Understanding Solution Concentration Basics

In this video, we're going to focus on solution concentration. We're going to talk about things such as mass percent, volume percent, mole fraction, molality, density, and molarity. So, before we begin working on these practice problems, let's write down a few notes. So you need to be familiar with the difference between the solute and the solvent. In this example problem, we have sodium chloride dissolved in water. Which one of these is the solute and which one is the solvent? The solute is the stuff that's being dissolved. In this case, that would be salt or sodium chloride. The solvent is the stuff that is doing the dissolving. So, sodium chloride is dissolved by water. So, water is the solvent. The solute is dissolved by the solvent. Now when you combine these two, you get the solution. So remember, the solute plus the solvent equals the solution. Now the next equation that you want to write down is mass percent. Mass percent is equal to the mass of the solute divided by the mass of the solution times 100%. Now, volume percent, which you can write as V%, it's very similar. It's equal to the volume of the solute Divided by the volume of the solution times 100% Now let's talk about the next topic which is mole fraction The mole fraction of substance A is equal to the moles of that substance divided by the total moles. So let's say if we had two substances, let's say A and B. So the mole fraction of B It's going to be the moles of B divided top divided by the total moles, which is the moles of a plus the moles of B Now let's say if we have three substances a B and C The mole fraction of let's say substance C. It's going to be the moles of substance C Divided by the total number of moles. So that's na plus and B plus NC So those are the formulas that you need in order to calculate mole fraction. Now the next thing we need to talk about is molarity. And I want to distinguish this with molality. So let's spell it out. This is molarity with an R. Malarity, represented by the symbol capital M, is equal to the moles of the solute divided by the liters of solution. So it's moles over volume, but the volume has to be in liters, not milliliters. Malality, this is with an L instead of an R. malality is equal to the moles of the solute but here is how it differs from molarity its moles over solute divided by the kilograms not of the solution but of the solvent so you need to be aware of that distinction So both molarity and molality, they both have moles of solute on top. But for molarity, you're dividing it by the volume of the solution. And for molality, you're dividing it by the mass of the solvent. This is in liters, and this is in kilograms. Now the next thing that you need to know, which you already know, is the density. Density is equal to mass over volume. So in chemistry, density is typically represented in units of grams per milliliter, or sometimes you'll see it as grams per cubic centimeter. So what you need to know is that one milliliter is... equivalent to one cubic centimeter and also one liter is a thousand milliliters and one kilogram is equal to 1,000 grams so those are the most common conversions that you can be using when dealing with solution concentration problems So now we have everything that we need to know for the problems that we're going to work on in this video. So let's begin with number one. 15 grams of sodium chloride was dissolved in 225 grams of water. What is the mass percent of sodium chloride in the solution? So mass percent is equal to the mass of the solute divided by the mass of the solution times 100%. So we need to identify the solute and the solvent, and the solution as well. The solute is sodium chloride. So we have 15 grams of solute. What is the mass of the solution? We're given the mass of the solvent, not the mass of the solution. So remember, the solute plus the solvent makes up the solution. So we have 15 grams of solute, we have 225 grams of solvent, so 15 plus 225 means that we have 240 grams of solution. That's what we're going to put on the bottom here. So let's divide 15 by 240, and then let's multiply that result by 100%. So the answer for this problem is 6.25%. So that's the mass percent of certain chloride in a solution. Now let's move on to number two. 25 milliliters of methanol is mixed with 150 milliliters of water. What is the volume percent of methanol? And what is the mass percent of water? So let's start with part A. Volume percent is equal to the volume of the solute divided by the volume of the solution. Notice that both methanol and water can behave as solvents. However, in this example, we have a greater