Molarity: Study with Video Lessons, Practice Problems & Examples

Molarity, represented as M=moles of soluteliters of solution, quantifies the concentration of a solute in a solution. It is crucial for calculations involving solutions, where molarity serves as a conversion factor. For instance, a 5.8 M sodium chloride solution contains 5.8 moles of NaCl per liter. Understanding molarity enhances problem-solving skills in chemistry, particularly in stoichiometry and titration calculations, facilitating accurate quantitative analysis of chemical reactions.

Molarity

In these series of videos, we're going to take a look at the concept of molarity. Now, molarity itself is represented by the variable m. And in class, you'll hear molarity and concentration sometimes being used interchangeably. Remember, concentration represents the amount of a given solute within a certain amount of solution. Now, molarity is more specific. Molarity represents the number of moles of solute per liter of solution, and with this, we have our molarity formula. So, molarity, which is capital M, equals moles of soluteliters of solution. Now that we have a more fleshed out idea of what concentration is through molarity, let's take a look at some problems or ask to calculate molarity or to look at the components that help to make molarity.

Molarity Example 1

Here it says to calculate the molarity of a solution prepared by dissolving 23.7 grams of sodium hydroxide in enough water to make 2.50 liters of solution. So here we're looking for molarity. Molarity here equals our moles of solute which would be the NaOH and it is in a solution of 2.50 liters. So we need liters of solution. Alright, we already have the 2.50 liters of solution, so we just need to convert the grams of sodium hydroxide into moles. So one mole of sodium hydroxide goes on the bottom and then grams of sodium hydroxide go on the top. Sodium hydroxide is composed of 1 sodium, 1 oxygen, and 1 hydrogen. Multiply them by their atomic masses from the periodic table, and then add up the totals. This will give us the molar mass of sodium hydroxide. So when we add that all up, it's 39.998 grams. So that's going to go here. Here grams of sodium hydroxide cancel out, and now I'm going to have moles of sodium hydroxide. So that comes out to 0.5925 moles. Take that and plug it in, and we'll have the molarity of our solution. So this comes out to 0.237 molar for our solution.

Molarity

Now that we know what molarity is, we can use it to help us calculate unknowns. We're going to say problems with a molarity value present can use a given amount and conversion factors to isolate an end amount. Now, if you haven't watched our videos on dimensional analysis and conversion factors, I highly suggest you go back and take a look. Because it is just a continuation of this. We're now incorporating molarity into our calculations. The molarity value itself can represent one of those conversion factors because remember, conversion factors possess two different units connected together.

For example, we say 5.8 molar sodium chloride. 5.8 molar represents 5.8 moles of NaCl within 1 liter of solution. And because we're using two different units connected together, this is a conversion factor. So we can write it as 5.8moles of NaCl1liter of solution. So, this is what 5.8 molar really represents. Keep that in mind when we start doing calculations.

Molarity Example 2

In this example question, it says, how many grams of sodium phosphate, which has a molecular mass of 163.94 grams per mole, are present in 300 milliliters of a 0.550 molar sodium phosphate solution. Alright. So in this question, they're asking us to discover grams. Now before we start plugging in numbers and doing calculations, let's talk about what the sentence is saying and how we need to be on the lookout when we see it written in such a way. Now here we have a number, and here we have another number. And they are connected together by the word "of." Just remember, when "of" is in between two values, "of" is really telling us to multiply them together. Now we have molarity here, which remember molarity equals moles over liters. And realize this, that if I were to multiply both sides by liters, then we would see that moles=liters⋅molarity. Look at this. This is milliliters. If I were to convert those into liters and then multiply by the molarity, that would give me moles of sodium phosphate. And we know if we have moles of a compound, we can use its molecular mass or molar weight to help us determine the grams of that compound. So that's the key. So just remember here, when we have volume of molarity, that's really just telling us, hey, you have moles there.

