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:


  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.

  • 1
    $\begingroup$ With the density of NaCl at 2.16 g/cc, 50 ml should weigh 108 g. Bottom line is there is an error on your part somewhere. $\endgroup$ Jan 12, 2023 at 1:25
  • $\begingroup$ @ToddMinehardt I guess the error was that instead of 50ml of salt I've added 37ml, which makes the result even closer. As I've said, I used food measuring cups, so volumes are not too precise, that's why I've used weight to keep me honest. Bottom line, water volume went up by roughly the amount of salt added. I'm sure you have salt and water handy, give it a go. $\endgroup$
    – roman m
    Jan 12, 2023 at 5:13
  • $\begingroup$ @ToddMinehardt forget the experiment, how much should the water volume go up if I add 50ml of salt to 500ml of water? Is there a reference table for something like this? I'd expect there to be one. $\endgroup$
    – roman m
    Jan 12, 2023 at 5:21
  • $\begingroup$ Why, you can look up the solution density tables. $\endgroup$ Jan 12, 2023 at 5:38
  • $\begingroup$ How do you measure the volume of the salt ? If you measure it by filling a graduated glass column, you obtain the volume of the salt plus the volume of the air between the salt crystals. This measured value is NOT the volume of the salt. The volume of the salt cannot be measured. It must be calculated from the mass and the density. $\endgroup$
    – Maurice
    Jan 12, 2023 at 9:45

1 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
13.815 %w/w
- Volume 526.7 mL
- Density 1.09947 g/mL
- Mass 579.1 g
- Volume
-10.2 mL
  • $\begingroup$ youtube.com/watch?v=rLTn5jKqipU the level clearly went up way more than 3 ml, what am I missing? $\endgroup$
    – roman m
    Jan 23, 2023 at 1:13
  • $\begingroup$ I have fixed the error due somehow wrong value of tabelated solution density. Volume contraction is less than 1/3 of volume of added salt. $\endgroup$
    – Poutnik
    Jan 23, 2023 at 6:28

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