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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:
start text, M, o, l, a, r, i, t, y, end text, equals, start fraction, start text, m, o, l, space, s, o, l, u, t, e, end text, divided by, start text, L, space, o, f, space, s, o, l, u, t, i, o, n, end text, end fraction
• 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, percent 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

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, percent nitrogen gas, 21, percent oxygen gas, and 1, percent 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!):
start text, M, o, l, a, r, i, t, y, end text, equals, start fraction, start text, m, o, l, space, s, o, l, u, t, e, end text, divided by, start text, L, space, o, f, space, s, o, l, u, t, i, o, n, end text, end fraction
Molarity has units of start fraction, start text, m, o, l, end text, divided by, start text, l, i, t, e, r, end text, end fraction, which can be abbreviated as molar or start text, M, end text (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 open bracket, start text, C, l, end text, start superscript, minus, end superscript, close bracket. 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, percent copper mixed with 12, percent 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, point, 355, start text, g, end text of sulfuric acid, start text, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, in water. The total volume of the solution is 50, point, 0, start text, m, L, end text. What is the molar concentration of sulfuric acid, open bracket, start text, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, close bracket?
To find open bracket, start text, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, close bracket 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, point, 08, start fraction, start text, g, end text, divided by, start text, m, o, l, end text, end fraction:
start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, equals, 2, point, 355, start cancel, start text, g, end text, end cancel, start text, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, times, start fraction, 1, start text, m, o, l, end text, divided by, 98, point, 08, start cancel, start text, g, end text, end cancel, end fraction, equals, 0, point, 02401, start text, m, o, l, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript
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{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 start text, H, end text, start superscript, plus, end superscript ions in a 4, point, 8, start text, M, space, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript 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.
start text, m, o, l, space, s, o, l, u, t, e, end text, equals, start text, M, o, l, a, r, i, t, y, end text, times, start text, L, space, o, f, space, s, o, l, u, t, i, o, n, end text
For example, let's say we want to make 0, point, 250, start text, L, end text of an aqueous solution with open bracket, start text, N, a, C, l, end text, close bracket, equals, 0, point, 800, start text, M, end text. What mass of the solute, start text, N, a, C, l, end text, would we need to make this solution?
We can use the rearranged molarity equation to calculate the moles of start text, N, a, C, l, end text needed for the specified concentration and volume:
\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, point, 44, start fraction, start text, g, end text, divided by, start text, m, o, l, end text, end fraction, to convert from moles to grams of start text, N, a, C, l, end text:
start text, M, a, s, s, space, o, f, space, N, a, C, l, end text, equals, 0, point, 200, start cancel, start text, m, o, l, end text, end cancel, times, start fraction, 58, point, 44, start text, g, end text, divided by, 1, start cancel, start text, m, o, l, end text, end cancel, end fraction, equals, 11, point, 7, start text, g, space, N, a, C, l, end text
In practice, we could use this information to make our solution as follows:
Step 1, point, space Weigh out 11, point, 7, start text, g, end text of sodium chloride.
Step 2, point, space Transfer the sodium chloride to a clean, dry flask.
Step 3, point, space Add water to the start text, N, a, C, l, end text until the total volume of the solution is 250, start text, m, L, end text.
Step 4, point, space Stir until the start text, N, a, C, l, end text 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:
start text, M, o, l, a, r, i, t, y, end text, equals, start fraction, start text, m, o, l, space, s, o, l, u, t, e, end text, divided by, start text, L, space, o, f, space, s, o, l, u, t, i, o, n, end text, end fraction
• 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 start text, P, b, left parenthesis, N, O, end text, start subscript, 3, end subscript, right parenthesis, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis and start text, K, I, end text, left parenthesis, a, q, right parenthesis. When these two solutions are combined, bright yellow start text, P, b, I, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis precipitates out of solution. The balanced equation for this reaction is:
start text, P, b, left parenthesis, N, O, end text, start subscript, 3, end subscript, right parenthesis, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis, plus, 2, start text, K, I, end text, left parenthesis, a, q, right parenthesis, right arrow, start text, P, b, I, end text, start subscript, 2, end subscript, left parenthesis, s, right parenthesis, plus, 2, start text, K, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis
If we have 0, point, 1, start text, L, end text of 0, point, 10, start text, M, space, P, b, left parenthesis, N, O, end text, start subscript, 3, end subscript, right parenthesis, start subscript, 2, end subscript, what volume of 0, point, 10, start text, M, space, K, I, end text, left parenthesis, a, q, right parenthesis should we add to react with all the start text, P, b, left parenthesis, N, O, end text, start subscript, 3, end subscript, right parenthesis, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis?

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.
• 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 ?