quantity of water than methanol. So, methanol is going to be the solute, water is the solvent. So, we can say that methanol will be dissolved by water. And also, we're looking for the volume percent of methanol. So based on this question, the way it's worded, we're going to treat methanol as the solute. But in part B, we're looking for the mass percent of water. Based on the way that question is worded, we're going to treat water as the solute for the second part. Because we don't want the mass percent of methanol, we want the mass percent of water. So sometimes the way the question is worded tells you which substance you should treat as a solute. Because if we had a greater quantity of methanol, we could say that water is dissolved by methanol. So this problem is a little bit tricky. But for part A, methanol is chosen to be the solute. So we have 25 milliliters of solute. Let me not forget to put 100% here. And then the volume of the solution. The solution is the solute plus the solvent. So we need to add 25 plus 150. So we have 175 milliliters of solution So the volume percent is going to be 14.29 So that's the answer for part a Now let's move on to part B. So what is the mass percent of water? So this is going to be the mass of the solute divided by the mass of the solution times 100%. So we want the mass percent of water. Therefore, we're going to treat water as a solute based on the question. Now, we don't have the mass of the solute or the mass of the solution, so we need to use density to help us out. Density is mass over volume. So if we multiply both sides by volume, we get that mass is density times volume. So this is the formula we need to use, but we're going to use it with a conversion process. So let's start with the volume of methanol. So we have 25 milliliters, and the chemical structure of methanol is CH3OH. Now the density of methanol is 0.792 grams per cubic centimeter. And keep in mind, one cubic centimeter is the same as one milliliter. So if the density is reported as 0.792 grams per cubic centimeter, it's the same as 0.792 grams per cubic milliliter. So I'm going to put one milliliter on the bottom. So the unit milliliters will cancel. 25 times 0.792, that's 19.8. Now let's start with 150 milliliters of water. And if we multiply by the density of water, which is 1 gram per milliliter, we're going to get 150 grams of water. Now, to get the mass of the solution, we need to add the solute and the solvent together. So that's going to be, in this case, we're treating water as the solute. But regardless of the way you add them, it's going to be the same. 150 plus 19.8 is 169.8 grams of solution. So now let's go ahead and finish the problem. So the mass percent of water is going to be the mass of water, which is 150 grams, divided by the total, which is 169.8 grams. That's the mass of the solution, times 100%. So this will be 88.3%. Now, if the problem were to ask for the mass percent of methanol, then we would put 19.8 grams on top. If they were to say what is just the mass percent for the solution, and they didn't specify methanol or water, then you need to identify which one will most likely be the solute. Because both methanol and water can dissolve things. So typically the solute will be the one that's usually in lower quantity. We have much less of methanol than water. So if you have to choose, I would go with methanol as being the solute. But in this case, we specified what the solute was for each part of the problem. So be careful with questions like that. Number three, two moles of potassium chloride is dissolved in eight moles of water. What is the mole fraction of KCl? To calculate the mole fraction of KCl, it's going to be the moles of KCl divided by the total moles. That is the moles of KCl plus the moles of water. So it's going to be 2 moles divided by 2 plus 8, so that becomes 2 over 10, and 2 over 10 is 0.2. So that's the mole fraction of potassium chloride for this problem. Now let's move on to number 4. 