Understanding Molarity & Molality

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concept

## Molarity vs. Molality

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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 collective properties. Um 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 um units that we discussed when covering solutions. These include the units dealing with polarity and morality. Remember polarity which is capital, M is moles of our salute, divided by leaders of our solution. Remember your salute is a smaller portion. So this is the smaller amount and then we have our solvent which is our large amount and together they form our solution. A solution is just a homogeneous or homogeneous mixture in which the solvent has successfully dissolved at least to an appreciable amount. The amount of solute and polarity is one of the terms that we associate with concentrations of solutions. Also we have morality. Remember morality is moles of solute divided by kilograms of solvent, although there are different terms they do share in common moles of solute. Knowing this will be key to answering questions that we're gonna see below. So just remember some of the things, 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 polarity of a given solution.

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example

## Molarity & Molality

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So here it says a solution is prepared by mixing 20 g of cadmium chloride. Here we're given the molecular weight of cadmium chloride is being 183.317 g per mole with 80 g of water has a density of 20 degrees at 20 degrees Celsius of 1.1988 g times centimeters cubed inverse. Here it says compute the polarity of cadmium chloride in this solution. Alright, so with any type of word problem, but we should always do first is write down what exactly are they asking me to solve? They're asking me to find the polarity of cadmium chloride solution. So here I am of cadmium chloride equals moles of my salute. Remember your salute is a smaller portion within your given solution here, we're told that we have 20 g of cadmium chloride and 80 g of water. Here, the 20 g of cadmium chloride would represent my smaller amount. Therefore it's my salute. And water. Here's the larger amount. So it is my saw vent. That would mean that we're looking for moles of my solute which is cadmium chloride divided by leaders of my solution. Now, from the given information, Let's bring that all down. We have 20 g of cadmium chloride. We have 80 g of water. We have the density of my solution as 1.1988 g of solution per one cm cubed of solution. Now remember anything they're asking us to find is always somewhere within the given information. We need moles of cadmium chloride And here we have grams of cadmium chloride right there. So what we're gonna do first is we're gonna convert grams of cadmium chloride into moles of cadmium chloride. Bring that down. We have 20 g cadmium chloride. We're already given the massive cadmium chloride right from the beginning as this. 183.317 g per mole. So we put one mole of cadmium chloride on top. 183.317g of cadmium chloride on the bottom Here. That's gonna give me .109101 moles of cadmium chloride. Now remember analytical is the chemistry of precision which means we should not do any types of rounding until we get to the very end. So if you have a bunch of string of numbers within your answer at this point, you have to bring all of it into your calculations. So that's the moles that we're gonna plug up here. Now I need leaders of solution, realize here we have centimeters of solution. I need to isolate those centimeters of solution by canceling out these grams of solution to get to cancel out those grams of solution. I need to multiply it um with something else that has grams of solution. Remember your solution is saul you plus solvent? We have 20 g of cadmium chloride, 80 g of water together. That's 100 g of solution. Bring those 100 g down of solution. We want them to cancel out with the g of solution from the density. So that's 1.1988 g of solution on the bottom. One centimeters cubed of solution on top. Remember that one millimeter is the same thing as one centimeters cubed And that one leader is equal to 1000 ml. So at this point we're gonna have our leaders of solution which comes out to be .083417 L of solution. So take those leaders of solution and plug him here on the bottom. All right. So at this point we'll have 1.3079 Moeller as my answer. Let's look at the given information. Look how many significant figures they've given to us Here. This 20 has four significant figures within it. This 80 also has four significant figures. And this number here, 1.1988 has five significant figures. So at the end, we want our answer to have four significant figures surrounding this. Seven up because it's next to a nine gives me 1.308 moller for my cadmium chloride solution. So based on our understanding of polarity, this is the set up, we need to um to create in order to find our polarity of our cadmium chloride solution. Now that you've seen this one. Look to see if you can attempt example to attempted on your own If you get stuck, don't worry, just come back and take a look at how I approach that same type of question, and you'll see the techniques I employ in order to get the answer. So, good luck, guys.

