Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and comcalculations related to molarity. 

Key points

  • Mixtures with uniform composition are called homogeneous mixtures or solutions.
  • Mixtures with non-uniform composition are heterogeneous mixtures.
  • The chemical in the mixture that is present in the largest amount is called the solvent, and the other components are called solutes.
  • Molarity or molar concentration is the number of moles of solute per liter of solution, which can be calculated using the following equation:
Molarity=mol soluteL of solution\text{Molarity}= \dfrac{\text{mol solute}}{\text{L of solution}}
  • Molar concentration can be used to convert between the mass or moles of solute and the volume of the solution.

Introduction: Mixtures and solutions

In real life, we often encounter substances that are mixtures of different elements and compounds. One example of a mixture is the human body. Did you know that the human body is approximately 57%57\% water by mass? We are basically an assortment of biological molecules, gases, and inorganic ions dissolved in water. I don't know about you, but I find that pretty mind-boggling!
An photograph of an oceanside beach. The yellowish sand is covered with people on beach towels, and there are also some swimmers in the blue-green ocean. The beach is also surrounded by houses from a small town.
Besides the bodies of the beach-goers, beach sand and ocean water are both mixtures as well! Photo of Bondi Beach by penreyes on flickr, CC BY 2.0
If substances are mixed together in such a way that the composition is the same throughout the sample, they are called homogeneous mixtures. In contrast, a mixture that does not have a uniform composition throughout the sample is called heterogeneous.
Homogeneous mixtures are also known as solutions, and solutions can contain components that are solids, liquids and/or gases. We often want to be able to quantify the amount of a species that is in the solution, which is called the concentration of that species. In this article, we'll look at how to describe solutions quantitatively, and discuss how that information can be used when doing stoichiometric calculations.

Molar concentration

Nope!
Molar concentration is the same as molarity, but molarity and molality are not the same thing. They are different ways to quantify the amount of solute in a solution, and the concentration of a solution in molarity is not interchangeable with its concentration in molality. In this article we are only discussing molarity.
For more on the difference between the two definitions, see this video on molarity vs. molality.
The component of a solution that is present in the largest amount is known as the solvent. Any chemical species mixed in the solvent is called a solute, and solutes can be gases, liquids, or solids. For example, Earth's atmosphere is a mixture of 78%78\% nitrogen gas, 21%21\% oxygen gas, and 1%1\% argon, carbon dioxide, and other gases. We can think of the atmosphere as a solution where nitrogen gas is the solvent, and the solutes are oxygen, argon and carbon dioxide.
The molarity or molar concentration of a solute is defined as the number of moles of solute per liter of solution (not per liter of solvent!):
To learn more about how moles are used in chemistry, see this video on the mole and Avogadro's number.
Molarity=mol soluteL of solution\text{Molarity}= \dfrac{\text{mol solute}}{\text{L of solution}}
You might think that it is because, well, we didn't take into account the volume of the solute! But actually, even if we added the volume of the solute to the volume of the solvent, we still might end up with a total volume that is different from the volume of the solution.
This is because the volume of a mixture depends on the intermolecular forces between the molecules in the mixture. This is especially true for a liquid solution, where the molecules are closer together and thus have strong interactions than in a solution of gases. Intermolecular forces between the solvent molecules are likely to be different than the intermolecular forces between the solvent and solute molecules. Depending on the nature of these interactions, the volume of the solution can be more or less than the volume of the solvent plus the volume of the solute!
Pretty crazy, but that's chemistry for you.
Molarity has units of molliter\dfrac{\text {mol}}{\text {liter}}, which can be abbreviated as molar or M\text M (pronounced "molar"). The molar concentration of the solute is sometimes abbreviated by putting square brackets around the chemical formula of the solute. For example, the concentration of chloride ions in a solution can be written as [Cl][\text{Cl}^-]. Molar concentration allows us to convert between the volume of the solution and the moles (or mass) of the solute.
Concept check: Bronze is an alloy that can be thought of as a solid solution of ~88%88\% copper mixed with 12%12\% tin. What is the solute and solvent in bronze?
Since there is more copper than tin in bronze, we can think of the copper as the solvent and tin as the solute.

