Chemical Concentrations - Online Tutor, Practice Problems & Exam Prep

In analytical chemistry, understanding solution properties is crucial. Molarity (M) is defined as the moles of solute per liter of solution, while molality (m) is moles of solute per kilogram of solvent. These concepts are essential for calculating concentrations and understanding colligative properties, which depend on the number of dissolved particles. A solution is a homogeneous mixture where the solute is dissolved in the solvent, impacting physical properties like boiling point and freezing point. Mastery of these terms is vital for solving related problems effectively.

Molarity vs. Molality

So recall in general chemistry, when we talked about the properties of solutions, we saw that the number of ions dissolved within a solution could have a direct impact on the different types of physical properties of those solutions. And remember, we also talked about terms such as colligative properties. Now we don't have to worry about that so much in terms of analytical chemistry, but we still need to be aware of certain units that we discussed when covering solutions. These include the units dealing with molarity and molality. Remember, molarity, which is capital M, is moles of our solute divided by liters of our solution.

Remember, your solute is the smaller portion. So, this is the smaller amount. And then we have our solvent, which is our larger amount. And together, they form our solution. A solution is just a homogeneous mixture in which the solvent has successfully dissolved, at least to an appreciable amount, the amount of solute.

And molarity is one of the terms that we associate with concentrations of solutions. Also, we have molality. Remember, molality is moles of solute divided by kilograms of solvent. Although they are different terms, they do share in common moles of solute.

Knowing this will be key to answering questions that we are going to see below. So just remember, some of the concepts that we learned in the past in general chemistry are now being reintroduced again when dealing with analytical chemistry. So come back to the next video and see how I approach the following example question that asks us to solve the molarity of a given solution.

Molarity & Molality

So here it says, a solution is prepared by mixing 20 grams of cadmium chloride. Here, we're given the molecular weight of cadmium chloride as being 183.317 grams per mole with 80 grams of water, which has a density of 20 degrees Celsius of 1.1988 grams per cubic centimeter. Here it says compute the molarity of cadmium chloride in this solution. Alright. So with any type of word problem, what we should always do first is write down what exactly they are asking me to solve. They're asking me to find the molarity of the cadmium chloride solution. So here M of CadmiumChloride = moles of my solute. Remember, your solute is the smaller portion within your given solution. Here we're told that we have 20 grams of cadmium chloride and 80 grams of water. Here, the 20 grams of cadmium chloride would represent my smaller amount, therefore it's my solute. And water her

Molarity & Molality

So here we're told that in order to sterilize our drinking water, chlorine is routinely added to our water supply. If the water fountains at a park have a chlorine level of 185 parts per million, calculate the molarity in micromolar. Now, they're asking us to figure out micromolar. What's going to be easier for us to do is to first calculate the molarity of this given solution and then change it into micromolar. So we want the molarity of our solution.

So that means we want moles of Cl2 because remember, chlorine is a diatomic molecule that exists naturally as Cl2. So when it dissolves in our water supply, it's also in Cl2 form, divided by liters of solution. We're told here that we have 185 parts per million. Recall that I said that 1 part per million is equal to 1 milligram per liter. So when they tell us 185 PPM, they're really saying that we have 185 milligrams of Cl2 divided by 1 liter of solution.

So all we really have to do here is convert milligrams into grams and then grams into moles. We'll have moles over liters at the end which gives us our molarity. So we're going to say here that 1 milligram is equal to 10e-3 grams. Milligrams cancel out, I have grams now. We're told the molecular weight of one chlorine is 35.453 grams per mole.

So here we're dealing with Cl2 which is two chlorines. So if we multiply this by 2, that gives us the weight of Cl2. So 1 mole of Cl2 when we multiply this number times 2, that gives me 70.906 grams of Cl2. So grams of Cl2 cancel out and now I have moles of Cl2.

I have moles over liters. At this point, that's going to give me 2.609×10-3 moles over liters which is just simply molarity. Now remember, we don't want molarity, we want micromolar. So we're going to say 2.609×10-3 molar. So we're going to say for every 1 micro, it's 10e-6.

Molarities cancel out and I'll have micromolar at the end. That comes out to being 2.61 and we do 2.61 because here 185 has 3 significant figures times 103 micromolars of Cl2. We're applying some of the concepts that we learned earlier on when we talked about parts per thousand, parts per million, and parts per billion. We use that information to find the molarity of our solution, and then from there we had to convert it into micromolar. So keep in mind when we're talking about concentrations of solutions, we tend to use the terms of molarity and molality when it comes to certain types of calculations.

