# What is the difference between “adding a liquid” and “diluting to a liquid”?

1. Place $\pu{20.0 g}$ $\ce{NaOH(s)}$ in a flask and dilute to $\pu{1.00 L}$ with water.
2. Place $\pu{20.0 g}$ $\ce{NaOH(s)}$ in a flask and add $\pu{1.00 L}$ of water.

How exactly do these two statements differ? Wouldn't adding $\pu{1.00 L}$ of water to $\pu{20.0 g}$ of $\ce{NaOH}$ dilute it anyways?

• There is no difference between adding and diluting. There is, however, a difference between "adding 1 L" and "adding 0.5 L". – Ivan Neretin Jun 3 '18 at 10:41

The gist here is that adding a volume $x$ of solid and a volume $y$ of liquid does not result in a solution of volume $y$ nor $x+y$. The volume of the solution is unpredictable.

Place $\pu{20.0 g}$ $\ce{NaOH (s)}$ in a flask and dilute to $\pu{1.00 L}$ with water.

Chemists do this frequently. Typically by using a volumetric flask to be the container. Since the molecular mass of $\ce{NaOH}$ is $\pu{40.00 g/mol}$, this will yield a $0.500$ molar solution, $c(\ce{NaOH}) = \pu{0.500 mol/L}$.

Place $\pu{20.0 g}$ $\ce{NaOH (s)}$ in a flask and add $\pu{1.00 L}$ of water.

In this case the volume of the solution is unpredictable with only the information given. However the volume will be a little more than $1.00$ liter.

In chemistry there are also molal solutions. The units are moles per kilogram of solvent. In this case there are $\pu{0.02 kg}$ of $\ce{NaOH}$ ($\pu{0.500 mol}$) per $1.00$ kilogram solution, so the molality of the $\ce{NaOH}$ is $b(\ce{NaOH}) = \pu{0.500 mol/kg}$.

"Dilute to" means add solvent until the designated volume is reached. That is, the final volume is known and definite.
"Add x liters of water" means to add the designated volume of water with no pre-defined final volume. The final volume will be whatever it happens to be when the solute and solvent are mixed.

If you are diluting to 1.00 L with water, you are adding water until the solution equals 1.00 L. You will be adding less than 1.00 L of water to bring the volume up to 1.00 L. Therefore the final volume will be 1.00 L of solution.

However, if you are adding 1.00 L of water then the final volume will be more than 1.00 L. That final volume will depend on the density of the solution that results, and how much material that was already present.