25 grams of sodium fluoride is mixed with 200 grams of water. What is the mole fraction of sodium fluoride in the solution? So we're going to use the same formula. So we're looking for the mole fraction of sodium fluoride. Thus, we need to calculate the moles of sodium fluoride and the moles of water. So we need to convert grams to moles. So let's start with 25 grams of sodium fluoride. Let's convert it to moles. So we need to find the molar mass of sodium fluoride. If we look at the periodic table, sodium is approximately, it's like 22.99. and fluorine is 19. So if you want to round it to 23 you could. So let's say 23 plus 19, we're going to say approximately 42 grams per mole. So that's the molar mass of sodium fluoride. So we have one mole of NAF over 42 grams. So we can cross out the unit grams of NAF. 25 divided by 42 is 0.5952 moles. Now let's do the same thing with water. So we're going to start with 200 grams of water. and let's convert it to moles. So for water we have two hydrogen atoms so it's two times 1.008 and then oxygen is 16 so you get 18.016 grams per mole. So it's gonna be one mole of water over 18.016 grams of H2O. So we're going to divide 200 by 18.016. And we're going to get 11.1 moles of water. So now we can plug everything into our original equation. So it's 0.5952 over 0.5952 plus 11.1. You can put this in parentheses if you're going to type it in the way you see it. So the mole fraction that I got is 0.051. So that's the mole fraction of sodium fluoride in this solution. Number 5. 15 grams of sodium bromide is dissolved in 400 milliliters of solution. What is the molarity of the solution? Molarity is equal to the moles of the solute. divided by the liters of solution. So, for this problem, we can do a conversion process to get the answer. Let's start with 15 grams of sodium bromide over 1. And let's convert it to moles. The atomic mass of sodium is approximately 23, and for bromine, it's 79.9. So the molar mass of N-ABR is going to be 102.9 grams per mole. So what this means is that one mole of sodium bromide has a mass of 102.9 grams. So we can cross out grams of NaBr. So now we have moles of this solute, and we need to divide it by the liters of solution. Let's convert 400 milliliters into liters. To convert milliliters into liters, you need to divide by 1,000. One liter is 1,000 milliliters. So you can divide by 1,000 or move the decimal three units to the left. Either case, you're going to get 0.4 liters. So we need to divide moles by liters. So I'm going to write 1 over 0.4 liters of solution. So this will give us the molarity. We got moles on top, liters of solution on the bottom. So it's going to be 15 divided by 102. 0.9, that will give us the moles of NaBr, and then take that, divide it by 0.4 liters of solution. So the answer for this problem is 0.364m. So that is the molarity of the solution. Number 6. 10 grams of sodium hydroxide is dissolved in 500 grams of water. What is the molality of the solution? Molality, lowercase m, is equal to the moles of the solute divided by the kilograms of the solvent. So let's use the conversion process to get our answer, kind of like what we did in number 5. Now we need to identify the solute and the solvent. The solute is sodium hydroxide. The solute is typically a salt, and water, for most of these problems, will usually be the solvent. Now, when we mix methanol and water, it could be tricky determining which one is the solvent and which one is the solute. So you may have to look at the relative quantities of the two, or if the problem specifies which one is the solute. As we saw, the problem could say what is the volume percent of water, or what is the mass percent of NaOH, or what is the volume percent of methanol. Sometimes the problem will specify which one is the solute. Or which one should be treated as a solute, rather. But for this particular problem, the solute can easily be identified. It's typically some type of salt like sodium chloride, sodium hydroxide, potassium iodide. That's usually the solute. If you were to make sodium chloride with methanol, methanol will be the solvent. Methanol can dissolve a small amount of sodium chloride. So here's some examples of common solutes and common solvents. So most solutes are salts like sodium chloride, potassium iodide, sodium hydroxide, and so forth. These are typical solutes. Water is a solvent. Methanol is a solvent. Ethanol, that's another solvent. So if you were to combine these two, it's clear to see that methanol is a solvent. But if you were to mix two solvents together, then that's where you have to be careful. If the problem specifies what is the volume percent of methane, methanol, then you want to treat methanol as a solute. If it specifies what is the mass percent of water, treat water as a solute. If it's not specified, then you need to look at the relative quantities. So if I dissolve, let's say if I were to mix 10 milliliters of methanol with 100 milliliters of water. I would treat this as the solvent. But now let's say if I were to mix 50 milliliters of methanol with 5 milliliters of ethanol. Both of these are solvents, but relative to each other, I would select this one as the solvent. Because it exists in much greater quantity than ethanol So just want to clear that up for those of you who might be confused with that topic