Rank the contents of the following containers in order of increasing entropy: [Section 13.1] (a) (b) (c)
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Textbook Solutions for Chemistry: The Central Science
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 \(25\ ^{\circ} \mathrm{C}\), and their solubilities in water at \(25\ ^{\circ} \mathrm{C}\) and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Explain why the molarity of each of the solutions should be very close numerically to the molality. (c) Based on their molecular structures, account for the differences in solubility of the four fluorocarbons. (d) Calculate the Henry’s law constant at \(25\ ^{\circ} \mathrm{C}\) for \(\mathrm{CHCIF}_{2}\), and compare its magnitude to that for \(\mathrm{~N}_{2}\left(6.8 \times 10^{-4} \mathrm{~mol} / \text {L-atm }\right)\). Suggest a reason for the difference in magnitude.
Fluorocarbon |
Solubility (mass %) |
\(\mathrm{CF}_{4}\) |
0.0015 |
\(\mathrm{CClF}_{3}\) |
0.009 |
\(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) |
0.028 |
\(\mathrm{CHClF}_{2}\) |
0.30 |
Solution
The first step in solving 13 problem number trying to solve the problem we have to refer to the 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 \(25\ ^{\circ} \mathrm{C}\), and their solubilities in water at \(25\ ^{\circ} \mathrm{C}\) and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Explain why the molarity of each of the solutions should be very close numerically to the molality. (c) Based on their molecular structures, account for the differences in solubility of the four fluorocarbons. (d) Calculate the Henry’s law constant at \(25\ ^{\circ} \mathrm{C}\) for \(\mathrm{CHCIF}_{2}\), and compare its magnitude to that for \(\mathrm{~N}_{2}\left(6.8 \times 10^{-4} \mathrm{~mol} / \text {L-atm }\right)\). Suggest a reason for the difference in magnitude.
Fluorocarbon
Solubility (mass %)
\(\mathrm{CF}_{4}\)
0.0015
\(\mathrm{CClF}_{3}\)
0.009
\(\mathrm{CCl}_{2} \mathrm{~F}_{2}\)
0.028
\(\mathrm{CHClF}_{2}\)
0.30
From the textbook chapter Properties of Solutions you will find a few key concepts needed to solve this.
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