calculations involving colligative properties prentice-hall chapter 16.4 dr. yager
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Calculations Involving Calculations Involving Colligative PropertiesColligative Properties
Prentice-Hall Chapter 16.4Prentice-Hall Chapter 16.4Dr. YagerDr. Yager
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ObjectivesObjectives
SolveSolve problems related to molality and problems related to molality and mole fraction of a solutionmole fraction of a solution
DescribeDescribe how freezing-point depression how freezing-point depression and boiling point elevation are related to and boiling point elevation are related to molalitymolality
CalculateCalculate magnitudes of freezing-point magnitudes of freezing-point depression and boiling point elevation depression and boiling point elevation
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The DifferenceThe Difference
solution of liters
solute of moles (M) Molarity
solvent of kilograms
solute of moles (m) Molality
molality (m)- new term
Molarity (M) - term from 16.2
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Key PointKey Point
MolalityMolality and and mole fractionsmole fractions are two additional are two additional ways in which chemists express the concentration ways in which chemists express the concentration of a solution.of a solution.
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The molality (m) is the number of moles of solute dissolved in 1 kilogram (1000 g) of solvent. Molality is also known as molal concentration.
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To make a 0.500m solution of NaCl, use a balance to measure 1.000 kg of water and add 0.500 mol (29.3 g) of NaCl.
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How many grams of sodium fluoride are needed to prepare a 0.400m NaF solution that contains 750 g of water?
solvent of kilograms
solute of moles molality
g 13.g 12.6 solute of gramsmol
g42.0 x mol 0.30 solute of grams
mol 0.30 solute of moles
kg .75
solute of moles m 0.400
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In a solution containing nA mol of solute A and nB mol of solvent B, the mole fraction of solute A (XA) and the mole fraction of solvent B (XB) can be expressed as follows:
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Mole Fraction can be of Mole Fraction can be of Solute or SolventSolute or Solvent
solvent of moles solute of moles
solvent of moles solvent of mf
solvent of moles solute of moles
solute of moles solute of mf
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Adding Ethlylene Glycol (EG) to water as antifreeze.
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A solution contains 50.0g of carbon tetrachloride (CCl4) and 50.0g 0f chloroform (CHCl3). Calculate the mole fraction of each component in the solution.
mol 0.419
molg
119.3
g 50.0CHCl of moles
mol 0.325
molg
153.8
g 50.0CCl of moles
3
4
0.563 0.419mo 0.325mol
mol 0.419CHCl of fraction mole
0.437 0.419mol 0.325mol
mol 0.325CCl of fraction mole
3
4
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The magnitudes of the freezing-point depression and the boiling-point elevation of a solution are directly proportional to the molal concentration (m), when the solute is molecular, not ionic.
Freezing-Point Depression and Freezing-Point Depression and Boiling-Point ElevationBoiling-Point Elevation
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The constant, Kf, is the molal freezing-point depression constant, which is equal to the change in freezing point for a 1-molal solution of a nonvolatile molecular solute.
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The constant, Kb, is the molal boiling-point elevation constant, which is equal to the change in boiling point for a 1-molal solution of a nonvolatile molecular solute.
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What is the freezing point depression of an aqueous solution of 10.0 g of glucose (C6H12O6) in 50.0 g H2O?
C2.071.11m x m
C1.86m x KT
1.11mkg .0500
mol .0555glucose of molality
mol 0.0555
molg
180.
10.0g glucose of moles
oo
ff
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What mass of NaCl would have to be dissolved in 1.000 kg of water to raise the boiling point by 2.00oC?
g 114 mol
g58.5 x mol 1.96 NaCl of grams
NaCl mol 1.96 NaCl mol 1 x particles mol 2
particles mol 3.91 NaCl mol
NaCl mol 1
particles mol 2x
1.000kg
NaCl mol
1.000kg
particles 3.91mol 3.91m
3.91m
mC
.512
C2.00particles of molality
C2.00salt water of molality x m
C0.512m x KT
0
0
00
bb
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1. What is the mole fraction of He in a 1. What is the mole fraction of He in a gaseous solution containing 4.0 g of gaseous solution containing 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne?He, 6.5 g of Ar, and 10.0 g of Ne?
a.a. 0.600.60
b.b. 1.51.5
c.c. 0.200.20
d.d. 0.110.11
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1. What is the mole fraction of He in a 1. What is the mole fraction of He in a gaseous solution containing 4.0 g of gaseous solution containing 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne?He, 6.5 g of Ar, and 10.0 g of Ne?
a.a. 0.600.60
b.b. 1.51.5
c.c. 0.200.20
d.d. 0.110.11
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2. The freezing point depression caused 2. The freezing point depression caused by a given concentration of a by a given concentration of a nonvolatile molecular solute nonvolatile molecular solute
a.a. depends on the solute.depends on the solute.
b.b. depends on the solvent.depends on the solvent.
c.c. is always the same.is always the same.
d.d. cannot be determined.cannot be determined.
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2. The freezing point depression caused 2. The freezing point depression caused by a given concentration of a by a given concentration of a nonvolatile molecular solute nonvolatile molecular solute
a.a. depends on the solute.depends on the solute.
b.b. depends on the solvent.depends on the solvent.
c.c. is always the same.is always the same.
d.d. cannot be determined.cannot be determined.
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3. What are the freezing and boiling 3. What are the freezing and boiling points of a 0.1points of a 0.1mm solution of CaCl solution of CaCl22 in in
water? water?
a.a. -0.2°C, 100.1°C-0.2°C, 100.1°C
b.b. -0.6°C, 100.1°C-0.6°C, 100.1°C
c.c. -0.6°C, 100.2°C-0.6°C, 100.2°C
d.d. -0.6°C, 99.8°C -0.6°C, 99.8°C
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3. What are the freezing and boiling 3. What are the freezing and boiling points of a 0.1points of a 0.1mm solution of CaCl solution of CaCl22 in in
water? water?
a.a. -0.2°C, 100.1°C-0.2°C, 100.1°C
b.b. -0.6°C, 100.1°C-0.6°C, 100.1°C
c.c. -0.6°C, 100.2°C-0.6°C, 100.2°C
d.d. -0.6°C, 99.8°C -0.6°C, 99.8°C
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4. Compared to the freezing point 4. Compared to the freezing point depression by ethylene glycol (Cdepression by ethylene glycol (C22HH66OO22,) ,)
for a given solvent, the freezing point for a given solvent, the freezing point depression caused by the same molal depression caused by the same molal concentration of CaClconcentration of CaCl22 would be would be
• exactly the same.exactly the same.
• twice as large.twice as large.
• three times as large.three times as large.
• four times as largefour times as large
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4. Compared to the freezing point 4. Compared to the freezing point depression by ethylene glycol (Cdepression by ethylene glycol (C22HH66OO22,) ,)
for a given solvent, the freezing point for a given solvent, the freezing point depression caused by the same molal depression caused by the same molal concentration of CaClconcentration of CaCl22 would be would be
• exactly the same.exactly the same.
• twice as large.twice as large.
• three times as large.three times as large.
• four times as largefour times as large