Now let's go ahead and finish this problem So we're going to use the conversion process like we did in number 5. Let's start with the solute. So we have 10 grams of sodium hydroxide. And let's convert it to moles. So for sodium hydroxide, the molar mass is going to be, it's 23 for Na, 16 for oxygen. For hydrogen, you could use 1.008, but I'm going to use 1 because it gives us a nice round number of 40 for NaOH. Sometimes I'm in the habit of using 1.008 for H. But if you were to round it to 40, your answer won't change much. So we have one mole of sodium hydroxide for every 40 grams of NaOH. So we can cross out those units. So right now we have moles of solute. All we need to do is divide it by the kilograms of solvent. So we have 500 grams of water. We need to convert it to kilograms. So let's convert it. From grams to kilograms, simply divide by 1,000, or move the decimal three units to the left. So 500 divided by 1,000 is 0.5. So we're going to divide moles of NaOH by 0.5 kilograms of the solvent, which is water. So moles over kilograms, that's going to give us the molality. 10 divided by 40, that's 0.25 moles of NaOH, divided by 0.5 kilograms of water, gives us a molality of 0.5. So that is the molality of the solution. Now let's move on to some harder problems. Number 7. The mass percent of AlCl3 in water is 15%. The density of the solution is 1.17 grams per milliliter. What is the molarity of the solution? So what is the concentration of aluminum chloride in water? So feel free to pause the video if you want to try this problem. Now we're looking for molarity, which means we need to find the moles of solute, the solute being aluminum chloride, and we also need to get the liters of solution. In order to get the moles of aluminum chloride, we need the mass of aluminum chloride, which we don't have, and we don't have the volume of the solution. So how can we get these two things in order to calculate molarity? Well, let's go to mass percent. We know that mass percent is the mass of the solute. divided by the mass of the solution times 100%. The mass percent can give us some numbers that we can use. In this problem, we have a 15% solution by mass of aluminum chloride. So what that means is that out of 100 grams of solution, there are 15 grams of aluminum chloride. Now what matters is not the individual number that we have here, but the ratio is important. Because a 15% solution could be 30 grams of aluminum chloride dissolve in 200 grams of solution. And so we can have an infinite number of ratios. We can have 45 grams of aluminum chloride over 300 grams of solution. But to keep things simple, I'd like to use 100 grams of solution because these two numbers are the same. So if I were to have, let's say, a 30% solution, I would have 30 grams of solute over 100 grams of solution. So if we select this number, we have to use this number. If we select 30 as the grams of aluminum chloride, we have to use 200. As long as the ratio is correct, you're going to get the right answer. So in this problem, we could say that if we have 15 grams of aluminum chloride, then we have 100 grams of solution. So now we could use this number to get the moles of aluminum chloride. Right now we need to get the liters of solution. That's what we're missing. But we do have the density of the solution. So let's start with 100 grams of solution. And let's convert that to milliliters using the density. So, one milliliter of solution has a mass of 1.17 grams, according to this value. So, we can cancel the unit grams. And then we can convert milliliters into liters. So, there's 1,000 milliliters per one liter of solution. So now we have the volume in liters. So it's 100 divided by 1.17 divided by 1,000. And that's 0.08547 liters. So now that we have the volume of the solution, we can calculate the molarity. So let's start with the mass of the solute. That is 15 grams of aluminum chloride over 1. And let's convert it to moles. So we need the molar mass of AlCl3. Aluminum has an atomic mass of 26.98, and chlorine, there's three of them, each of which has an atomic mass of 35.45. You can get those numbers using the periodic table. So this is going to be 133.33 grams per mole. So that means that one mole of aluminum chloride has a mass of 133.33 grams. So we can cancel these. Now that we have moles of solute on top, all we need to do is put the liters of solution on the bottom. So 15 divided by 133.33, that's 0.1125 moles. Divide that by 0.08547, and this will give you the molarity, which is 1.316. So that's how you can calculate the molarity of the solution if you're given the mass percent of the solution and the density of the solution.

