If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

AP®︎/College Chemistry

Course: AP®︎/College Chemistry>Unit 7

Lesson 3: Mixtures and solutions

Molarity

Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and how to calculate 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:
$\text{Molarity}=\frac{\text{mol solute}}{\text{L of solution}}$
• 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 $57\mathrm{%}$ 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!
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

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 $78\mathrm{%}$ nitrogen gas, $21\mathrm{%}$ oxygen gas, and $1\mathrm{%}$ 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!):
$\text{Molarity}=\frac{\text{mol solute}}{\text{L of solution}}$
Molarity has units of $\frac{\text{mol}}{\text{liter}}$, which can be abbreviated as molar or $\text{M}$ (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 $\left[{\text{Cl}}^{-}\right]$. 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 ~$88\mathrm{%}$ copper mixed with $12\mathrm{%}$ tin. What is the solute and solvent in bronze?

Example 1: Calculating the molar concentration of a solute

Let's consider a solution made by dissolving $2.355\phantom{\rule{0.167em}{0ex}}\text{g}$ of sulfuric acid, ${\text{H}}_{2}{\text{SO}}_{4}$, in water. The total volume of the solution is $50.0\phantom{\rule{0.167em}{0ex}}\text{mL}$. What is the molar concentration of sulfuric acid, $\left[{\text{H}}_{2}{\text{SO}}_{4}\right]$?
To find $\left[{\text{H}}_{2}{\text{SO}}_{4}\right]$ 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, $98.08\phantom{\rule{0.167em}{0ex}}\frac{\text{g}}{\text{mol}}$:
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:
$\begin{array}{rl}\left[{\text{H}}_{2}{\text{SO}}_{4}\right]& =\frac{\text{mol solute}}{\text{L of solution}}\\ \\ & =\frac{0.02401\phantom{\rule{0.167em}{0ex}}\text{mol}}{0.050\phantom{\rule{0.167em}{0ex}}\text{L}}\\ \\ & =0.48\phantom{\rule{0.167em}{0ex}}\text{M}\end{array}$
Concept check: What is the molar concentration of ${\text{H}}^{+}$ ions in a $4.8\phantom{\rule{0.167em}{0ex}}{\text{M H}}_{2}{\text{SO}}_{4}$ solution?

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.
$\text{mol solute}=\text{Molarity}×\text{L of solution}$
For example, let's say we want to make $0.250\phantom{\rule{0.167em}{0ex}}\text{L}$ of an aqueous solution with $\left[\text{NaCl}\right]=0.800\phantom{\rule{0.167em}{0ex}}\text{M}$. What mass of the solute, $\text{NaCl}$, would we need to make this solution?
We can use the rearranged molarity equation to calculate the moles of $\text{NaCl}$ needed for the specified concentration and volume:
$\begin{array}{rl}\text{mol NaCl}& =\left[\text{NaCl}\right]×\text{L of solution}\\ & =0.800\phantom{\rule{0.167em}{0ex}}\frac{\text{mol}}{\overline{)\text{L}}}×0.250\phantom{\rule{0.167em}{0ex}}\overline{)\text{L}}\\ & =0.200\phantom{\rule{0.167em}{0ex}}\text{mol NaCl}\end{array}$
We can then use the molecular weight of sodium chloride, $58.44\phantom{\rule{0.167em}{0ex}}\frac{\text{g}}{\text{mol}}$, to convert from moles to grams of $\text{NaCl}$:
$\text{Mass of NaCl}=0.200\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol}}×\frac{58.44\phantom{\rule{0.167em}{0ex}}\text{g}}{1\phantom{\rule{0.167em}{0ex}}\overline{)\text{mol}}}=11.7\phantom{\rule{0.167em}{0ex}}\text{g NaCl}$
In practice, we could use this information to make our solution as follows:
Step Weigh out $11.7\phantom{\rule{0.167em}{0ex}}\text{g}$ of sodium chloride.
Step Transfer the sodium chloride to a clean, dry flask.
Step Add water to the $\text{NaCl}$ until the total volume of the solution is $250\phantom{\rule{0.167em}{0ex}}\text{mL}$.
Step Stir until the $\text{NaCl}$ 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).

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:
$\text{Molarity}=\frac{\text{mol solute}}{\text{L of solution}}$
• Molar concentration can be used to convert between the mass or moles of solute and the volume of the solution.

