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The vapor pressure of benzene is $100.0 \mathrm{mmHg}$ at $26.1^{\circ} \mathrm{C} .$ Calculate the vapor pressure of a solution containing $24.6 \mathrm{g}$ of camphor $\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\right)$ dissolved in $98.5 \mathrm{g}$ of benzene. (Camphor is a low-volatility solid.)

$P_{1}=\left(\frac{1.26 \mathrm{mol}}{1.26 \mathrm{mol}+0.162 \mathrm{mol}}\right) \times 100.0 \mathrm{mm} \mathrm{Hg}=88.6 \mathrm{mmHg}$

Chemistry 102

Chapter 12

Physical Properties of Solutions

Solutions

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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The vapor pressure of benz…

02:24

What is the vapor pressure…

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A solution of 75.0 g of be…

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The vapour pressure of pur…

03:06

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03:19

to computer vapor pressure of a solution. We gun to use Raoul's low, which says that the pressure off some substance say a is equal the pressure of the substance when it's pure, that would be our pure solvent times the Moeller ratio of it. So how many moles you have that species compared to the total. So for this problem, we have two suspicious with we have come for and we have been zing kcom for is dissolved in benzene. So we want to compute the vapor pressure of benzene. So to do that, we we are already have the pressure of benzene. When, um when it's pure, we have the data. So all we need is the molar ation. Okay, So in order to compute that first, we're going to start by computing the number of malls off can for so gonna say in seat for Ken for in we know that we have 24 0.6 grams. Okay, So to convert grams to moles, we used the molar mass so you can look up the more massive compute the mule. Muller. Matthew. With the formula, it's given in the problem too. So this is the molar Mass. 152.2 for the camp. For So using this, you can convert the mass, and you should be able to get 0.16 to mold. All right, now we do the same for the benzene. We have the same data for benzene. Oops. Compute the number of moles. So we know that we have 98.5 grams off things in in the Mormon benzene is 78 0.1 grams. Okay, so, um oh, I'm doing here is converting mass to more mess, but you should be able to do this easily, just dividing by the more mess. And in this case, you get 1.26 mol. There we go. So we have both numbers that we need. So now we just need to put everything together. Okay, So I'm gonna repeat the formula here. Okay? So pressure off the benzene, because being the thing that we want will be equals the beings in pure times, the Moeller ratio off benzene. Okay, now let's substitute things. So the pressure of being zines, but we want now we do have this pressure when it's pure. That is 100. That is given on the problem. 100 millimeters of mercury have that. Now we multiply by the molar ratio. So how do we compute molar ratio? Why was the moderation of benzene? So first, with Puto here in the top of dis fraction, the number of moles of beings in 1.26 more. Okay, and here on the bottom, we gonna add all the number of months. In this case, we have two components and our solution. That is again the benzene 1.26 and they came for, which will be 0.162 So the more the units off mole should cancel. The molar ratio is dimension. List has no units, so you just get a number out of them in, um, when you compute this, you should get 88.6 millimeters off mercury, and that's your final and

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