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colligative key

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Name: ________________________
Hour: ____ Date: ___________
Chemistry: Colligative Properties – Freezing Point Depression
1. How much will the freezing point be lowered if enough sugar is dissolved in water to make a 0.50 molal
solution?
2. What is the freezing point of a solution of a nonelectrolyte dissolved in water if the concentration of the solution
is 0.24 m?
3. What is the freezing point of a solution that contains 68.4 g of sucrose, C 12H22O11, dissolved in 1.00x102 g of
water?
4. Suppose that 98.0 g of a nonelectrolyte is dissolved in 1.00 kg of water. The freezing point of this solution is
found to be –0.465oC. What is the molecular mass of the solute?
5. A researcher places 53.2 g of an unknown nonelectrolyte in 505 g napthalene. The nonelectrolyte lowers
napthalene’s freezing point by 8.8oC. What is the molar mass of the unknown substance?
ANSWERS:
1. 0.93oC
2. –0.446 oC
3. –3.72 oC
4. 392 amu
5. 81.4 g/mol
Name: ________________________
Hour: ____ Date: ___________
Chemistry: Colligative Properties – Boiling Point Elevation
1a. What is the molal boiling point constant for carbon tetrachloride?
b. The boiling point of carbon tetrachloride is 76.8oC. A given solution contains 8.10 g of a nonvolatile electrolyte in
300 g of CCl4. It boils at 78.4oC. What is the gram molecular mass of the solute?
2. The molal boiling point constant for ethyl alcohol is 1.22 oC/molal. Its boiling point is 78.4oC. A solution of
14.2 g of a nonvolatile nonelectrolyte in 264 g of the alcohol boils at 79.8 oC. What is the gram molecular
mass of the solute?
3. Calculate the mass of a mole of a compound that raises the boiling point of water to 100.78 oC at 101.3 kPa
when 51 g of the compound is dissolved in 500 g of water.
4. Suppose that 13 g of a nonelectrolyte is dissolved in 0.50 kg of benzene. The boiling point of this solution is
80.61oC. What is the molecular mass of the solute?
ANSWERS:
1b. 84.9 g/mol
2. 46.7 g/mol
3. 67 g/mol
4. 129 amu
Name: _________________________
Hour: ____ Date: ___________
Chemistry: Electrolytes and Colligative Properties
1. Write the dissociation for the following:
a. strontium nitrate
b. calcium chloride
c. magnesium sulfate
d. potassium iodide
Also, calculate the concentrations of the products assuming each reactant is 0.1 molal.
2. What is the expected freezing point for a solution that contains 2.0 mol of magnesium sulfate dissolved in
500 mL of water?
3. What is the expected change in the freezing point of water in a solution of 62.5 g of barium nitrate, Ba(NO 3) 2, in
1.00 kg of water?
4. What is the expected boiling point of water for a solution that contains 150 g of sodium chloride dissolved in
1.0 kg of water?
ANSWERS:
1a/b. 0.3 molal
1c/d. 0.2 molal
2. – 14.9 oC
3. – 1.3 oC
4. 102.6 oC
Name: _________________________
Hour: ____ Date: ___________
Chemistry: Review Colligative Properties
1. A water solution containing an unknown quantity of a nonelectrolyte solute is found to have a freezing point of
– 0.23oC. What is the molal concentration of the solution?
2. What is the boiling point of a solution of ethyl alcohol, C2H5OH, that contains 20.0 g of the solute dissolved in
250 g of water?
3. A solution contains 4.50 g of a nonelectrolyte dissolved in 225 g of water and has a freezing point of –0.310o C.
What is the gram formula mass of the solute?
4. How many grams of ethylene glycol, C2H4(OH) 2, must a researcher add to 500 g of water to yield a solution
that will freeze at –7.44oC? [Hint: you need to find the molal freezing constant, Kf, for ethylene glycol]
ANSWERS:
1. 0.124 molal
2. 100.89 oC
3. 120 g/mol
4. 24 g
5. Which of the following two solutes will raise the boiling point of water in a car’s radiator more: 1.00 mol of
ethylene glycol or 1.00 mol of ethyl alcohol?
Explain.
6. The freezing point of an aqueous sodium chloride solution is –20.0oC. What is the molarity of the solution?
7. What is the expected boiling point of a solution prepared by dissolving 117 g of NaCl in 463 mL of acetic acid?
(Assume the density of acetic acid is 1.08 g/mL)
ANSWERS:
5.
??
6. 5.4 molal
7. 142.5oC
Colligative Properties
Useful Equations:
Tf = Kf  m
Solvent
acetic acid
acetone
benzene
camphor
carbon tetrachloride
chloroform
ether
ethyl alcohol
ethylene glycol
methyl alcohol
formic acid
napthalene
phenol
toluene
water
Tb = Kb  m
Normal
Freezing Point
(oC)
16.6
- 95.4
5.5
178.8
- 23.0
- 63.5
- 116.3
- 117.3
- 13.5
- 97.8
8.3
80.2
40.9
- 94.5
0.0
molality =
Molal Boiling
Point Constant
Kb (oC/molal)
3.07
1.71
2.53
5.61
5.03
3.63
2.02
1.22
0.83
3.56
3.56
3.33
0.512
mol of solute
kg of solvent
Normal Boiling
Point (oC)
117.9
56.2
80.1
207.4
76.7
61.2
34.6
78..3
197.2
64.5
100.8
217.7
181.8
110.7
100.0
MM =
grams
moles
Molal Freezing
Point Constant
Kf (oC/molal)
- 3.90
- 4.90
- 39.7
- 1.79
- 2.77
- 6.8
- 7.40
- 1.86
................................................................................................................................................................. C U T H E R E ..................................................................................................................................................................
Colligative Properties
Useful Equations:
Tf = Kf  m
Solvent
acetic acid
acetone
benzene
camphor
carbon tetrachloride
chloroform
ether
ethyl alcohol
ethylene glycol
methyl alcohol
formic acid
napthalene
phenol
toluene
water
Tb = Kb  m
Normal
Freezing Point
(oC)
16.6
- 95.4
5.5
178.8
- 23.0
- 63.5
- 116.3
- 117.3
- 13.5
- 97.8
8.3
80.2
40.9
- 94.5
0.0
molality =
Molal Boiling
Point Constant
Kb (oC/molal)
3.07
1.71
2.53
5.61
5.03
3.63
2.02
1.22
0.83
3.56
3.56
3.33
0.512
mol of solute
kg of solvent
Normal Boiling
Point (oC)
117.9
56.2
80.1
207.4
76.7
61.2
34.6
78..3
197.2
64.5
100.8
217.7
181.8
110.7
100.0
MM =
grams
moles
Molal Freezing
Point Constant
Kf (oC/molal)
- 3.90
- 4.90
- 39.7
- 1.79
- 2.77
- 6.8
- 7.40
- 1.86
KEY
Chemistry: Colligative Properties – Freezing Point Depression
1. How much will the freezing point be lowered if enough sugar is dissolved in water to make a 0.50 molal
solution?
ΔTf = K f  m


