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• 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:

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

• $V$ is volume
• $\rho$ is density
• $m$ is mass

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:

• 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:

• 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:

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

• $V$ is volume
• $\rho$ is density
• $m$ is mass

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 $\pu{36.9 mL}.$

Here are calculates values for you particular case:

• 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:

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

• $V$ is volume
• $\rho$ is density
• $m$ is mass

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:

• 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:

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

• $V$ is volume
• $\rho$ is density
• $m$ is mass

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 $\pu{36.9 mL}.$*

Here are calculates values for you particular case:

• 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:

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

• $V$ is volume
• $\rho$ is density
• $m$ is mass

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 $\pu{36.9 mL}.$

Here are calculates values for you particular case: