Molality is depicted as moles of solute per kilograms of solvent. Molality, which is denoted by lowercase m, involves moles of our solute, the smaller part of our solution, divided by kilograms of the solvent, the larger portion of our solution. This is similar to molarity, which uses a capital M. It too has moles of solute as the numerator, but the denominator is actually liters of solution. Because of their similarities, there are questions which at times will ask you to interchange between the two. Now, in the same way we can expand molarity, the same approach can be applied to molality. So let's say that we have 0.30 moles of sodium chloride. What does this translate into? Well, this just means that we have 0.30 moles of the solute which is sodium chloride divided by 1 kilogram of our solvent which is usually water. Let's say, for example, that we had a 0.25 molar solution of glucose in an aqueous solution. Now, again, 0.25 molar. The number itself means that's how many moles we have of the solute, so it'd be 0.25 moles of glucose. Here we say aqueous solution. Aqueous means that our solvent is water, so that would be 1 kilogram of water. So the takeaway from this is to remember, if they give you molality of a compound or solution, it just means that number in terms of moles divided by 1 kilogram of our solvent. This will be important to remember if we're trying to convert molality to let's say molarity or mole fraction or even mass percent. Now that we've learned the basics of molality, click on to the next video and take a look at Molality with Oz.
Molality - Online Tutor, Practice Problems & Exam Prep
Molality (m) represents the amount of solute dissolved per kilogram of solvent.
Understanding Molality
Molality
Video transcript
Osmolality
Video transcript
Alright, so ionic molality or osmolality represents the molality of dissolved ions within a solution. Now, for example, if we're given 0.30 molar of sodium chloride, sodium chloride dissociates into sodium ion and chloride ion. That's a total of 2 ions involved. If we want the osmolality, that's equal to the number of ions which we said was 2 times the molality of the compound, which is 0.30 molal. This will give us an ionic molality or osmolality of 0.60. So remember, when we're talking about osmolality or ionic molality, we have to take into account the number of ions that dissociate within our given solution. Now that you've seen this, move on to the example left on the bottom of the page and see if your answer matches mine.
osmolality = 2 × 0.30 molalThis calculation results in an osmolality of 0.60.
Calculate Molality Example
Video transcript
A solution is prepared by dissolving 43 grams of Potassium Chlorate in enough water to make 100 ml of solution. If the density of the solution is 1.760 grams per milliliter, what is the molality of potassium chlorate in the solution? We're told the molecular weight of potassium chlorate is 122.55 grams per mole. Alright, so we need to determine the molality. That means we need moles of potassium chlorate divided by kilograms of our solvent. From the given information, we have 43 grams of potassium chlorate, we have 100 ml of solution, the density of our solution, and the molecular weight of potassium chlorate. Now, the molecular weight of potassium chlorate can be used to convert our grams of potassium chlorate into moles. So, we're going to say here for every 1 mole of potassium chlorate, the mass is 122.55 grams. So grams cancel out and now we're going to have moles. When we do that, we get 0.350877 moles. So that equals 0.350877 moles. Now, we need to determine our kilograms of solvent. The only other piece of information that we have left to use is the volume of our solution and the density of our solution. We can multiply them together to cancel out the volumes. When we do that, that gives us 176 grams of solution. Now remember, a solution is composed of solute plus solvent. We already know the grams of solute, so we can subtract it out. Realizing that when we subtract out the grams of solute, we'll have our grams of solvent. Here it isn’t important what the identity of the solvent is, but it is water because we said that this was in water. So, now we just have to convert those 133 grams into kilograms. For every 1 kilogram, it's 1,000 grams. That's 0.133 kilograms. Take that and plug it below the moles and we'll have our molality as being 2.64 moles for a potassium chlorate solution. So, like we always do, we approach this like a dimensional analysis question. We write down the information they're asking us to find first, then we write all the given information. From there we just need to manipulate it and arrange it in a way to isolate the variables that we need which in this case helps us to find the molality. So keep this in mind whenever facing any questions dealing with determining the molality of a solution or an ion.
Molality Calculations
Molality Calculations Example 1
Video transcript
If the molality of glucose in an aqueous solution is 2.56, what is the molarity? Here they tell us that the density of the solution is 1.530 grams per milliliter. Alright, so we're looking for the molarity. Molarity here would equal moles of glucose divided by liters of solution. The given information is that we have 2.56 molal of glucose, and we're also given the density of the solution as 1.530 grams of solution per 1 milliliter of solution. Alright. So we have to expand on molality. 2.56 molal, what does that really mean? Well, that's equal to 2.56 moles of glucose per 1 kilogram of solvent, which in this case would have to be water because it's an aqueous solution. Aqueous means water. Right now, we can see that we have the moles of glucose, which we can plug in here. So we have 2.56 moles of glucose, which we can use to calculate the grams.
Converting moles to grams, we use the formula: 2.56 moles of glucose × 180.156gper mole. This gives us 461.119 grams of glucose. We also have 1 kilogram of water, which equals 1,000 grams of water. Here we have our solute, and combining it with our solvent gives us the total grams of solution. Using the given density to find the total volume, we calculate: 1,461.199 grams of solution × 1 milliliterof solution1.530 grams then convert the result to liters by dividing by 1,000.
This results in 0.955032 liters of solution. Using these values, we calculate the molarity: 2.56 moles of glucose0.955032 liters of solution which simplifies to approximately 2.68 molar. So in this example, we are converting from molality to molarity using the density of our solution. As always, write down the information that they ask us to find first, then manipulate the given information to obtain your desired answer at the end.
Now that we've done this example, move on to the second one and see if you can get the correct answer. Come back as always, and check if your answer matches up with mine.
Molality Calculations Example 2
Video transcript
So, here it says what is the ionic molality of nitrate ions in 0.305 molal of lead (IV) nitrate? Alright, so here we're just looking for the molality of our nitrate ions. Here, lead (IV) nitrate breaks up into 1 lead (IV) ion plus 4 nitrate ions. So, the ionic molality or osmolality would equal the 4 nitrate ions times the molality of the entire compound. So that comes out to being 1.22 molal. So just remember, with ionic molality, they can ask you for the total ionic molality which takes into account all the ions, so that would be 5 ions total times the molality of the compound or they can ask for the ionic molality of individual ions. In this case, there are 4 nitrate ions, so we'd multiply the total molality by 4 to get our final answer.
Now that we've seen examples 1 and 2, tackle the practice question left on the bottom of the page.
What is the mass percent of NH3 of a 1.25 m aqueous solution of NH3?