ΔTf = -1.86 o C/molal  0.50 molal 
ΔTf = -0.93 C
o
2. What is the freezing point of a solution of a nonelectrolyte dissolved in water if the concentration of the solution
is 0.24 m?
ΔTf = K f  m

Normal freezing point of water is 0.0 o C.

ΔTf = -1.86 o C/molal  0.24 molal 
 freezing point = - 0.446 o C
ΔTf = -0.446 C
o
3. What is the freezing point of a solution that contains 68.4 g of sucrose, C 12H22O11, dissolved in 1.00x102 g of
water?
 1 mol C12H22 O11 
x mol C12H22 O11 = 68.4 g C12H22 O11 
  0.2 molal
 342 g C12H22 O11 
m=
mol solute
0.2 mol

kg solvent
0.1 kg
ΔTf = K f  m


ΔTf = -1.86 o C/molal  2 molal 
m = 2 molal
 freezing point = - 3.72 o C
ΔTf = - 3.72 C
o
4. Suppose that 98.0 g of a nonelectrolyte is dissolved in 1.00 kg of water. The freezing point of this solution is
found to be –0.465oC. What is the molecular mass of the solute?
m=
ΔTf = K f  m


- 0.465 o C = -1.86 o C/molal  m 
mol solute
kg solvent
0.25 molal =
x mol
1 kg
Molar Mass =
g
mol
Molar Mass =
98 g
0.25 mol
m = 0.25 molal
x = 0.25 mol solute
Molar Mass = 392 g/mol  392 amu
5. A researcher places 53.2 g of an unknown nonelectrolyte in 505 g napthalene. The nonelectrolyte lowers
napthalene’s freezing point by 8.8oC. What is the molar mass of the unknown substance?
m=
ΔTf = K f  m


