Find Molality of Ethanol in Solution
This page has all of the required homework for the material covered in the second exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
- What is the mole fraction of ethanol in this solution?
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( 58.3 mL ) ( ) ( ) = 1 mole C2H5OH ( 500 mL H2O ) ( ) ( ) = 27.8 mole H2O The total moles = 1 mole C2H5OH + 27.8 mole H2O = 28.8 moles
The mole fraction = moles C2H5OH total moles = 1 mole C2H5OH 28.8 total moles = 0.035 The mole percent would be 3.5%.
- What is the weight fraction?
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( 58.3 mL ) ( ) = 46 g C2H5OH ( 500 mL H2O ) ( ) = 500 g H2O The total mass = 46 g C2H5OH + 500 g H2O = 546 g
The mass fraction = = 46 g C2H5OH 546 total grams = 0.084 The mass percent would be 8.4%.
- What is the molarity?
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The molarity = moles C2H5OH L of solution = = 1.79 M - What is the molality?
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The molality = moles C2H5OH kg of solvent = 1 mole C2H5OH 0.5 kg H2O = 2 m
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- What is the mole fraction and mole percentage of water in this solution?
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( 54.94 g C3H7OH ) ( 1 mole C3H7OH 60 g C3H7OH ) = 0.916 mole C3H7OH ( 30 g H2O ) ( ) = 1.67 mole H2O The total moles = 0.916 mole C3H7OH + 1.67 mole H2O = 2.586 moles
The mole fraction = = 1.67 mole H2O 2.586 total moles = 0.646 The mole percent would be 64.6%.
- What is the mass fraction and mass percent of water?
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The total mass = 54.94 g C3H7OH + 30 g H2O = 84.94 g
The mass fraction = = 30 g C2H2O 84.94 total grams = 0.353 The mass percent would be 35.3%.
- What is the molarity of water?
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To get the liters of solution we use the total mass and the density. The total mass is 54.94 g plus 30 g or 84.94 g.
( 84.94 g solution ) ( ) = 100 mL of solution ( 30 g H2O ) ( ) = 1.67 mole H2O The molarity = moles H2O L of solution = = 16.7 M - What is the molality of water?
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The molality = moles H2O kg of solvent = 1.67 mole H2O 0.05494 kg C3H7OH = 30.4 m
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( 5 mL C6H14 ) ( 0.658 g C6H14 1 mL C6H14 ) ( 1 mole C6H14 86 g C6H14 ) = 0.0383 mole C6H14 ( 100 mL C6H6 ) ( 0.8765 g C6H6 1 mL C6H6 ) ( ) = 0.08765 kg C6H6 The molality = 0.0383 mole C6H14 0.08765 kg C6H6 = 0.437 m
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- ΔT = kfm ⇒ 20 °C = (1.86 °C/m)m ⇒
The molality = = 10.75 m = 10.75 moles C2H6O2 kg of water ( 3 L H2O ) ( ) ( 10.75 mole C2H6O2 1 kg H2O ) = 32.25 mole C2H6O2 ( 32.25 mole C2H6O2 ) ( 62 g C2H6O2 1 mole C2H6O2 ) ( 1 mL C2H6O2 1.11 g C2H6O2 ) = 1800 mL C2H6O2 = 1.8 L
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( 0.5 mole Ca(NO3)2 1 kg water ) ( 3 moles of particles 1 mole Ca(NO3)2 ) = 1.5 m in particles ( 0.8 mole sucrose 1 kg water ) ( 1 moles of particles 1 mole sucrose ) = 0.8 m in particles ( 0.6 mole LiF 1 kg water ) ( 2 moles of particles 1 mole LiF ) = 1.2 m in particles ΔTf = kfm suggests that the change in freezing point is directly related to the molality of the particles in the solution. The larger the molality, the larger the ΔT and the lower the freezing point. The highest freezing point will be the solution with the smallest molality, which is sucrose. The LiF solution will be next and the Ca(NO3)2 solution will have the lowest freezing point.
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The molality = = = 0.0587 m = 0.0587 moles nandrolone kg of CCl4 (500 mL CCl4)(1.59 g/mL) = 795 g CCl4 = 0.795 kg CCl4
( 0.795 kg CCl4 ) ( 0.0587 mole nandrolone 1 kg CCl4 ) = 0.04667 mole nandrolone The molecular weight = 20 g nandrolone 0.04667 mole nandrolone = 428.5 g/mole (C28H44O3)
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- Using Henry's Law: k = Cg/Pg = (6.5 x 10-5 M)/(0.225 atm) = 0.00029 M/atm. Under the new conditions the concentration would be: Cg = (0.00029 M/atm)(0.18 atm) = 5 x 10-5 M
The fish will have plenty of oxygen!
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In this case the solute is non-volatile and so P°solute = 0.
Psoln = XsolventP°solvent + XsoluteP°solute = XsolventP°solvent
Xsolvent = Psoln / P°solvent = (80 torr)/(120 torr) = 0.667
Xsolvent + Xsolute = 1 ⇒ Xsolute = 1 - 0.667 = 0.333
Calculate the total vapor pressure above a solution at 20 °C when 100 moles of C6H12 are combined with 10 moles C7H16. Here is some data for these substances that can be used in this problem and the next two problems:
P° (at 20°C) | Density | Molec.Wt. | |
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C6H12 | 77.7 Torr | 0.779 g/mL | 84.16 g/mole |
C7H16 | 40 Torr | 0.684 g/mL | 100.21 g/mole |
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- Ptot = XC6H12 P°C6H12 + XC7H16 P°C7H16 = PC6H12 + PC7H16
PC6H12 = ( 100 mole C6H12 110 Total Moles ) ( 77.7 Torr ) = 70.64 Torr (C6H12) PC7H16 = ( 10 mole C7H16 110 Total Moles ) ( 40 Torr ) = 3.636 Torr (C7H16)
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Get the moles of each:
( 500 mL C6H12 ) ( ) ( 1 mole C6H12 84.16 g C6H12 ) = 4.628 mole C6H12 ( 100 mL C7H16 ) ( ) ( 1 mole C6H12 100.21 g C7H16 ) = 0.6826 mole C7H16 PC6H12 = ( 4.628 mole C6H12 5.311 Total Moles ) ( 77.7 Torr ) = 67.7 Torr (C6H12) PC7H16 = ( 0.6826 mole C7H16 5.311 Total Moles ) ( 40 Torr ) = 5.14 Torr (C7H16)
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- Ptot = XC6H12 P°C6H12 + XC7H16 P°C7H16 ; 1 = XC6H12 + XC7H16 70 Torr = (1 - XC7H16 )(77.7 Torr) + (XC7H16 )(40 Torr) Solve to get XC7H16 = 0.2 and C6H12 = 0.8
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- Ptot = XC6H12O6 P°C6H12O6 + XH2OP°H2O ; P°C6H12O6 = 0 XH2O = Ptot/P°H2O = (9 torr)/(9.21 torr) = 0.977 = (mole H2O)/(total moles)
( 500 mL H2O ) ( ) ( ) = 27.77 mole H2O
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(a) An ideal solution, all intermolecular interactions are the same.
(b) Sovent-solvent and solute-solute intermolecular forces are greater than the solvent-solute intermolecular forces. It makes it harder to dissolve and more likely to go into the vapor phase.
(c) Solvent-solvent and solute-solue intermolecular forces are less than the solvent-solute intermolecular forces. The solute is attracted into the solvent and there is less vapor above the solution.
Find Molality of Ethanol in Solution
Source: https://www2.rivier.edu/faculty/dburgess/web/genchem/chem2HomeworkEx2.htm