Example 1: Calculating the molar concentration of a solute

Let's consider a solution made by dissolving 2.355g2.355\,\text g of sulfuric acid, H2SO4\text H_2 \text {SO}_4, in water. The total volume of the solution is 50.0mL50.0\,\text{mL}. What is the molar concentration of sulfuric acid, [H2SO4][\text H_2 \text{SO}_4]?
To find [H2SO4][\text H_2 \text{SO}_4] we need to find out how many moles of sulfuric acid are in solution. We can convert the mass of the solute to moles using the molecular weight of sulfuric acid, 98.08gmol98.08\,\dfrac{\text g}{\text {mol}}:
mol H2SO4=2.355g H2SO4×1mol98.08g=0.02401mol H2SO4\text{mol H}_2\text{SO}_4=2.355\,\cancel{\text g} {\text{ H}_2\text{SO}_4}\times \dfrac{1\,\text {mol}} {98.08\,\cancel{\text {g}}} = 0.02401\,\text{mol H}_2\text{SO}_4
We can now plug in the moles of sulfuric acid and total volume of solution in the molarity equation to calculate the molar concentration of sulfuric acid:
[H2SO4]=mol soluteL of solution=0.02401mol0.050L=0.48M\begin{aligned} [\text H_2 \text{SO}_4]&= \dfrac{\text{mol solute}}{\text{L of solution}}\\ \\ &=\dfrac{0.02401\,\text{mol}}{0.050\,\text L}\\ \\ &=0.48 \,\text M\end{aligned}
Concept check: What is the molar concentration of H+\text H^+ ions in a 4.8M H2SO44.8\,\text {M H}_2 \text{SO}_4 solution?
Answering this question requires thinking about what is happening in solution. Since sulfuric acid is a strong acid, it dissociates completely into its constituent ions in solution, H+(aq)\text H^+(aq) and SO4(aq)\text {SO}_4^-(aq). This can also be written as follows:
H2SO4(aq)2H+(aq)+SO42(aq)\text H_2 \text{SO}_4(aq) \rightarrow 2\text H^+(aq)+\text{SO}_4^{2-}(aq)
From this balanced equation, we can see for every one mole of sulfuric acid in solution, we end up with 22 moles of H+(aq)\text H^+(aq). Using this ratio, the molar concentration of H+(aq)\text H^+(aq) will be:
[H+(aq)]=4.8M H2SO4×2mol H+1mol H2SO4=9.6M H+(aq)[\text H^+(aq)]=4.8\,\text {M H}_2 \text{SO}_4\times \dfrac{2\,\text {mol H}^+}{1\,\text{mol H}_2\text{SO}_4 }=9.6\,\text {M H}^+(aq)
Therefore, our molar concentration of H+(aq)\text H^+(aq) is twice the molar concentration of H2SO4\text {H}_2 \text{SO}_4.

Example 2: Making a solution with a specific concentration

Sometimes we have a desired concentration and volume of solution, and we want to know how much solute we need to make the solution. In that case, we can rearrange the molarity equation to solve for the moles of solute.
mol solute=Molarity×L of solution\text{mol solute}= {\text{Molarity}}\times{\text{L of solution}}
For example, let's say we want to make 0.250L0.250\,\text {L} of an aqueous solution with [NaCl]=0.800M[\text{NaCl}]=0.800\,\text {M}. What mass of the solute, NaCl\text{NaCl}, would we need to make this solution?
We can use the rearranged molarity equation to calculate the moles of NaCl\text{NaCl} needed for the specified concentration and volume:
mol NaCl=[NaCl]×L of solution=0.800molL×0.250L=0.200mol NaCl\begin{aligned}\text{mol NaCl}&= [\text{NaCl}]\times{\text{L of solution}}\\ &=0.800\,\dfrac{\text{mol}}{\cancel{\text L}} \times 0.250\,\cancel{\text{L}}\\ &=0.200\,\text {mol NaCl}\end{aligned}
We can then use the molecular weight of sodium chloride, 58.44gmol58.44\,\dfrac{\text g}{\text {mol}}, to convert from moles to grams of NaCl\text{NaCl}:
Mass of NaCl=0.200mol×58.44g1mol=11.7g NaCl\text {Mass of NaCl}=0.200\,\cancel{\text {mol}} \times \dfrac {58.44\,\text g}{1\,\cancel{\text {mol}}} = 11.7\,\text {g NaCl}
In practice, we could use this information to make our solution as follows:
Step 1. 1.~ Weigh out 11.7g11.7\,\text g of sodium chloride.
Step 2. 2.~ Transfer the sodium chloride to a clean, dry flask.
Step 3. 3.~ Add water to the NaCl\text{NaCl} until the total volume of the solution is 250mL250\,\text {mL}.
Step 4. 4.~ Stir until the NaCl\text{NaCl} is completely dissolved.
The accuracy of our molar concentration depends on our choice of glassware, as well as the accuracy of the balance we use to measure out the solute. The glassware determines the accuracy of our solution volume. If we aren't being too picky, we might mix the solution in a Erlenmeyer flask or beaker. If we want to extremely precise, such as when making a standard solution for an analytical chemistry experiment, we would probably mix the solute and solvent in a volumetric flask (see picture below).
A picture of a volumetric flask, which has a wide pear-shaped base with a very thin, straight neck on top. The flask is filled with a deep-blue solution that goes partially up the thin neck of the flask.
A volumetric flask containing a solution of methylene blue, a dye. Photo by Amanda Slater on flickr, CC BY-SA 2.0

