Interactions (Solutions)

The discussion of solutions is treated separately because solutions are important to all fields of chemistry. Generally solutions are liquids, but it is possible for metals (solids) to form solid solutions. This section will focus on liquid solutions.

The solvent is a liquid that will dissolve an added compound named the solute. The resulting mixture is a solution. The solute will usually be a solid or liquid. Gaseous solutions do exist, oxygen dissolved in water is an example. This discussion will focus on solutions comprising a liquid solvent and either a solid or liquid solute.

 Solubility is a measure of the amount of solute that will dissolve in the solvent.   Click on the link  Solubility.  Then find the solubility of the following compounds: NaCl, CaCl2, CaSO4, FeCl3, AgCl, AgNO3 to see the solubility values.

Solubility is determined by the statement “Like dissolves like.”
This means when the solute and solvent have the same kind of interactions the solute will be more soluble.  A polar solvent will dissolve a polar solute.  A non-polar solvent will dissolve a non-polar solute.  Ions can interact with polar solvents and most ionic compounds can be dissolved in polar solvents.  Non-polar solvents cannot dissolve ionic solutes. Likewise polar solvents will not dissolve non-polar solutes

It is important to know the amount of solute dissolved in a given amount of solvent. This information is referred to as the “concentration of the solution”. There are many ways to express the concentration; mass of solute/mass of solution,  mass of solute/volume of solution,  volume of solute/volume of solvent,  moles of solute/volume of solution.  NOTE: Concentrations are a ratio of the amount of solute to the amount of solvent or solution.

It is common to express the concentration of solutions in one of the following percentages:

Percent mass/mass – This expresses the mass of solute in 100 g of solvent
Percent Mass/Volume – This expresses the mass of solute in 100 ml of solvent
Percent Volume/ Volume – This expresses the volume of solute in 100 ml of solvent

This following discussion will focus on mass of solute/volume of solution and moles of solute/volume of solution concentration calculations.

PROBLEM 1: A solution of sodium chloride (NaCl) is prepared by adding 5 grams of solid NaCl to approximately 25 mL of water and stirred until the solid has dissolved. More water is added to get a total volume of 100 mL. What is the percent (%) mass concentration of the NaCl solution.

SOLVING: Percent mass/volume is defined as the mass of the solute / 100 ml of solution. The answer to  this problem is quite easy since the final volume is 100 ml.

The solution is 5 % w/v

PROBLEM 2: A solution of CaBr2 is prepared by adding 15.5 g of solid CaBr2 to water and stirring to dissolve the solute.  Water is added to get a total volume of 250 mL of solution. What is the % w/v CaBr2 concentration?

SOLVING: The ratio of the solute to solvent is  15.5 g CaBr2 to 250 mL solution.  Multiply this ratio be 100 to get the % w/v.


Concentration of solute (w/v %)   =       15.5 g CaBr2      x 100   =   % 6.2 w/v
                                     250 mL solution

Molarity is defined as the mole amount of  solute in one liter of solution. It is commonly written as moles/Liter. The symbol M is commonly used to substitute for moles/L.

Before starting you might visit the Metric section in Math Skills.

A solution of glucose (C6H12Omolar mass = 180 g/mole) in water is prepared by weighing 3.75 g of glucose on a balance and transferring it to a beaker.

The sketch below is to represent the 3.75 g of glucose in the beaker.


How many moles of glucose are in the beaker?

                           3.75g C6H12O x   1 mole  =  0.0208  moles  C6H12O6
                                                                 180 g

In the sketch, the 20 red ovals represent 0.0208 moles of C6H12O6

A small amount of water is added to the solute (glucose in this example) and the mixture is stirred until the solute dissolves.

The Visual 1 is attempting to show water molecules interacting with glucose molecules since they both polar compounds(Like dissolves Like).

When the solute has completely dissolved additional water is added until the final volume of the solution is 50 mL. The Visual 1 represents this step.

Visual 1


How many moles of glucose are in this solution?  0.0208 moles of C6H12O6. There are still 20 ovals representing the solute. The only change is they are found in a volume of liquid.

The ratio of solute to volume of solution is:


This is an expression of concentration because it tells the amount of solute in a given volume of final solution. But it is not the molar concentration of the solution because it does not express the volume of solution as 1 liter.

Molarity is calculated by converting mL in the concentration above to liter.  The conversion factor used is 1000 mL/ 1L to cancel out the mL units and leave the L unit in the denominator.


Molarity could also be calculated in one step by using the following conversion set-up.


In the calculation above the conversion factors 1 mole/180 g and 1000 mL/ 1L are selected.

If additional water is added to the original glucose in the Example above, it will be diluted.

What happens to the amount of solute when a solution is diluted by adding solvent to the solution? The moles of solute remains the same, but the volume has increased.

Since molarity concentration is the ratio of moles of solute to the total volume, a larger volume will give a lower molarity.

In the Visual 2 below, the volume of solution in beaker 1 has been increased by 30 mL to a total of 80 mL.

Visual 2


There are still the same number of moles of solute molecules, but they are in a larger volume.

The diluted molarity would be:


Visual 3 shows the  two beakers placed side by side.

Visual 3

Glusolution1            Gludilsoln80

Compare and contrast the two solutions. What is the same? (moles of solute) What is different? (more water and more solution).

In the glucose solution prepared from solid, the molarity was expressed as 0.417 M. Chemists use the capital letter M to represent the units of moles/liter. Thus the molarity of the glucose solution would also be 0.417 moles of glucose/liter of solution.

The student needs to be able to recognize the letter M as a substitute for the units moles solute/liter solution.  This is useful when considering chemical reactions that occur in solutions. In these reactions, a certain volume of solution one is added to another volume of solution two.  The actual moles of each chemicals reacting depends on the molarity and the volume taken.