In this example problem, we have sodium chloride dissolved in water. Which one of these is the solute and which one is the solvent? The solute is the stuff that's being dissolved. In this case, that would be salt or sodium chloride.

The solvent is the stuff that is doing the dissolving. So, sodium chloride is dissolved by water. So, water is the solvent.

The solute is dissolved by the solvent. Now when you combine these two, you get the solution. So remember, the solute plus the solvent equals the solution. Now the next equation that you want to write down is mass percent. Mass percent is equal to the mass of the solute divided by the mass of the solution times 100%.

Now, volume percent, which you can write as V%, it's very similar. It's equal to the volume of the solute Divided by the volume of the solution times 100% Now let's talk about the next topic which is mole fraction The mole fraction of substance A is equal to the moles of that substance divided by the total moles. So let's say if we had two substances, let's say A and B.

So the mole fraction of B It's going to be the moles of B divided top divided by the total moles, which is the moles of a plus the moles of B Now let's say if we have three substances a B and C The mole fraction of let's say substance C. It's going to be the moles of substance C Divided by the total number of moles. So that's na plus and B plus NC So those are the formulas that you need in order to calculate mole fraction.

Now the next thing we need to talk about is molarity. And I want to distinguish this with molality. So let's spell it out. This is molarity with an R. Malarity, represented by the symbol capital M, is equal to the moles of the solute divided by the liters of solution.

So it's moles over volume, but the volume has to be in liters, not milliliters. Malality, this is with an L instead of an R. malality is equal to the moles of the solute but here is how it differs from molarity its moles over solute divided by the kilograms not of the solution but of the solvent so you need to be aware of that distinction So both molarity and molality, they both have moles of solute on top.

But for molarity, you're dividing it by the volume of the solution. And for molality, you're dividing it by the mass of the solvent. This is in liters, and this is in kilograms. Now the next thing that you need to know, which you already know, is the density.

Density is equal to mass over volume. So in chemistry, density is typically represented in units of grams per milliliter, or sometimes you'll see it as grams per cubic centimeter. So what you need to know is that one milliliter is...

equivalent to one cubic centimeter and also one liter is a thousand milliliters and one kilogram is equal to 1,000 grams so those are the most common conversions that you can be using when dealing with solution concentration problems So now we have everything that we need to know for the problems that we're going to work on in this video. So let's begin with number one. 15 grams of sodium chloride was dissolved in 225 grams of water.

What is the mass percent of sodium chloride in the solution? So mass percent is equal to the mass of the solute divided by the mass of the solution times 100%. So we need to identify the solute and the solvent, and the solution as well. The solute is sodium chloride.

So we have 15 grams of solute. What is the mass of the solution? We're given the mass of the solvent, not the mass of the solution. So remember, the solute plus the solvent makes up the solution. So we have 15 grams of solute, we have 225 grams of solvent, so 15 plus 225 means that we have 240 grams of solution.

That's what we're going to put on the bottom here. So let's divide 15 by 240, and then let's multiply that result by 100%. So the answer for this problem is 6.25%.

So that's the mass percent of certain chloride in a solution. Now let's move on to number two. 25 milliliters of methanol is mixed with 150 milliliters of water. What is the volume percent of methanol?

And what is the mass percent of water? So let's start with part A. Volume percent is equal to the volume of the solute divided by the volume of the solution.

Notice that both methanol and water can behave as solvents. However, in this example, we have a greater quantity of water than methanol. So, methanol is going to be the solute, water is the solvent. So, we can say that methanol will be dissolved by water.

And also, we're looking for the volume percent of methanol. So based on this question, the way it's worded, we're going to treat methanol as the solute. But in part B, we're looking for the mass percent of water.

Based on the way that question is worded, we're going to treat water as the solute for the second part. Because we don't want the mass percent of methanol, we want the mass percent of water. So sometimes the way the question is worded tells you which substance you should treat as a solute. Because if we had a greater quantity of methanol, we could say that water is dissolved by methanol. So this problem is a little bit tricky.

