Why water volume went up by almost the same amount after adding salt?

Question

Seems like it's a well-known fact that adding salt to water will not raise the water volume by much (e.g.: Why there is no change in water level when salt is added? )

I've done this experiment at home and saw unexpected results:

Setup:

1. Start with 500 ml of tap water in measuring cup (500 grams)
2. Measure 50 ml of iodized salt in measuring cup (80 grams)
3. Add salt to water, stir until dissolved

Result: around 540 ml (give or take) of salt water in measuring cup weighing 580 grams

Weight was measured with kitchen scale, volume measure by eye, but ~40 ml increase in total volume seems way too much.

Followed up by replacing salt with sand, volume went up by about the same amount.

Video: https://www.youtube.com/watch?v=rLTn5jKqipU

What am I missing here?

p.s.: the commenter pointed out that my salt volume is off, I should have looked up the density to go by weight rather than by measuring cup notches. Going by weight, I've added 37 ml or salt, making the result event closer. Volumes were judged by eye in food measuring cups, not exactly the proper lab equipment.

Answer 1

• You have water volume.
• Calculate salt volume from its mass and density.
• Calculate their total volume before dissolution.
• Calculate solution mass concentration in mass %.
• From tabulated data or online calculators, obtain its density.
• Calculate solution volume from its total mass and density.
• Subtract from it the total volume of separate water and salt.
• You have the volume difference.

Here is the simple formula to calculate the difference:

$$\Delta V = V_\mathrm{solution} - V_\mathrm{solute} - V_\mathrm{solvent} = \\ \frac{m_\mathrm{solute} + m_\mathrm{solvent}}{\rho_\mathrm{solution}} - \frac{m_\mathrm{solute}}{\rho_\mathrm{solute}} - \frac{m_\mathrm{solvent}}{\rho_\mathrm{solvent}}$$

For our particular $\ce{NaCl}$ case:

$$\Delta V = \frac{m_{\ce{NaCl}} + m_{\ce{H2O}}}{\rho_\mathrm{solution}} - \frac{m_{\ce{NaCl}}}{\rho_{\ce{NaCl(s)}}} - \frac{m_{\ce{H2O}}}{\rho_{\ce{H2O}}}= \\ \frac{m_{\ce{NaCl}} + V_{\ce{H2O}}\cdot \rho_{\ce{H2O}} }{\rho_\mathrm{solution}} - \frac{m_{\ce{NaCl}}}{\rho_{\ce{NaCl(s)}}} - V_{\ce{H2O}}$$

• $V$ is volume [mL]
• $\rho$ is density [g/mL]
• $m$ is mass [g]

Volume changes during dissolution are common, usually contractions. $\ce{NaCl}$ solution volume decreases, as water molecules as electric dipoles are attracted to ions $\ce{Na+}$ and $\ce{Cl-}$. As the result, ions and molecules in solution are packed in average better than salt and water apart.

It can be somewhat illustrated on mechanical macro analogy of 2 jars with marbles and sand. If mixed together, their total loose volume decreases as they better utilized the unused space.

There happened an error in obtaining density for given salt mass fraction, which is smaller than originally provided, therefore the volume of solution is larger.

The volume contraction $\pu{10.2 mL}$ is less than 1/3 of the (real) volume of added salt $\pu{36.9 mL}.$

Using the link at the solution density (or other density data sources), you can predict volume contraction for any salt/water mass ratio (within salt solubility)

Here are calculates values for you particular case:

System	Quantity	Value	Unit
Water	Volume	500	mL
-	Density	0.9982	g/mL
-	Mass	499.1	g
Salt	Volume	36.9	mL
-	Density	2.17	g/mL
-	Mass	80	g
Total	Volume	536.9	mL
Solution	Mass concentration	13.815	%w/w
-	Volume	526.7	mL
-	Density	1.09947	g/mL
-	Mass	579.1	g
-	Volume difference	-10.2	mL