- 8.8 o C = -6.8 o C/molal  m 
mol solute
kg solvent
1.294 molal =
x mol
0.505 kg
Molar Mass =
g
mol
Molar Mass =
53.2 g
0.654 mol
m = 1.294 molal
x = 0.654 mol solute
Molar Mass = 81.4 g/mol
KEY
Chemistry: Colligative Properties – Boiling Point Elevation
5.03 oC/molal
1a. What is the molal boiling point constant for carbon tetrachloride?
b. The boiling point of carbon tetrachloride is 76.8oC. A given solution contains 8.10 g of a nonvolatile electrolyte in
300 g of CCl4. It boils at 78.4oC. What is the gram molecular mass of the solute?
m=
ΔTb = K b  m
78.4 o C


1.6 o C = 5.03 o C/molal  m 
o
- 76.8 C
T = 1.6 o C
mol solute
kg solvent
0.3181 molal =
x mol
0.3 kg
Molar Mass =
g
mol
Molar Mass =
8.10 g
0.0954 mol
m = 0.3181 molal
x = 0.0954 mol solute
Molar Mass = 84.9 g/mol
2. The molal boiling point constant for ethyl alcohol is 1.22 oC/molal. Its boiling point is 78.4oC. A solution of
14.2 g of a nonvolatile nonelectrolyte in 264 g of the alcohol boils at 79.8 oC. What is the gram molecular
mass of the solute?
mol solute
g
m=
Molar Mass =
kg solvent
mol
ΔTb = K b  m
79.8 o C


1.4 o C = 1.22 o C/molal  m 
o
- 78.4 C
T = 1.4 o C
1.51 molal =
x mol
0.264 kg
Molar Mass =
14.2 g
0.304 mol
m = 1.51 molal
x = 0.304 mol solute
Molar Mass = 46.7 g/mol
3. Calculate the mass of a mole of a compound that raises the boiling point of water to 100.78 oC at 101.3 kPa
when 51 g of the compound is dissolved in 500 g of water.
m=
100.78 o C
ΔTb = K b  m


- 100.00 C
0.78 o C = 0.52 o C/molal  m 
T = 0.78 o C
m = 1.5 molal
o
mol solute
kg solvent
1.5 molal =
x mol
0.5 kg
x = 0.75 mol solute
Molar Mass =
g
mol
Molar Mass =
51 g
0.75 mol
Molar Mass = 67 g/mol
4. Suppose that 13 g of a nonelectrolyte is dissolved in 0.50 kg of benzene. The boiling point of this solution is
80.61oC. What is the molecular mass of the solute?
m=
80.61o C
o
- 80.10 C
T = 0.51o C
ΔTb = K b  m


0.51o C = 2.53 o C/molal  m 
mol solute
kg solvent
0.2016 molal =
x mol
0.5 kg
Molar Mass =
g
mol
Molar Mass =
13 g
0.1008 mol
m = 0.20 molal
x = 0.1008 mol solute
Molar Mass = 129 g/mol
or 129 amu
KEY
Chemistry: Electrolytes and Colligative Properties
1. Write the dissociation for the following:
a. strontium nitrate
b. calcium chloride
c. magnesium sulfate
Sr(NO3) 2 (aq)
Sr2+(aq)
0.1 molal
0.1 m
CaCl 2 (aq)
Ca2+(aq)
0.1 m
MgSO4 (aq)
Mg2+(aq)
0.1 molal
0.1 m
KI (aq)
K1+(aq)
0.1 molal
0.1 m
d. potassium iodide
2 (0.1 molal)
0.2 m
2 Cl1- (aq)
+
0.1 molal
2 NO31- (aq)
+
0.2 m
+
SO42- (aq)
0.1 m
I1- (aq)
+
0.1 m
Also, calculate the concentrations of the products assuming each reactant is 0.1 molal.
2. What is the expected freezing point for a solution that contains 2.0 mol of magnesium sulfate dissolved in
500 mL of water?
m=
mol solute
2.0 mol

kg solvent
0.5 kg
m = 4 molal 
 2 "ions"  
MgSO4  Mg
2+
4m
4m
ΔTf = K f  m


ΔTf = -1.86 o C/molal  8 molal "ions" 
8 molal
+ SO 4
ΔTf = - 14.9 C
o
2
4m
3. What is the expected change in the freezing point of water in a solution of 62.5 g of barium nitrate, Ba(NO 3) 2, in
1.00 kg of water?
 1 mol Ba NO3 2
x mol Ba NO3 2 = 62.5 g Ba NO3 2 
 261.34 g Ba NO3 