But for part A, methanol is chosen to be the solute. So we have 25 milliliters of solute. Let me not forget to put 100% here.

And then the volume of the solution. The solution is the solute plus the solvent. So we need to add 25 plus 150. So we have 175 milliliters of solution So the volume percent is going to be 14.29 So that's the answer for part a Now let's move on to part B. So what is the mass percent of water?

So this is going to be the mass of the solute divided by the mass of the solution times 100%. So we want the mass percent of water. Therefore, we're going to treat water as a solute based on the question.

Now, we don't have the mass of the solute or the mass of the solution, so we need to use density to help us out. Density is mass over volume. So if we multiply both sides by volume, we get that mass is density times volume.

So this is the formula we need to use, but we're going to use it with a conversion process. So let's start with the volume of methanol. So we have 25 milliliters, and the chemical structure of methanol is CH3OH.

Now the density of methanol is 0.792 grams per cubic centimeter. And keep in mind, one cubic centimeter is the same as one milliliter. So if the density is reported as 0.792 grams per cubic centimeter, it's the same as 0.792 grams per cubic milliliter.

So I'm going to put one milliliter on the bottom. So the unit milliliters will cancel. 25 times 0.792, that's 19.8. Now let's start with 150 milliliters of water.

And if we multiply by the density of water, which is 1 gram per milliliter, we're going to get 150 grams of water. Now, to get the mass of the solution, we need to add the solute and the solvent together. So that's going to be, in this case, we're treating water as the solute.

But regardless of the way you add them, it's going to be the same. 150 plus 19.8 is 169.8 grams of solution. So now let's go ahead and finish the problem. So the mass percent of water is going to be the mass of water, which is 150 grams, divided by the total, which is 169.8 grams.

That's the mass of the solution, times 100%. So this will be 88.3%. Now, if the problem were to ask for the mass percent of methanol, then we would put 19.8 grams on top. If they were to say what is just the mass percent for the solution, and they didn't specify methanol or water, then you need to identify which one will most likely be the solute.

Because both methanol and water can dissolve things. So typically the solute will be the one that's usually in lower quantity. We have much less of methanol than water.

So if you have to choose, I would go with methanol as being the solute. But in this case, we specified what the solute was for each part of the problem. So be careful with questions like that.

Number three, two moles of potassium chloride is dissolved in eight moles of water. What is the mole fraction of KCl? To calculate the mole fraction of KCl, it's going to be the moles of KCl divided by the total moles. That is the moles of KCl plus the moles of water.

So it's going to be 2 moles divided by 2 plus 8, so that becomes 2 over 10, and 2 over 10 is 0.2. So that's the mole fraction of potassium chloride for this problem. Now let's move on to number 4. 25 grams of sodium fluoride is mixed with 200 grams of water. What is the mole fraction of sodium fluoride in the solution? So we're going to use the same formula.

So we're looking for the mole fraction of sodium fluoride. Thus, we need to calculate the moles of sodium fluoride and the moles of water. So we need to convert grams to moles.

So let's start with 25 grams of sodium fluoride. Let's convert it to moles. So we need to find the molar mass of sodium fluoride. If we look at the periodic table, sodium is approximately, it's like 22.99. and fluorine is 19. So if you want to round it to 23 you could.

So let's say 23 plus 19, we're going to say approximately 42 grams per mole. So that's the molar mass of sodium fluoride. So we have one mole of NAF over 42 grams.

So we can cross out the unit grams of NAF. 25 divided by 42 is 0.5952 moles. Now let's do the same thing with water. So we're going to start with 200 grams of water. and let's convert it to moles.

So for water we have two hydrogen atoms so it's two times 1.008 and then oxygen is 16 so you get 18.016 grams per mole. So it's gonna be one mole of water over 18.016 grams of H2O. So we're going to divide 200 by 18.016.