To determine the moles of chemical in a given volume of a solution with a known molarity, one merely multiplies the volume (expressed as a liter quantity) of the solution times the molarity.

In the first solution prepared from solid glucose, the moles of solute could be calculated by:


The moles of solute in the diluted solution prepared by adding 30 more mL of water to get a total volume of 80 would be:


The general formula used for these types of calculation is:


This equation will be applied in the next section.

A  10 mL aliquot (aliquot is a term which means a portion) of the original solution in Beaker 1 is removed from the beaker and transferred to a second beaker.  The following questions are

a.  What will be the concentration of the solution remaining in the first beaker?
b.  How many moles of solute will remain in the first beaker?
c.  How many moles of solute are in the pipet?
d. What is the molarity of the solution in the pipet?
e.  How many moles of solute will be in the second beaker?
f.   What will be the concentration of the solution in the second beaker?
g.  Water is added to the second beaker to get a total volume of 100 mL, what will be the concentration of the resulting solution?

The beaker in Visual 1 will be used to assist the student in answering parts a to g in the previous paragraph.

Visual 4 represents a 10 mL pipet lowered into the solution of Beaker 1 to remove 10 mL of it.

Visual 4


Visual 5 shows the pipet containing 10 ml of solution removed from the beaker. Notice the volume remaining in the beaker.

Visual 5


Visual 6 shows the solution transferred to Beaker 2.

Visual 6



Visual 7 shows the solution diluted to 100 ml.

Visual 7



a.  Has the molarity of the solution remaining in Beaker 1 been changed by removing the 10 mL aliquot? No!  In Visual 1, the molarity is represented as 20 ovals per 50 mL for a solute to solution ratio of 2 to 5.  The solution remaining in the beaker after removal of 10 mL has 16 ovals in 40 mL. This is still a ratio of solute to solution of 2 to 5, so the remaining solution has the same concentration

b.  The moles of solute remaining in beaker one would be:


c.  How many moles of solute are in the pipet?


d. What is the molarity of the solution in the pipet?
The number of solute molecules in the pipet is 4 and the volume is 10 ml.  This is a ratio of 2 to 5 the same as the original solution.  The molarity is the same as the original solution

e.  The moles of solute in the second beaker that received the 10 ml form the original solution is that same as the number of moles in the pipet, 0.00417 moles C6H12O .

f.  The concentration of the solution in the Visual 6 before dilution is:


g.  The molarity of the solution in Visual 7  that has been diluted to 100 ml will be:



Obviously the solution is more dilute because there is no longer a ratio of solute to solution of 2 to 5. The ratio is now 4 to 100.

This solution has 0.00417 moles of C6H12O6 but in a larger volume (100 ml).

There is a very important observation to be made. The moles of solute in the aliquot (volume taken from the original solution) equals the moles of solute in the diluted solution.

There is a streamlined calculation which can be used to calculate dilutions. It is:

Md * Vd = Mc * Vc

Mc is the molarity of the concentrated solution

Vc is the volume of the concentrated solution

Md is the molarity of the diluted solution

Vd is the volume of the diluted solution

Using this formula to solve Dilution 2 and converting ml volumes to L:

Md x 0.1 L =  0.417M  x 0.01L
Divide both sides by 0.01 L to get
Md = 0.0417 M

The important thing to remember about the dilution equation is that the moles of solute in the aliquot [Mc*Vc] equals the moles of solute in the diluted solution [Md*Vd].


Some solutes dissolve in water to form solutions that do not conduct electricity. These solutes are called non-electrolytes and are compounds whose basic unit consists of molecules. Glucose, ethanol, and sucrose are examples of non-electrolyte solutes.

Other solutes dissolve in water to form solutions that conduct electricity. These solutes are called electrolytes and are compounds whose basic units are ions. NaCl, Ca(NO3)2, and FeBr3 are examples of electrolyte solutes.

When an ionic solute such as FeBr3 dissolves, the solution contains primarily the ions of that solute.  In this case the ions are Fe+3 and Br-1 .  There is a very small concentration of FeBr3. For our purposes in this discussion, the solution contains only Fe+3 and Br-1 ions.

When a solution is prepared using an ionic solute such as FeBr3, it is common practice to express the concentration of the solution in terms of the formula of the solute. The student of chemistry needs to understand the solution contains ions and not formula units of the solute. Therefore the concentration of the ions can also be expressed.

The example of a solution prepared by dissolving 8.13 g of FeBr3 in water to prepare 250 mL of solution will demonstrate these ideas.
The molar mass of FeBr3 is 295.58 g/mole. The molarity of this solution in terms of FeBr3 will be:


When this solute is placed into solution, there will be one Fe+3 ion and three Br-1 ions for each formula unit of FeBr3 placed in solution.

Therefore the molarity of Fe+3 will be:


The molarity of Br-1 will be:


Visual 8 of the FeBr3 solution is shown in the following graphic.

Visual 8


This sketch shows the correct ratio of Fe+3 ions to Br-1 ions. In this sketch, 4 diamond shapes represents the 0.1100 M Fe+3 ions. Therefore, there must be 12 ovals to represent the 0.3300 M Br-1 ions.

Molarity of water

An interesting calculation is the molarity of water.  Water has a GFW of 18.02 g/mole.  The density of water at 25o C is 1.0 g/ mL.  Converting this to liters gives the value of 1000g/L.  So the calculation for the molarity of water is:

Water Molar calc

This value will be used in a later section.

Molality is another way of expressing concentration: moles of solute in kilograms of solvent. You may encounter this term, but it will not be discussed in this site.

The Homework link has problem sets to practice calculating % Wt/Vol, molarity concentration, and dilution calculations.