{NaOH}$ crystalizes.
If all water evaporates and then condenses, the initial volumes are obtained, if we neglect experimental troubles of removing all water.
The difference of volumes during dissolution (assuming the same initial and final temperature) can be determined from experimental or tabulated values of volume, mass or density:
Many solutions, including $\ce{NaOH}$ have tabulated densities that can be interpolated and the volume difference calculated.
Here is the simple formula to calculate the difference:
$$\Delta V = V_\mathrm{solution} - V_\mathrm{solute} - V_\mathrm{solvent} = \\ \frac{m_\mathrm{solute} + m_\mathrm{solvent}}{\rho_\mathrm{solution}} - \frac{m_\mathrm{solute}}{\rho_\mathrm{solute}} - \frac{m_\mathrm{solvent}}{\rho_\mathrm{solvent}}= \\ \frac{m_{\ce{NaOH}} + m_{\ce{H2O}}}{\rho_\mathrm{solution}} - \frac{m_{\ce{NaOH}}}{\rho_{\ce{NaOH(s)}}} - \frac{m_{\ce{H2O}}}{\rho_{\ce{H2O}}}= \\ \frac{m_{\ce{NaOH}} + V_{\ce{H2O}}\cdot \rho_{\ce{H2O}} }{\rho_\mathrm{solution}} - \frac{m_{\ce{NaOH}}}{\rho_{\ce{NaOH(s)}}} - V_{\ce{H2O}}$$