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example

## Molarity & Molality

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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 polarity and micro moller. Now they're asking us to figure out micro moller what's gonna be easier for us to do is to first calculate the polarity of this given solution and then change it into micro molar. So we want polarity of our solution. So that means we want moles of cl two. Because remember chlorine is a diatonic molecule that exists naturally a cl two. So when it dissolves in our in our water supply, it's also in cl to form divided by leaders of solution. We're told here that we have 100 and 85 parts per million. Recall that? I said that one parts per million is equal to one mg per one liter. So when they tell us 100 and 85 ppm They're really saying that we have 185 mg of cl two Divided by one L of solution. So all we really have to do here is convert milligrams into grams and then grams into moles will have moles over leaders at the end which gives us our polarity. So we're gonna say here that one mg Is equal to 10 to the negative three g milligrams cancel. And I have grams. Now we're told the molecular weight of one chlorine is 35.453 g per mole. So here we're dealing with cl two which is two chlorine. So if we multiply this by two that gives us the weight of cl two, so one mole of cl two. When we multiply this number of times two, that gives me 70.90 and then six g of cl two, so grams of cl two cancel out. And now I have moles of cl two. So at the end now, what do I have? I have moles over leaders At this point. That's gonna give me 2.609 times 10 to the -3 moles over L, which is just simply polarity. Now remember we don't want more clarity, we want micro molar. So we're gonna say 2.609 times 10 to the minus three molar. So we're gonna say for everyone micro It's 10 to the -6. So polarities cancel out and I'll have micro moller at the end. That comes out to being two 61 And we do 2.61 because here 185 has three significant figures times 10 to the three Micro Muellers of Cl two. So we're applying some of the concepts that we learned earlier on when we talked about parts per 1000 parts per million parts per billion. We use that information to find the polarity of our solution and then from there we had to convert it into micro Moeller. So keep in mind when we're talking about concentrations of solutions. Um we tend to use the terms of polarity and morality when it comes to certain types of calculations, so we'll continue with this understanding of solutions as we tackle more concepts dealing with parts per 1000 parts per million, parts per billion, more clarity and morality.

Molarity & Molality Calculations

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example

## Molarity & Molality 1

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we continue with our discussion of malaria and morality of solutions. In this first question, it says a solution is prepared by dissolving 41.33 g of nitric acid in enough water to make 100 ml of solution, we're told if the density of the solution is 1.380 g per milliliter, what is the morality of nitric acid in the solution? Now they're asking us to find morality. So remember, morality of this would just be the moles of nitric acid divided by kilograms of my solvent. Next we're gonna write down all the given information. We're told that we have 41.33 g of nitric acid. We're told that we have 100 MLS of solution. We're told that our solution is 1.380 g of solution for everyone. Middle liter of solution from this information, we need to calculate the morality 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 63.18 g per mole. So that's gonna go on the bottom For every one mole of nitric acid. So when we do that grams cancel out and what we'll get here for our moles is .655844 moles of nitric acid. Now we want to find the kilograms of our solvent which we assume is water. But based on the information that we have left, what do we see? We have the volume of our solution and we have the density of our solution. Realize here that if I can multiply those two together, I can isolate the mass of my solution. So I'm gonna take the 100 mls of solution and I'm gonna multiply it by the density of my solution And here middle years of solution cancel out. And now I have grams of solution which comes out 238 g of my solution. Remember that a solution is made up of sol ute plus solvent? I just want the solvent portion. We're told from the very beginning how many grams of salt we have. We're told that we have 41.33 g of nitric acid. So subtract that amount from the total amount of your solution. That's gonna give me the amount of solvent I have. Now, we really don't need to know what the solvent is in this case. But um we know that it's water because it says it's water. All we have to do at this point is just change these grams into kilograms. So remember here that one kg is equal to 1000 g. So that's gonna give me .09667 kg at the end. Which you can plug on the bottom here. So at this point I'll have 6.784 36 molo we look at the numbers given to us in the question. So 41.33 has 46 fix Here. This also has four sig figs and this has 46 fix. The molecular way we don't include that because that could have not been given to us. We could have calculated by looking at the periodic table. So we have four sig figs amongst all these numbers. So we want our answer at the end. I have four significant figures as well. So that means that we have to change this to 6.784 molo for my nitric acid solution. So that represents the morality of this asset solution. Now that we've seen this one, we have in addition example to let's see if you guys can approach this question and solve for the morality and polarity that's being asked again. Don't worry if you can't get the answer, just come back and see how I approach the same question in order to find those two variables