And we're going to get 11.1 moles of water. So now we can plug everything into our original equation. So it's 0.5952 over 0.5952 plus 11.1. You can put this in parentheses if you're going to type it in the way you see it. So the mole fraction that I got is 0.051.

So that's the mole fraction of sodium fluoride in this solution. Number 5. 15 grams of sodium bromide is dissolved in 400 milliliters of solution. What is the molarity of the solution?

Molarity is equal to the moles of the solute. divided by the liters of solution. So, for this problem, we can do a conversion process to get the answer. Let's start with 15 grams of sodium bromide over 1. And let's convert it to moles. The atomic mass of sodium is approximately 23, and for bromine, it's 79.9.

So the molar mass of N-ABR is going to be 102.9 grams per mole. So what this means is that one mole of sodium bromide has a mass of 102.9 grams. So we can cross out grams of NaBr.

So now we have moles of this solute, and we need to divide it by the liters of solution. Let's convert 400 milliliters into liters. To convert milliliters into liters, you need to divide by 1,000. One liter is 1,000 milliliters.

So you can divide by 1,000 or move the decimal three units to the left. Either case, you're going to get 0.4 liters. So we need to divide moles by liters. So I'm going to write 1 over 0.4 liters of solution. So this will give us the molarity.

We got moles on top, liters of solution on the bottom. So it's going to be 15 divided by 102. 0.9, that will give us the moles of NaBr, and then take that, divide it by 0.4 liters of solution. So the answer for this problem is 0.364m.

So that is the molarity of the solution. Number 6. 10 grams of sodium hydroxide is dissolved in 500 grams of water. What is the molality of the solution?

Molality, lowercase m, is equal to the moles of the solute divided by the kilograms of the solvent. So let's use the conversion process to get our answer, kind of like what we did in number 5. Now we need to identify the solute and the solvent. The solute is sodium hydroxide. The solute is typically a salt, and water, for most of these problems, will usually be the solvent.

Now, when we mix methanol and water, it could be tricky determining which one is the solvent and which one is the solute. So you may have to look at the relative quantities of the two, or if the problem specifies which one is the solute. As we saw, the problem could say what is the volume percent of water, or what is the mass percent of NaOH, or what is the volume percent of methanol. Sometimes the problem will specify which one is the solute.

Or which one should be treated as a solute, rather. But for this particular problem, the solute can easily be identified. It's typically some type of salt like sodium chloride, sodium hydroxide, potassium iodide.

That's usually the solute. If you were to make sodium chloride with methanol, methanol will be the solvent. Methanol can dissolve a small amount of sodium chloride. So here's some examples of common solutes and common solvents. So most solutes are salts like sodium chloride, potassium iodide, sodium hydroxide, and so forth.

These are typical solutes. Water is a solvent. Methanol is a solvent. Ethanol, that's another solvent. So if you were to combine these two, it's clear to see that methanol is a solvent.

But if you were to mix two solvents together, then that's where you have to be careful. If the problem specifies what is the volume percent of methane, methanol, then you want to treat methanol as a solute. If it specifies what is the mass percent of water, treat water as a solute. If it's not specified, then you need to look at the relative quantities. So if I dissolve, let's say if I were to mix 10 milliliters of methanol with 100 milliliters of water.

I would treat this as the solvent. But now let's say if I were to mix 50 milliliters of methanol with 5 milliliters of ethanol. Both of these are solvents, but relative to each other, I would select this one as the solvent. Because it exists in much greater quantity than ethanol So just want to clear that up for those of you who might be confused with that topic Now let's go ahead and finish this problem So we're going to use the conversion process like we did in number 5. Let's start with the solute. So we have 10 grams of sodium hydroxide.

And let's convert it to moles. So for sodium hydroxide, the molar mass is going to be, it's 23 for Na, 16 for oxygen. For hydrogen, you could use 1.008, but I'm going to use 1 because it gives us a nice round number of 40 for NaOH.