We'll continue with this understanding of solutions as we tackle more concepts dealing with parts per thousand, parts per million, parts per billion, molarity, and molality.

Molarity & Molality 1

We continue with our discussion of molarity and molality of solutions. In this first question, it says a solution is prepared by dissolving 41.33 grams of nitric acid in enough water to make 100 ml of solution. We're told if the density of the solution is 1.380 grams per milliliter, what is the molality of nitric acid in the solution? Now, they're asking us to find molality. So remember, molality of this would just be the moles of nitric acid divided by kilograms of my solvent.

Next, we're going to write down all the given information. We're told that we have 41.33 grams of nitric acid, 100 ml of solution, and our solution has a density of 1.380 grams per milliliter. From this information, we need to calculate the molality of our solution.

Well, first thing, we can say is we want moles of nitric acid, and I have grams of nitric acid. So if I change that into moles, I can plug that into my formula. We're told that the molecular mass of nitric acid is 63.018 grams per mole. That's going to go on the bottom. For every 1 mol of nitric acid.

So when we do that, grams cancel out, and what we'll get here for our moles is 0.655844 mol of nitric acid. Now, we want to find the kilograms of our solvent which we assume is water. Remember that a solution is made up of solute plus solvent. I just want the solvent portion. We're told from the very beginning how many grams of solute we have, 41.33 grams of nitric acid.

We need to find the mass of our solution: We multiply the 100 ml of solution by the density of our solution. Milliliters of solution cancel out and now we have grams of solution which comes out to 138 grams of my solution.

So subtract that amount from the total amount of your solution: 138-41.33=96.67g

Molarity & Molality 2

So in this question, it says if the mole fraction of ethanol in an aqueous solution is 0.090, what is the molality and molarity? Here we're told the density of the solution is also 1.35 grams per milliliter. Now, they're asking us to solve for molality and molarity. Let's just first focus on solving for molality since it's stated first. The molality of my ethanol solution would equal the moles of ethanol divided by kilograms of my solvent.

Here, they tell us that it's an aqueous solution. That would mean that the solvent has to be water, so it'd be kilograms of water on the bottom. They tell me here that the mole fraction of ethanol is this value here. Remember, the mole fraction, which is represented by x, would be the moles of that solute, which is ethanol, over the moles of the solution. So when they tell me 0.090, that really means I have 0.090 moles of ethanol over now here this is always over 1 mole of solution.

Now realize also that the solution is made up of solute and solvent. If we were to expand this further, it'd be these many moles of ethanol divided by the moles of solute plus solvent together. We already know how many moles of solute we have. We have this number of 0.090 of our solute, and remember, the solute and solvent together equal 1.

We already know the solute is 0.090, so subtract that out. And the difference will be the moles of my solvent. Which we said earlier is water because it's an aqueous solution. Now we need more room guys so let me take myself out of the image. All right.

For molality, we know that our moles of our solute are 0.090 moles. All we have to do now is figure out our kilograms of water. We already have our moles of water, so take that, change that into grams, and then kilograms. So 1 mole of water. If you look within your book, you'll see that the atomic masses of the elements, hydrogen is 1.00794 grams, and there's 2 of them in water so multiply by 2.

And oxygen is 15.99994 grams. Again, remember, analytical chemistry, we have to be as precise as possible so I know it may be a hassle to write all these numbers out but if you want the best possible answer, you should write it all the way through. Multiply hydrogen times 2, add it to the amount of oxygen. That gives us a mass for water of 18.0153 grams of water. And then we're going to say here, we want to change that to kilograms, so just remember here that 1,000 grams are equal to 1 kilogram.

So when we solve for that, we'll get kilograms which I'm just going to write over here, write it underneath, but 0.016394 mol of water. That gives me 5.48984 molar. Here, this number here has 2 significant figures in it. This has 3 significant figures in it. So with 2 significant figures, that comes out to 5.5 molal.

Now that we've done molality, let's see if we can figure out the molarity. Molarity would be the moles of my ethanol as my solute divided by liters of the solution. So we already know the moles from earlier. It's 0.090 moles of ethanol. Now though, we need our liters of solution.

