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Ch.13 - Properties of Solutions

Chapter 13, Problem 106c

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 °C, and their solubilities in water at 25 °C and 1 atm fluorocarbon pressure are given as mass percentages. (c) Infants born with severe respiratory problems are sometimes given liquid ventilation: They breathe a liquid that can dissolve more oxygen than air can hold. One of these liquids is a fluorinated compound, CF3(CF2)7Br. The solubility of oxygen in this liquid is 66 mL O2 per 100 mL liquid. In contrast, air is 21% oxygen by volume. Calculate the moles of O2 present in an infant’s lungs (volume: 15 mL) if the infant takes a full breath of air compared to taking a full “breath” of a saturated solution of O2 in the fluorinated liquid. Assume a pressure of 1 atm in the lungs.

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Hey everyone, we're told that carbon dioxide is previously used as a refrigerant, raw carbon dioxide gas is a byproduct during the fermentation of sugars to lessen the amount of carbon dioxide that is released into the atmosphere. Raw carbon dioxide gas undergoes treatment and can be liquefied. The treated carbon dioxide gas is stored in tanks A and the liquefied carbon dioxide is stored in tanks B. The ability of oxygen gas in the liquefied carbon dioxide is 0.1 mL of oxygen per 100 mL liquefied carbon dioxide. Whereas the treated carbon dioxide gas contains 0.0 200% oxygen gas by volume listed in the following table are the parameters of tank and tank B calculate the moles of oxygen gas, President Tank A and Tank B. To answer this question, we need to use our ideal gas equation and that's going to be our pressure times volume is equal to the number of moles times r gas constant times our temperature. But since our question is asking us to come calculate the number of moles of oxygen gas, we're going to have to rearrange this equation. So we're going to get the number of moles is equal to our pressure times volume divided by our gas constant times our temperature. Starting with tank A. Let's first go ahead and determine our volume of oxygen gas. So taking that 250 liters of carbon dioxide that's listed in our table. We're going to use dimensional analysis and we were told that we had 0.0 200 Leaders of oxygen gas per 100 liters of carbon dioxide. So we converted that percentage into leaders by assuming that we had liters of carbon dioxide. So when we calculate this out and cancel out our units, we end up with 0.5 liters of oxygen gas. Now let's go ahead and write our temperature. We were told that we're at 35 degrees Celsius. Now we want to convert this into kelvin. So we're going to add 273.15 and this will get us to a temperature of 308.15 kelvin. Now let's go ahead and plug in these values into our equation, which is the number of moles equals our pressure times volume divided by our gas constant, times our temperature. So looking at our table, we can see that our pressure is 1.0 atmospheric pressure and we're going to multiply this by our volume, which is 0.5 liters of oxygen. Now, we're going to go ahead and divide this By R. Gas Constant, which is 0.08206 leaders times atmospheric pressure, divided by Mole Times Kelvin. Now we're going to multiply this value by our temperature which is 308. kelvin. Now, when we calculate this out and cancel out all of our units, we end up with 1.98 times 10 to the negative three mole of oxygen gas. Now, let's go ahead and look at tank B. Now for tank be, they told us that our solid ability is equal to 0.0150 ml of oxygen gas per 100 ml of carbon dioxide. Now we want to change this into leaders so this can be easily changed into leaders by saying that we have 0. liters of oxygen gas per 100 liters of carbon dioxide. Now, let's go ahead and look at our volume for our volume of oxygen gas. We're going to take our 250 liters of carbon dioxide from our table and we're going to use that conversion that we just listed. So using dimensional analysis, we know that we have 0.0150 L of oxygen gas per 100 L of carbon dioxide. Now, when we calculate this out, we end up with 0.0375 L of oxygen gas. Now, looking at our temperature, our table tells us that we have 20°C. So we're going to convert this into Kelvin by adding 273.15 and this gets us to 293.15 Kelvin. Now let's go ahead and plug in these values into our equation. So the number of moles is going to equal our pressure, which was said to be 75 atmospheric pressure Times are volume, which is 0.0375 L. And we're going to divide this all by our gas constant, which is 0.08206l times atmospheric pressure divided by Mole Times Kelvin. And we're going to multiply this value by our temperature of 293. Kelvin. Now, when we calculate this out and cancel out all of our units, We end up with 1.17 times 10 to the - mole of oxygen gas. Now, these are going to be our final answers. Now, I hope this made sense and let us know if you have any questions.
Related Practice
Textbook Question

Two beakers are placed in a sealed box at 25 °C. One beaker contains 30.0 mL of a 0.050 M aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains 30.0 mL of a 0.035 M aqueous solution of NaCl. The water vapor from the two solutions reaches equilibrium. (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?

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Textbook Question

Carbon disulfide (CS2) boils at 46.30 °C and has a density of 1.261 g/mL. (a) When 0.250 mol of a nondissociating solute is dissolved in 400.0 mL of CS2, the solution boils at 47.46 °C. What is the molal boiling-point-elevation constant for CS2?

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Textbook Question

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 °C, and their solubilities in water at 25 °C and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution.

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Textbook Question

At ordinary body temperature (37 °C), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N2 from air in blood at this pressure?

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Textbook Question
A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (d) Tetrabutylammonium, (CH3CH2CH2CH2)4N + is a bulky cation. Which anion, when paired with the tetrabutylammonium cation, would lead to a salt that will be most soluble in nonpolar solvents?
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Textbook Question

A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (a) What is the central atom and the number of electronpair domains around the central atom in each of these anions?

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