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Home / Finding Properties of a Solution Using Density and Mass Percent

Finding Properties of a Solution Using Density and Mass Percent

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.

  • The density of ethanol is 0.789 g/mL and the density of water is 1.0 g/mL. A solution is made where 58.3 mL of ethanol, C2H5OH, is dissolved in 500 mL of water. Assume the volumes are additive.
    • What is the mole fraction of ethanol in this solution?
      • Answer
      ( 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?
      • Answer
      ( 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?
      • Answer
      The molarity =
      moles C2H5OH
      L of solution
      		  = = 1.79 M
    • What is the molality?
      • Answer
      The molality =
      moles C2H5OH
      kg of solvent
      		  =
      1 mole C2H5OH
      0.5 kg H2O
      		= 2 m
  • Isopropyl alcohol at the drug store is usually 70% isopropyl alcohol, C3H7OH, and 30% water by volume. This corresponds to 54.94 g C3H7OH and 30 g H2O in 100 mL. The density of this solution is 0.8494 g/mL. In this problem assume water is the solute and the alcohol is the solvent.
    • What is the mole fraction and mole percentage of water in this solution?
      • Answer
      ( 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?
      • Answer

      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?
      • Answer

      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?
      • Answer
      The molality =
      moles H2O
      kg of solvent
      		  =
      1.67 mole H2O
      0.05494 kg C3H7OH
      		= 30.4 m
  • What will the melting point and the boiling point be for a solution where 5 mL of hexane, C6H14, (density = 0.658 g/mL) is placed into 100 mL of benzene, C6H6 (density = 0.8765 g/mL)? The normal boiling point of benzene is 80.1 °C and the normal freezing point of benzene is 5.5 °C. For benzene the boiling point elevation constant, kb, is 2.53 °C/m and the freezing point depression constant, Kf, is 5.12 °C/m.
    Answer
    ( 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
    ΔTf = kfm =(5.12 °C/m)(0.437 m) = 2.24 °C New f.p. = 5.5 °C - 2.24 °C = 3.26 °C ΔTb = kbm =(2.53 °C/m)(0.437 m) = 1.1 °C New b.p. = 80.1 °C + 1.1 °C = 81.2 °C
  • How many liters of ethylene glycol, C2H6O2, should be added to 3 L of water to make a solution that will freeze at -20 °C (-4 °F)? The density of C2H6O2 is 1.11 g/ml and the density of water is 1 g/ml.
    Answer
    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
  • List the following aqueous solutions in the order of increasing freezing point: 0.5 m Ca(NO3)2, 0.8 m sucrose, 0.6 m LiF.
    Answer
    (
    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.

  • Nandrolone is an anabolic steroid (a muscle-building chemical) which occurs naturally in the human body, but only in tiny quantities. A 20 g sample of nandrolone was placed in 500 mL of CCl4 (density = 1.59 g/mL) and the freezing point of the solution was found to be 1.75 °C lower than the normal freezing point. What is the molecular weight of nandrolone? kf for CCl4 is 29.8 °C/m.
    Answer
    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)
  • The partial pressure of oxygen, O2, at the highest recorded atmospheric pressure was 0.225 atm and the concentration in water under those conditions was 6.5 x 10-5 M. What was the concentration in water at the lowest recorded atmospheric pressure when the partial pressure of oxygen was 0.18 atm? Fish need 1.56 x 10-7 M O2 in order to survive. Do they have a high enough concentration at the lowest recorded atmospheric pressure?
    Answer
    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!

  • The vapor pressure above a pure solvent is 120 torr. After a non-volatile solute is added to the solvent the pressure above the solution is 80 torr. What is the mole fraction of the solute in this solution?
    Answer

    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.
    C6H12 77.7 Torr 0.779 g/mL 84.16 g/mole
    C7H16 40 Torr 0.684 g/mL 100.21 g/mole
    Answer
    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)
    PTotal = 74.28 Torr
  • What is the vapor pressure above a solution made by combining 500 mL C6H12 and 100 mL C7H16? See the data table in the previous problem.
    Answer

    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)
    PTotal = 72.84 Torr
  • What are the mole fractions of C6H12 and C7H16 if the total vapor pressure above the solution is 70 Torr? See the data table in the next to last problem.
    Answer
    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
  • A mixture of sugar, C6H12O6, and 500 mL of water has a total vapor pressure of 9.0 torr at 10 °C. How much sugar was added to the water? At 10 °C the vapor pressure of pure water is 9.21 torr and the density of water is 0.9997 g/mL.
    Answer
    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
    ⇒ 0.977 = (27.77 mole H2O)/(x mole C6H12O6 + 27.77 mole H2O) Solve for x to get 0.654 mole C6H12O6 (0.654 mole C6H12O6)(180 g/mole) = 118 g C6H12O6
  • Describe the intermolecular forces for each of the solutions (a), (b), and (c) that would cause the behavior shown in each case.
    Answer

    (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.

    • Finding Properties of a Solution Using Density and Mass Percent

    Source: https://www2.rivier.edu/faculty/dburgess/web/genchem/chem2HomeworkEx2.htm