Summary

  • Mixtures with uniform composition are called homogeneous solutions.
  • Mixtures with non-uniform composition are heterogeneous mixtures.
  • The chemical in the mixture that is present in the largest amount is called the solvent, and the other components are called solutes.
  • Molarity or molar concentration is the number of moles of solute per liter of solution, which can be calculated using the following equation:
Molarity=mol soluteL of solution\text{Molarity}= \dfrac{\text{mol solute}}{\text{L of solution}}
  • Molar concentration can be used to convert between the mass or moles of solute and the volume of the solution.

Attributions

This article was adapted from the following articles:
  1. Molarity” from Boundless Chemistry, CC BY-SA 4.0.
  2. "Molarity" from UC Davis ChemWiki, CC BY-NC-SA 3.0 US.
The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Other references

Atmosphere of Earth” from Wikipedia
"Lab glassware" from Wikimedia Commons, CC BY 3.0

Try it: The stoichiometry of a precipitation reaction

Molarity is a useful concept for stoichiometric calculations involving reactions in solution, such precipitation and neutralization reactions. For example, consider the precipitation reaction that occurs between Pb(NO3)2(aq)\text{Pb(NO}_3)_2 (aq) and KI(aq)\text{KI} (aq). When these two solutions are combined, bright yellow PbI2(s)\text{PbI}_2 (s) precipitates out of solution. The balanced equation for this reaction is:
Pb(NO3)2(aq)+2KI(aq)PbI2(s)+2KNO3(aq)\text{Pb(NO}_3)_2(aq) + 2\text{KI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)
If we have 0.1L0.1\,\text{L} of 0.10M Pb(NO3)20.10\,\text {M Pb(NO}_3)_2, what volume of 0.10M KI(aq)0.10\,\text {M KI}(aq) should we add to react with all the Pb(NO3)2(aq)\text {Pb(NO}_3)_2(aq)?
Choose 1 answer:
Choose 1 answer:
We can start by determining how many moles of Pb(NO3)2\text {Pb(NO}_3)_2 are present using the molarity equation:
mol Pb(NO3)2=0.100L×0.10mol1L=0.010mol Pb(NO3)2\text{mol Pb(NO}_3)_2=0.100\,\cancel{\text L} \times \dfrac{0.10\,\text {mol}}{1\,\cancel{\text L}} = 0.010\,\text{mol Pb(NO}_3)_2
Based on the stoichiometry of the balanced chemical equation, we know it takes 22 moles of KI\text{KI} for every 1mol Pb(NO3)21\,\text{mol Pb(NO}_3)_2 that reacts. If we use this stoichiometric ratio to calculate the moles of KI\text{KI}, we get:
0.010mol Pb(NO3)2×2mol KI1mol Pb(NO3)2=0.020mol KI0.010\,\cancel{\text{mol Pb(NO}_3)_2} \times \dfrac{2\,\text{mol KI}}{1\,\cancel{\text {mol Pb(NO}_3)_2} }= 0.020\,\text{mol KI}
That means we need 0.020mol KI0.020\,\text{mol KI}.
To determine what volume of KI\text{KI} solution we need, we can rearrange the molarity equation to solve for the total volume of solution:
mol solute=Molarity×L of solution\text{mol solute}= {\text{Molarity}}\times{\text{L of solution}}
If we divide both sides by the molarity of the solution, we get:
L of solution=mol soluteMolarity\text{L of solution}= \dfrac{\text{mol solute}}{\text{Molarity}}
We can plug in our known values for the moles of solute and the molarity:
L of solution=mol soluteMolarity=0.020mol0.10M KI=0.20L KI\begin{aligned}\text{L of solution}&= \dfrac{\text{mol solute}}{\text{Molarity}}\\ \\ &=\dfrac{0.020\,\text{mol}}{0.10\,\text{M KI}}\\ \\ &=0.20\,\text{L KI}\end{aligned}
Thus, we need 0.20L0.20\,\text L of the 0.10M KI0.10\,\text {M KI} solution to completely react with 100mL100\,\text {mL} of 0.10M Pb(NO3)20.10\,\text{M Pb(NO}_3)_2.
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