Alright. So we're going to first take the 300 milliliters and I'm going to convert it into liters. Remember, this is a metric prefix conversion, so milliliters go on the bottom, liters go on top. Remember that the metric prefix is on the same side with 1. So 1 milli is 10 to the minus 3. Milliliters cancel out, now I have liters. Also remember that molarity is a conversion factor. So that 0.550 molar is really saying that I have 0.550 moles of that compound over 1 liter. So I need liters to cancel out so I'll put the 1 liter here, and then we have 0.550 moles of sodium phosphate. Here, liters cancel out and look, I have moles of sodium phosphate. Finally, I convert those moles into grams. So one mole of sodium phosphate, we're told that the molecular weight is 163.94 grams. So plug that in. So moles cancel out and I'll be left with grams at the end. So initially what I'll get is 27.0501 grams of sodium phosphate. Looking at the question, 300.0 has 4 significant figures, 0.550 has 3 significant figures, and this number here has 5 significant figures. We go with the least number of significant figures, so this comes out to be 27.1 grams of sodium phosphate. So this would be the grams of our compound. So just remember, when they're giving us the volume of molarity, they're telling us what the moles are. Here in this case, we just had to find grams. So take those moles that you've isolated and convert them into grams, and you'll have your final answer.

What volume in (µL) of 0.125 M HBr contains 0.170 moles HBr?

• A. 7.35x10³ μ L
• B. 7.35x10⁶ μ L
• C. 1.36x10³ μ L
• D. 1.36x10⁹ μ L
• E. 1.36x10⁶ μ L

Hypernatremia is a medical condition where a patient has high levels of sodium in their blood, and is the result of the body containing too little water. A patient has a measured sodium level of 165 mM. If 30.0 mL of their blood were drawn, what mass (in ng) of sodium would be present?

• A. 1.14x10⁷ ng
• B. 1.14x10⁶ ng
• C. 1.14x10⁸ ng
• D. 4.18x10⁸ ng
• E. 4.18x10⁶ ng

2.64 grams of an unknown compound was dissolved in water to yield 150 mL of solution. The concentration of the solution was 0.075 M. What was the molecular weight of the substance?

• A. 190 g/mol
• B. 30 g/mol
• C. 90 g/mol
• D. 230 g/mol
• E. 320 g/mol

A solution with a final volume of 750.0 mL was prepared by dissolving 30.00 mL of benzene (C₆H₆, density = 0.8787 g/mL) in dichloromethane. Calculate the molarity of benzene in the solution.

• A. 0.45 M
• B. 0.035 M
• C. 0.35 M
• D. 0.71 M
• E. 0.091 M

Here’s what students ask on this topic:

What is molarity and how is it calculated?

Molarity (M) is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity is:

$M=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{liters} \mathrm{of} \mathrm{solution}}$

For example, if you dissolve 1 mole of sodium chloride (NaCl) in 1 liter of water, the molarity of the solution is 1 M. This concept is crucial for various calculations in chemistry, such as stoichiometry and titration.

How do you calculate the molarity of a solution given the moles of solute and volume of solution?

To calculate the molarity of a solution, you need to know the number of moles of solute and the volume of the solution in liters. The formula is:

$M=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{liters} \mathrm{of} \mathrm{solution}}$

For example, if you have 2 moles of solute dissolved in 0.5 liters of solution, the molarity is:

$M=\frac{2 \mathrm{moles}}{0.5 \mathrm{liters}}=4 M$

What is the difference between molarity and concentration?

While molarity and concentration are often used interchangeably, they are not exactly the same. Concentration is a general term that describes the amount of solute in a given amount of solution. Molarity, on the other hand, is a specific type of concentration that measures the number of moles of solute per liter of solution. The formula for molarity is:

$M=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{liters} \mathrm{of} \mathrm{solution}}$

For example, a 1 M solution of NaCl means there is 1 mole of NaCl in 1 liter of solution.

How can molarity be used as a conversion factor in chemical calculations?

Molarity can be used as a conversion factor in chemical calculations because it relates moles of solute to liters of solution. For example, if you have a 5.8 M solution of NaCl, this means there are 5.8 moles of NaCl per liter of solution. You can use this relationship to convert between moles and liters. For instance, if you need to find out how many moles are in 2 liters of this solution, you can set up the following calculation:

$5.8 \mathrm{moles}/\mathrm{liter} \times 2 \mathrm{liters}=11.6 \mathrm{moles}$

This makes molarity a powerful tool for stoichiometry and other quantitative analyses in chemistry.

What are some common applications of molarity in chemistry?

Molarity is widely used in various applications in chemistry, including:

• Stoichiometry: Molarity helps in calculating the amounts of reactants and products in chemical reactions.
• Titration: Molarity is used to determine the concentration of an unknown solution by reacting it with a solution of known molarity.
• Solution Preparation: Molarity is essential for preparing solutions with precise concentrations for experiments.
• Concentration Calculations: Molarity aids in converting between moles and liters, facilitating various concentration-related calculations.

Understanding molarity is crucial for accurate quantitative analysis in these and other chemical processes.