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A solution of 2.50 g of a compound having the empirical formula $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}$ in $25.0 \mathrm{g}$ of benzene is observed to freeze at $4.3^{\circ} \mathrm{C}$. Calculate the molar mass of the solute and its molecular formula.

$4.3 \times 10^{2} \mathrm{g} / \mathrm{mol} . \mathrm{C}_{24} \mathrm{H}_{20} \mathrm{P}_{4}$

Chemistry 102

Chapter 12

Physical Properties of Solutions

Solutions

Carleton College

University of Kentucky

University of Toronto

Lectures

03:58

In chemistry, a solution is a homogeneous mixture composed of two or more substances. The term "solution" is also used to refer to the resultant mixture. The solution is usually a fluid. The particles of a solute are dispersed or dissolved in the solvent. The resulting solution is also called the solvent. The solvent is the continuous phase.

05:06

In physics and thermodynamics, the natural tendency of a system to change its state is its tendency to increase the entropy of the system. It is a measure of the disorder in a system.

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A solution of $2.50 \mathr…

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A solution of 2.50 g of a …

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The molecular weight of an…

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What are the molecular mas…

Hello, everyone. And thanks for joining me, Miss Hallstrom, as we use this session as an opportunity to learn how to find molar mass, using freezing point, depression information and given an empirical formula, use the molar mass. We we determined to find the molecular formula. The substance we will be considering today is C six h five p. And if you look at this, you can see that this is an empirical formula. Will come back to this when we're finished with our freezing point depression calculations to see if the empirical formula is the molecular formula and I'll give you a hint. It isn't. Let's quickly review the four steps for finding bowler Mass from freezing point depression data. Our first step will be to find the change in freezing point. We find that from subtracting the freezing point of our solvent, our peer solvent from the freezing point of our solution, that'll be Step one, Step two. We're going to use the freezing point depression calculation where this is our value from step one. Que is a freezing point Depression constant and Emma's morality made stick and I on here the Vantaa factor. But the substance were dealing with the day is molecular, so we can ignore it and we will be solving formal ality will rearranges equation. Bring Monette morality. Third, morality is equal to bowls of Salyut per kilograms of solvent and will be solving for bulls. And last but not least, Moller Mass will be grams of Salyut divided by the moles of our Salyut. And as we look at this, we will be given the temperature of our salute. The freezing point of our solution. The freezing point of the solvent, the kilograms of solvent and the grams of Salyut. The constants are something we can look up or in this case, I'll give it to you and let me change color one more time. The moles right here will be used right here. So that's what we're going to dio. I'm not going to copy these steps down. Um, I guess I will, to a certain extent, but not too much on the next page. Okay, let's begin. Step 10 I better give you some data first. First we have given grams of Salyut and the grams of Salyut that were given is 2.50 We also know our grams of our solvent and were given 25.0 grands of our solvent, which is benzene. And let's go ahead right now and write that in kilograms because that's what we'll need to use. We're given the freezing point of temperature for the solution. And we know that the freezing point of the pure solvent benzene is 5.50 degrees C. And our solution waas 4.3 degrees c que for Bansi is 5.12 degrees C over morality. And that looks like everything we need. Okay, let's start step one. We're finding our delta t are changing freezing point. And to do that, we're taking this number minus this number. The temperature of her solution. The freezing point of our solution minus the freezing point of our peer solvent. Do our math on this one, and the answer is 1.2 degrees Celsius. And that's negative. Step two we're gonna use, um no, I will write it down, then rearrange it for you are changing. Freezing point is equal to K times am We're rearranging for em, which is Delta T. And this is a negative right there. This a negative over K. Okay, so let's sub in solve here I m equals negative. 1.2 degrees c Over and R K was 5.12 degrees c over morality. Do the math here and the answer is 0.234 for morality will worry about rounding later. For step three, morality is equal to moles of Salyut per kilogram of solve it and we're solving for moles. So that's gonna be morality. Times, kilograms and our morality was 0.2344 bowls. Times are kilograms was up here times 0.0 to 50 kilograms. Our answer for this is 5.8 59 times 10 to the minus third moles and our last and final step is for the molar mass, which is equal to grams of Salyut divided by moles of Salyut grams were given as 2.50 grams, divided by 5.85 nine times 10 to the minus three moles equals. Let's put this one in black rounded to correct significant figures 4.3 times 10 to the third second grams per mole or 430 grams per mole. When a jot this down right here. 430 grams per mole. And our empirical formula again was C six H five. Pete, we don't need to figure out our molar mass using, like, 12.1 for this. We can. You do? Sort of. Ah, quick, quick Moeller Mass calculation. And it's usually close enough for determining, um, molecular formulas. I'll stick this on here. Where did I write that? Because I have p and one. So I'm just gonna multiply these because they're all pretty close here. These air just the rounded bowler masses. So I get 72 I get five, I get 31. Add these together, I get 100 eight grams per mole as the molar mass for this. So if I take 4 30 divide that by 10 wait. I get four at exactly four, but you'll agree it's close enough. This means I need to multiply each one of the coefficients in my empirical formula by four. So the molecular formula will be see 24 h 20 p for and that is the second part of the answer. Here is one part of the answer. You can report this either way. Those were the same answer. There's the molecular, the molar mass. And here's the molecular formula. Thanks for joining me.

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