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 ${\text{Pb(NO}}_{3}{\right)}_{2}\left(aq\right)$ and $\text{KI}\left(aq\right)$. When these two solutions are combined, bright yellow ${\text{PbI}}_{2}\left(s\right)$ precipitates out of solution. The balanced equation for this reaction is:
${\text{Pb(NO}}_{3}{\right)}_{2}\left(aq\right)+2\text{KI}\left(aq\right)\to {\text{PbI}}_{2}\left(s\right)+2{\text{KNO}}_{3}\left(aq\right)$
If we have $0.1\phantom{\rule{0.167em}{0ex}}\text{L}$ of $0.10\phantom{\rule{0.167em}{0ex}}{\text{M Pb(NO}}_{3}{\right)}_{2}$, what volume of $0.10\phantom{\rule{0.167em}{0ex}}\text{M KI}\left(aq\right)$ should we add to react with all the ${\text{Pb(NO}}_{3}{\right)}_{2}\left(aq\right)$?

Want to join the conversation?

• Hi there,
I was just wondering shouldnt the answer in example 1 be 0.48 mol/Litre
and NOT 4.8 M? I get the same answer to the last step before the answer, but when i do the calculation i get 0.48 mol/litre.
thank you so much.
• I believe you're correct. There was likely a typographic error in the example. We see in the previous step the conversion was done correctly (50 mL = .050 L) so we have 0.02401 mol / .050 L. A quick check with the calculator shows that this is in fact 0.48 mol/L or 0.48 M.
• I was told in school that molarity should be moles/dm^3, but is this different from moles/litres?
• A liter is equal to a cubic decimeter, so it is the same.
• So this isn't quite the right place for my question, but I can't find the right place for the life of me... If someone could maybe point me to a video/article on converting between concentration units, especially molarity to ppt or ppm, that'd be great.

In the mean time, I've been asked to take a known molarity of a solution and convert it into parts per thousand. I tried Google and I /think/ I got the right formula but I'm not positive, so can someone check it for me please?

So what I did was start with my given molarity as mol/L. I assumed there wouldn't be enough solute to drastically affect density and so I changed 1 L to 1000g, so I now have mol/1000g. Then I multiply by the molar mass of the solute (NaOH - 39.998) so I'm now g NaOH/1000g solution. Then I multiply the whole thing by 1000 to get ppt, right? Sort of like calculating a percent?
• You did it almost perfectly.
A concentration of 1 g NaOH/1000 g solution is 1 g per 1000 g or one part per thousand (1 ppt) — no need to multiply by 1000.
In the same way, a concentration of 1 g per 100 g is one part per hundred (1 %).
• in hint one how do you know there is .1L of solute?
• There must have been a typo. I think in the description they meant 0.100L instead of 0.100mL.
• How would you find the molarity of SO2 if you have it dissolved in 100 grams of water at 85 degrees Celcius?
• Question: Is this just coincidence, or does this make sense...
In the equation, we have 1 Pb(NO3)2 + 2 KI...we have twice as many KI as Pb(NO3)2. Since we have 0.1L of 1Pb(NO3)2, can I just multiply the 0.1 L x 2, since we use twice as much KI as we do Pb(NO3)2? (Or if the equation happened to have 4KI, could we simply multiply 0.1L x 4)? Thanks for the help!
• What you suggest is fine just as long as the concentrations of the two solutions are the same. But if, say, the Pb(NO3)2 solution was twice the strength of the KI solution then you would only need 0.1 L of each to get the same number of moles.
• I understood what molarity is quite well......but what is normality, formality and molarity? If we have molarity why are they even needed then?
• how do you find the volume when given the mass and M value
(1 vote)
• We know that the formula to calculate the molarity of a substance is M = n/V (n = moles, and V = volume of the solution).
Rearranging the formula to make 'V' the subject allows us to figure out that V = n/M.
When given the mass in Analytical Chemistry, we should always seek to covert the mass (given in any units) first into grams (if it is, then do not worry about this). We should then convert these grams into moles, to do so we require the molar mass of the solute, and dividing the given mass (in grams) by the molar mass provides us with the moles of the substance.
Therefore, we have everything we need, we have calculated the moles, and we are already given the Molarity (M). Seek to substitute these values into their respective position within the rearranged equation above- V = n/M, calculating this value will output the volume.
(1 vote)
• What is the difference between molarity and molality?
(1 vote)
• Question1 :In a solution with 2 species "A" and "B" ,with "A" having a greater number of moles but the "B" having a bigger molecular mass in such a way that it exceeds the mass of "A", who is the solvent ?
Question 2 : when 2 species are in the same amount , what determines who is the solvent ?