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example

## Molarity & Molality 2

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So in this question, it says if the mole fraction of ethanol in an aqueous solution is 0.90, what is the morality and polarity here, we're told the density of the solution is also 1.35 g per milliliter. Now they're asking us to solve for morality and polarity. Let's let's just first focus on solving for morality since it's stated first. So morality of my ethanol solution would equal the moles of ethanol divided by kilograms of my solvent here. They're told they're telling us that it's an acquis 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. So remember 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 .090, that really means I have .090 moles of ethanol Over now here, this is always 1/1 mole of solution. Now realize also that solution is made up of solute and solvent. So if we were to expand this further, it would be this 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 .090 of our salyut. And remember the saw you've been solving together equal one. We already know the Salyut is .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. Alright, so for morality, we know that our moles of our salute is .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 g and then kg. So one mole of water. If you look within your book, you'll see that the atomic masses of the elements hydrogen is 1.794 g. And there's two of them in water. So multiply by two and oxygen is 15.9994 g. 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. So multiply hydrogen times two added to the amount of oxygen that gives us a mass for water of 18.0153 grams of water. And then we're gonna say here we want to change it to kg. So just remember here that 1000 g is equal to one kg. So when we saw for that we'll get kg which I'm just gonna right over here, right underneath. So .016394 moles of water That gives me 5.48984 molo Here. This number here has two significant figures in it. This has three significant figures in it. So with two significant figures that comes out to 5.5 molo. Now that we've done morality. Let's see if we can figure out the polarity polarity would be the moles of my ethanol as my solute divided by leaders of solution. So we already know the moles from earlier. It's .090 moles of ethanol. Now though we need our leaders of solution, We're told that the density of the solution is 1.35 g of solution per one middle liter 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 two g, add them up. And that will give us our grams of solution. Alright, so with water We have .91 moles of water. We found out it's way earlier Is 18.0153 g. So that comes out to 16.3939 g 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 at your periodic table, carbon will come out to 12.107 g and there's two of them. So you have to multiply by two, multiply the atomic mass of hydrogen by six, adding the mass of one oxygen. And that will give us the combined molecular mass of ethanol. So one mole of ethanol on the bottom. The mass would be 46.0684 g on top. So that would give me 4.14616 g of ethanol. Now if we add these two numbers together the grams of water with the grams of ethanol, that will give me my grams of solution. So that's what I'm doing. I'm adding these two numbers together to give me my grams of solution. When I do that I get 20.5401 g of solution. Now that I have grams of solution I bring in the density of the solution. So we have 1.35g of solution on the bottom one middle liter of solution on top grams of solution, cancel out Now I have middle liters of solution And then just remember 1000 ml is equal to one leader That comes out to being .015-15 L of solution, which which I can then take and plug it below here to get the polarity of my solution. So that comes out to 5.915 - seven moller. We want only two significant figures, so that comes out to 5.9 moller at the end. So based on the setup that we used in those situations were able to isolate the morality of the solution, and the polarity of the solution was simply looking at the mole fraction and the density of our solution.