CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS
Hauser- General Chemistry I
revised 8/03/08
13.21 The solubility of Cr(NO3)3 •9 H2O in water is 208 g per 100 g of water at 15 °C.
A solution of Cr(NO3)3 •9 H2O in water at 35 °C is formed by dissolving 324 g in 100 g
water. When this solution is slowly cooled to 15 °C, no precipitate forms.
(a) What term describes this solution?
SUPERSATURATED. THIS SOLUTION IS NOT AT EQUILIBRIUM AND
HOLDS MORE SOLUTE THAT IT SHOULD AT THIS TEMP.
(b) What action might you take to initiate crystallization? Use molecular-level processes
to explain how your suggested procedure works.
ADD A SEED CRYSTAL. THIS MIGHT PROVIDE AN ALIGNED, TEMPLATE
CRYSTAL THAT DISSOLVED SOLUTE CAN ADD TO AS IT STARTS TO
CRYSTALLIZE.
13.25 Water and glycerol, CH2(OH)CH(OH)CH2OH, are miscible in all proportions.
What does this mean?
SOLUBLE IN VIRTUALLY ALL PROPORTIONS.
How do the OH groups of the alcohol molecule contribute to this miscibility?
THE "OH" GROUPS OF THE GLYCEROL ALLOWS HYDROGEN-BONDING
(MIXING) WITH THE HYDROGEN-BONDING WATER MOLECULES.
13.27 (a) Would you expect stearic acid, CH3(CH2)16COOH, to be more soluble in water
or in carbon tetrachloride? Explain.
ALTHOUGH STEARIC ACID HAS A "COOH" PORTION THAT MIGHT
HYDROGEN-BOND TO WATER, THE LONG NONPOLAR "TAIL"
DOMINATES AND STEARIC ACID WOULD BE MORE SOLUBLE WITH THE
CARBON TETRACHLORIDE MOLECULE. CARBON TETRACHLORIDE
(CCl4) IS A SYMMETRIC TETRAHEDRAL MOLECULE WITH NO HBONDING POSSIBLE.
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(b) Which would you expect to be more soluble in water, cyclohexane or dioxane?
Explain.
DIOXANE IS SLIGHTLY MORE SOLUBLE WITH WATER. DIOXANE,
ALTHOUGH NOT CAPABLE OF HYDROGEN-BONDING, DOES PRESENT
LONE PAIRS ON ITS OXYGENS THAT COULD ATTRACT THE WATER
MOLECULE AS WATER SEEKS TO QUENCH ITS PARTIALLY POSITIVE
HYDROGEN ATOMS WITH NEGATIVE CHARACTER.
13.29 Which of the following in each pair is likely to be more soluble in hexane, C6H14:
(explain your answer)
HEXANE IS NONPOLAR (NO SIGNIFICANT DIPOLE; NO H-BONDING)
(b) benzene (C6H6) or glycerol, CH2(OH)CH(OH)CH2OH
BENZENE. BENZENE IS NONPOLAR (NO SIGNIFICANT DIPOLE; NO HBONDING). SHOULD MIX WITH HEXANE. GLYCEROL WILL NOT ALLOW
HEXANE TO BREAK UP THE STRONG GLYCEROL HYDROGEN BONDS.
(c) octanoic acid, CH3CH2CH2CH2CH2CH2CH2COOH, or acetic acid, CH3COOH.
OCTANOIC ACID. THE LONG NONPOLAR TAIL OF OCTANOIC ACID
WILL ALLOW IT TO INTERACT WITH THE NONPOLAR HEXANE. THE
COMPACT ACETIC ACID MOLECULE WILL MAINTAIN ITS HYDROGENBONDING WITH ITSELF AND NOT INTERACT WITH HEXANE.
13.31 (a) Explain why carbonated beverages must be stored in sealed containers.
NONPOLAR CO2 IS NOT VERY SOLUBLE IN POLAR WATER, SO HIGH
PRESSURE IS REQUIRED TO FORCE IT INTO SOLUTION.
(b) Once the beverage has been opened, why does it maintain more carbonation when
refrigerated than at room temperature? GASES ARE LESS SOLUBLE AT HOTTER
TEMPS, SO A COOLER BEVERAGE WILL CONTAIN HIGHER LEVELS OF
CARBONATION.
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13.37 A solution is made containing 14.6 g of CH3OH in 184 g H2O. Calculate the mole
fraction of CH3OH.
χ = MOLE FRACTION = MOL SUBSTANCE / TOTAL MOLES
Convert grams to moles of each substance.
14.6 g CH3OH ( 1 mole CH3OH/ 32.05 g ) =0.45553 mol CH3OH
184 g H2O ( 1 mol H2O / 18.02 g ) = 10.210 mol H2O
Beware of the denominator !
χ CH3OH = 0.45552 mol CH3OH / (0.45552 mol CH3OH + 10.210 mol H2O) =
0.042706 = 0.0427 (3 SF)
13.9 The figure shows two identical volumetric flasks containing the same solution at
two temperatures.
