A solution contains 50.0 g of heptane (C7 H16) and 50.0 g of octane (C8 H18) at 25^∘ C The vapor

A solution contains 50.0 g of heptane (C7 H16) and 50.0 g of...

A solution contains 50.0 $\mathrm{g}$ of heptane $\left(\mathrm{C}_{7} \mathrm{H}_{16}\right)$ and 50.0 $\mathrm{g}$ of octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ at $25^{\circ} \mathrm{C}$ The vapor pressures of pure heptane and pure octane at $25^{\circ} \mathrm{C}$ are 45.8 torr and 10.9 torr, respectively. Assuming ideal behavior, answer the following: $$\begin{array}{l}{\text { a. What is the vapor pressure of each of the solution compo- }} \\ {\text { nents in the mixture? }} \\ {\text { b. What is the total pressure above the solution? }} \\ {\text { c. What is the composition of the vapor in mass percent? }} \\ {\text { d. Why is the composition of the vapor different from the com- }} \\ {\text { position of the solution? }}\end{array}$$

A solution contains 50.0 $\mathrm{g}$ of heptane $\left(\mathrm{C}_{7} \mathrm{H}_{16}\right)$ and 50.0 $\mathrm{g}$ of octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ at $25^{\circ} \mathrm{C}$ The vapor pressures of pure heptane and pure octane at $25^{\circ} \mathrm{C}$ are 45.8 torr and 10.9 torr, respectively. Assuming ideal behavior, answer the following:
$$\begin{array}{l}{\text { a. What is the vapor pressure of each of the solution compo- }} \\ {\text { nents in the mixture? }} \\ {\text { b. What is the total pressure above the solution? }} \\ {\text { c. What is the composition of the vapor in mass percent? }} \\ {\text { d. Why is the composition of the vapor different from the com- }} \\ {\text { position of the solution? }}\end{array}$$

Key Concepts

Ideal Solutions

An ideal solution is one in which the interactions between unlike molecules are similar to those between like molecules, making the mixing process thermodynamically ideal. In such solutions, properties like vapor pressure can be accurately predicted by linear relationships such as Raoult's Law.

Raoult's Law

Raoult's Law is a principle that describes the vapor pressure of an ideal solution. It states that the partial vapor pressure of each component in the solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. This law is crucial when analyzing vapor-liquid equilibrium in binary mixtures.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture by dividing the number of moles of that component by the total number of moles. It is a key parameter in applying Raoult's Law to determine partial pressures in an ideal solution.

Partial Pressure and Dalton's Law

Partial pressure is the pressure contribution of an individual gas in a mixture, while Dalton's Law states that the total pressure of a mixture of gases is the sum of the partial pressures of each component. These concepts are used together to determine the overall vapor pressure of a solution based on its components.

Phase Equilibrium

Phase equilibrium refers to the condition where the rate of evaporation equals the rate of condensation, leading to stable compositions in the vapor and liquid phases. Differences in volatility between components cause the vapor phase to have a composition different from that of the liquid phase during equilibrium.

Volatility

Volatility is a measure of how readily a substance vaporizes, which is directly related to its vapor pressure. In mixtures, components with higher volatility (higher vapor pressures) tend to dominate the vapor phase, which explains the difference between the vapor and liquid phase compositions.

Transcript

00:01 When we're looking at this problem, we have to consider something called rules law, which basically says the pressure of solution is going to equal to, this is the mole fraction, multiplied against the pressure of, we're called the pure solvent, whatever it originally was before it got mixed into something.

00:25 So we're given that, let's see, that heptane is mixed with octane, 50 grams of heptanes mixed with 50 grams of octane at pretty much standard temperature.

00:43 So if we're looking at this part, what first we got to do is figure out the moles of these things.

00:47 We got 50 grams of the heptane, so c7 h16.

01:01 Now in order to convert this, we just have to look at a periodic table and figure out and it was a molecular weight of c7h16.

01:10 And if we calculate it out, what we get is for every one mole of heptane is equivalent to about 100 grams.

01:21 We're writing it out this way, so we can just cross out our units, be left with moles.

01:26 50 divided by 100 gets you about 0 .5 moles of the heptane.

01:41 From there, we need to calculate also the octane.

01:50 So c8h18.

01:56 Again, it's 50 grams.

01:58 The 50 grams of each substance multiplied for one mole is actually 114 grams.

02:13 Cancel our like units.

02:15 We get about 0 .439 moles of the octane.

02:32 This is the heptane, c7, h16.

02:38 To find the total pressure of this stuff, let's say just a pressure of the heptane, we would add these two together.

02:49 So it looked like this, so it would say pressure of the c7 h18, it's going to be equal to, and we're going to use this formula here.

03:01 We use the mole fraction, so 0 .5 is the number of moles for the heptane over 0 .5 plus the 0 .49 moles of the octane.

03:19 Let's run a calculator.

03:34 What we get is 0 .9385.

03:45 We take that, we multiply that against the number, the pressure of the heptane, the case the pressure of the heptane is 45 .8 tors.

04:04 Alright, so you run it through the calculator.

04:06 What she gets is about 24 .4 .4.

04:10 4 tors.

04:12 Now this is for c7h18, the heptane.

04:19 C7h16 the heptane.

04:21 Similarly, we can run the same thing for the octane pressure of the octane.

04:37 But instead of putting 0 .5, we just put 0 .439 on the top.

04:45 We already know it's calculated to be 0 .9385.

04:53 We multiply that against the pressure of the octane.

05:01 Pressure octane is different.

05:02 We've got 10 .9 tors.

05:10 Running that through calculator, we go about 5 .099 tors.

05:17 5 .5.

05:17 I want to say 5.

05:19 We should say really 5 .1...

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