Sometimes I'm in the habit of using 1.008 for H. But if you were to round it to 40, your answer won't change much. So we have one mole of sodium hydroxide for every 40 grams of NaOH.

So we can cross out those units. So right now we have moles of solute. All we need to do is divide it by the kilograms of solvent.

So we have 500 grams of water. We need to convert it to kilograms. So let's convert it.

From grams to kilograms, simply divide by 1,000, or move the decimal three units to the left. So 500 divided by 1,000 is 0.5. So we're going to divide moles of NaOH by 0.5 kilograms of the solvent, which is water. So moles over kilograms, that's going to give us the molality. 10 divided by 40, that's 0.25 moles of NaOH, divided by 0.5 kilograms of water, gives us a molality of 0.5.

So that is the molality of the solution. Now let's move on to some harder problems. Number 7. The mass percent of AlCl3 in water is 15%. The density of the solution is 1.17 grams per milliliter.

What is the molarity of the solution? So what is the concentration of aluminum chloride in water? So feel free to pause the video if you want to try this problem. Now we're looking for molarity, which means we need to find the moles of solute, the solute being aluminum chloride, and we also need to get the liters of solution.

In order to get the moles of aluminum chloride, we need the mass of aluminum chloride, which we don't have, and we don't have the volume of the solution. So how can we get these two things in order to calculate molarity? Well, let's go to mass percent. We know that mass percent is the mass of the solute.

divided by the mass of the solution times 100%. The mass percent can give us some numbers that we can use. In this problem, we have a 15% solution by mass of aluminum chloride.

So what that means is that out of 100 grams of solution, there are 15 grams of aluminum chloride. Now what matters is not the individual number that we have here, but the ratio is important. Because a 15% solution could be 30 grams of aluminum chloride dissolve in 200 grams of solution. And so we can have an infinite number of ratios.

We can have 45 grams of aluminum chloride over 300 grams of solution. But to keep things simple, I'd like to use 100 grams of solution because these two numbers are the same. So if I were to have, let's say, a 30% solution, I would have 30 grams of solute over 100 grams of solution. So if we select this number, we have to use this number. If we select 30 as the grams of aluminum chloride, we have to use 200. As long as the ratio is correct, you're going to get the right answer.

So in this problem, we could say that if we have 15 grams of aluminum chloride, then we have 100 grams of solution. So now we could use this number to get the moles of aluminum chloride. Right now we need to get the liters of solution. That's what we're missing.

But we do have the density of the solution. So let's start with 100 grams of solution. And let's convert that to milliliters using the density.

So, one milliliter of solution has a mass of 1.17 grams, according to this value. So, we can cancel the unit grams. And then we can convert milliliters into liters. So, there's 1,000 milliliters per one liter of solution.

So now we have the volume in liters. So it's 100 divided by 1.17 divided by 1,000. And that's 0.08547 liters.

So now that we have the volume of the solution, we can calculate the molarity. So let's start with the mass of the solute. That is 15 grams of aluminum chloride over 1. And let's convert it to moles. So we need the molar mass of AlCl3.

Aluminum has an atomic mass of 26.98, and chlorine, there's three of them, each of which has an atomic mass of 35.45. You can get those numbers using the periodic table. So this is going to be 133.33 grams per mole. So that means that one mole of aluminum chloride has a mass of 133.33 grams.

So we can cancel these. Now that we have moles of solute on top, all we need to do is put the liters of solution on the bottom. So 15 divided by 133.33, that's 0.1125 moles. Divide that by 0.08547, and this will give you the molarity, which is 1.316.

So that's how you can calculate the molarity of the solution if you're given the mass percent of the solution and the density of the solution.

Transcript for:Understanding Solution Concentration Basics

Transcript for:
Understanding Solution Concentration Basics