(a) Does the molarity of the solution change with the change in temperature? Explain.
MOLARITY IS mol of solute / L of solution. SINCE LIQUID VOLUMES
CHANGE WITH TEMP., MOLARITY IS IMPACTED BY TEMPERATURE.
(b) Does the molality of the solution change with the change in temperature? Explain.
MOLALITY IS mol of solute / kg of solvent. MASS IS NOT IMPACTED BY TEMP
CHANGES, SO MOLALITY VALUES ARE INDEPENDENT OF
TEMPERATURE.
13.41 Calculate the molality of the following solution:
(a) 8.66 g benzene (C6H6) dissolved in 23.6 g carbon tetrachloride (CCl4)
MOLALITY m = mole of solute / kg of solvent
Convert grams of benzene solute to moles
8.66 g C6H6 ( 1 mole C6H6 / 78.12 g) =0.11085 mol C6H6
Convert grams of solvent to kg.
m = 0.11085 mol / [(23.6 g solvent) ( 1kg / 1000 g)] = 4.6972 = 4.70 m
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13.59 Consider two solutions, one formed by adding 10 g of glucose (C6H12O6) to 1 L of
water and the other formed by adding 10 g of sucrose (C12H22O11) to 1 L of water. Are
the vapor pressures over the two solutions the same? Why or why not?
VAPOR PRESSURE IS IMPACTED BY THE # OF SOLUTE PARTICLES IN A
SOLUTION. SUCROSE HAS A GREATER MOLAR MASS, SO THE SUCROSE
SOLUTION HAS LESS PARTICLES (LOWER MOLE COUNT) INTERFERING
WITH THE VAPORIZATION PROCESS. THIS SUCROSE SOLUTION WILL
HAVE A HIGHER VAPOR PRESSURE.
13.65 (a) Why does a 0.10 m aqueous solution of NaCl have a higher boiling point than
a 0.10 m aqueous solution of C6H12O6?
ALTHOUGH THE LABELS ARE THE SAME, WE MUST TAKE INTO
ACCOUNT THE DISSOCIATION OF PARTICLES IN THE SOLUTION. NaCl
DISSOCIATES INTO TWO PARTS, EFFECTIVELY MAKING A 0.20 m
SOLUTION. COVALENT C6H12O6 WILL NOT DISSOCIATE. THE GREATER #
OF PARTICLE IN THE SODIUM CHLORIDE SOLUTION WILL CAUSE A
GREATER BP ELEVATION.
(b) Calculate the boiling point of each solution.
ΔT = Kb m For H2O, Kb is 0.51 °C/m (given)
DO NOT FORGET TO ADD ΔT TO THE BP OF PURE WATER
(assume 100.00 °C)
For 0.10 m NaCl: (0.51 °C/m) (0.20 m) = 0.102 °C + 100.00 = 100.10 °C
For 0.10 m C6H12O6 : (0.51 °C/m) (0.10 m) = 0.051 °C + 100.00 = 100.05 °C
13.67 List the following aqueous solutions in order of increasing boiling point: 0.120 m
glucose, 0.050 m LiBr, 0.050 m Zn(NO3)2.
TAKE INTO ACCOUNT DISSOCIATION !
GLUCOSE
LiBr
Zn(NO3)2
1 unit X 0.120 m = 0.120 m
2 units X 0.050 m = 0.100 m
3 units X 0.050 m = 0.150 m
HIGHER CONCENTRATION LEADS TO GREATER BP ELEVATION.
lower BP to higher BP
LiBr < glucose < Zn(NO3)2
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13.99 Calculate the freezing point of a 0.100 m aqueous solution of K2SO4 , ignoring
interionic attractions.
TAKE INTO ACCOUNT DISSOCIATION !
ΔT = Kf m For H2O, Kf is 1.86 °C/m (given)
0.100 m K2SO4 X three ions = 0.300 m particles
DO NOT FORGET TO ADJUST ΔT TO THE FP OF PURE WATER
(assume 0.00 °C)
(1.86 °C/m) (0.300 m) = 0.00 - 0.558 °C = 0.558 °C = -0.56 °C is FP of solution
13.73 What is the osmotic pressure formed by dissolving 44.2 mg of aspirin (C9H8O4)
in 0.358 L of water at 25 °C?
Π = MRT
Calculate M (MOLARITY)
MOLE ASPIRIN =
C 9 X 12.01 =108.09
H 8 X 1.01 = 8.08
O 4 X 16.00 = 64.00
sum = 180.17 g/mol
MANY CONVERSIONS NEEDED:
44.2 mg ( 1 g / 1000 mg) = 0.0442 g
25 °C + 273.15 = 298.15 K (a 3 SF K value)
[0.0442 g aspirin (1 mol asp / 180.17 g)]
[0.0821 L atm/ mol K] [298.15 K]
Π = ---------------------------------------------------- ------------------------------------------- =
[0.358 L]
0.016773 = 0.0168 atm (3 SF) OSMOTIC PRESSURE (Π)
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