We're told that the density of the solution is 1.35 grams of solution per 1 milliliter of solution. There goes milliliters of solution on the bottom, but we need to isolate it. So that means we need to cancel out those grams of solution on top. To be able to do that, I need to figure out how many grams of solution I have. And we can figure that out because we know the moles of ethanol.

We know the moles of water. Change them both to grams, add them up, and that'll give us our grams of solution. All right. So with water, we have 0.91 moles of water. We found out its weight earlier is 18.0153 grams.

So that comes out to 16.3939 grams of water. Next, I need to convert the moles that I have of ethanol. I need to convert those into grams as well. So ethanol has in it carbon, hydrogen, and oxygen. We already know the masses of hydrogen that I wrote earlier but if you look on your periodic table, carbon will come out to 12.0107 grams and there's 2 of them so you'd have to multiply by 2.

Multiply the atomic mass of hydrogen by 6. Add in the mass of 1 oxygen and that'll give us the combined molecular mass of ethanol. So 1 mole of ethanol on the bottom, the mass will be 46.0684 grams on top. So that would give me 4.14616 grams of ethanol. Now if we add these two numbers together, the grams of water with the grams of ethanol, that'll give me my grams of the solution.

So that's what I'm doing. I'm adding these two numbers together to give me my grams of the solution. When I do that, I get 20.5401 grams of the solution. Now that I have grams of the solution, I bring in the density of the solution. So we have 1.35 grams of solution on the bottom, 1 milliliter of solution on top.

Grams of the solution cancel out. Now, I have milliliters of the solution. And then just remember, 1,000 ml is equal to 1 liter. That comes out to being 0.015-215 liters of the solution which I can then take and plug it below here to get the molarity of my solution. So that comes out to 5.91527 molar.

We want only 2 significant figures, so that comes out to 5.9 molar at the end. Based on the setup that we used, in both situations, we were able to isolate the molality of the solution and the molarity of the solution simply by looking at the mole fraction and the density of our solution.

Here’s what students ask on this topic:

What is the difference between molarity and molality?

Molarity (M) is defined as the number of moles of solute per liter of solution. It is given by the formula:

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

Molality (m), on the other hand, is defined as the number of moles of solute per kilogram of solvent. It is given by the formula:

$m=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{kilograms} \mathrm{of} \mathrm{solvent}}$

The key difference is that molarity depends on the volume of the solution, while molality depends on the mass of the solvent. This distinction is important because volume can change with temperature, but mass does not.

How do you calculate the molarity of a solution?

To calculate the molarity (M) of a solution, you need to know the number of moles of solute and the volume of the solution in liters. The formula for molarity 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 1 liter of solution, the molarity would be:

$M=\frac{2 \mathrm{moles}}{1 \mathrm{liter}}=2 M$

Thus, the molarity of the solution is 2 M.

What are colligative properties and how are they related to solution concentration?

Colligative properties are physical properties of solutions that depend on the number of dissolved particles, not their identity. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. The relationship between colligative properties and solution concentration is direct; as the concentration of solute particles increases, the magnitude of these properties also increases. For example, adding more solute to a solution will lower its freezing point and raise its boiling point. This is because the solute particles disrupt the solvent's ability to form a solid or vaporize, requiring more extreme conditions to achieve these phase changes.

How does temperature affect molarity and molality?

Temperature affects molarity and molality differently. Molarity (M) is temperature-dependent because it is based on the volume of the solution, which can expand or contract with temperature changes. As temperature increases, the volume of the solution may increase, leading to a decrease in molarity. Conversely, as temperature decreases, the volume may decrease, leading to an increase in molarity. Molality (m), however, is not affected by temperature because it is based on the mass of the solvent, which remains constant regardless of temperature changes. Therefore, molality is a more reliable measure of concentration when dealing with varying temperatures.

What is the importance of understanding solution properties in analytical chemistry?

Understanding solution properties in analytical chemistry is crucial for several reasons. First, it allows for accurate preparation and standardization of solutions, which is essential for reproducible and reliable results. Second, knowledge of solution properties such as molarity and molality helps in calculating concentrations, which is vital for quantitative analysis. Third, understanding colligative properties enables chemists to predict and control the physical behavior of solutions, such as boiling and freezing points, which can impact experimental outcomes. Overall, mastery of these concepts ensures precision and accuracy in analytical measurements and experiments.