2
m=
mol solute
0.239 mol

kg solvent
1 kg
m = 239 molal 
3 "ions"  
ΔTf = K f  m


  0.239 mol Ba NO3 2



ΔTf = -1.86 o C/molal  0.717 molal "ions" 
0.717 molal
ΔTf = - 1.33 o C
4. What is the expected boiling point of water for a solution that contains 150 g of sodium chloride dissolved in
1.0 kg of water?
 1 mol NaCl 
x mol NaCl = 150 g NaCl 
  2.57 mol NaCl
 58.4 g NaCl 
m=
mol solute
2.57 mol

kg solvent
1 kg
m = 2.57 molal 
 2 "ions"  
ΔTb = K b  m


ΔTb = 0.512 o C/molal  5.13 molal "ions" 
5.13 molal
ΔTb = 2.6 C
o
 expected b.p. = 102.6 o C
KEY
Chemistry: Review Colligative Properties
1. A water solution containing an unknown quantity of a nonelectrolyte solute is found to have a freezing point of
– 0.23oC. What is the molal concentration of the solution?
ΔTf = K f  m


- 0.23 o C = -1.86 o C/molal  m 
m = 0.124 molal
2. What is the boiling point of a solution of ethyl alcohol, C2H5OH, that contains 20.0 g of the solute dissolved in
250 g of water?
m=
 1 mol 
x mol C2H5 OH = 20.0 g C2H5 OH 

 46 g 
x = 0.435 mol C2H5 OH
mol solute
kg solvent
ΔTb = K b  m


ΔTf = 0.512 o C/molal 1.74 molal 
0.435 mol
4 molal =
0.250 kg
x = 1.74 molal C2 H5 OH
ΔTf = 0.89 C
o
 boiling point is 100.89 o C
3. A solution contains 4.50 g of a nonelectrolyte dissolved in 225 g of water and has a freezing point of –0.310o C.
What is the gram formula mass of the solute?
m=
ΔTf = K f  m


- 0.310 o C = -1.86 o C/molal  m 
mol solute
kg solvent
0.167 molal =
x mol
0.225 kg
Molar Mass =
g
mol
Molar Mass =
4.50 g
0.0375 mol
m = 0.167 molal
x = 0.0375 mol solute
Molar Mass = 120 g/mol
4. How many grams of ethylene glycol, C2H4(OH) 2, must a researcher add to 500 g of water to yield a solution
that will freeze at –7.44oC?
m=
ΔTf = K f  m


- 7.44 o C = -1.86 o C/molal  m 
mol solute
kg solvent
4 molal =
x mol
0.5 kg
m = 4 molal
 62 g 
x g C2 H4 (OH)2 = 2 mol C2H4 (OH)2 

 1 mol 
x = 124 g C2H4 (OH)2
x = 2 mol solute
ANSWERS:
1. 0.124 molal
2. 100.89 oC
3. 120 g/mol
4. 124 g
5. Which of the following two solutes will raise the boiling point of water in a car’s radiator more: 1.00 mol of
ethylene glycol or 1.00 mol of ethyl alcohol?
Explain.
Both substances would initially have the same protective effect. A colligative property is only
dependant on the number of particles present – both the ethylene glycol and ethyl alcohol contain 1
mole or 6.02 x 1023 particles. However, ethyl alcohol is volatile and will rapidly evaporate (and is very
flammable) and should not be used in a car’s radiator. Therefore, ethylene glycol should be used as the
solute added to water to fill a car’s radiator!
6. The freezing point of an aqueous sodium chloride solution is –20.0oC. What is the molarity of the solution?
Sodium chloride, NaCl, dissociates into two ions: Na+(aq) + Cl1-(aq)
NaCl(aq) --> Na+(aq) + Cl1-(aq)
5.38 m
5.38 m
5.38 m
ΔTf = K f  m  i


- 20.0 o C = -1.86 o C/molal  m  (2)
m = 5.38 molal
7. What is the expected boiling point of a solution prepared by dissolving 117 g of NaCl in 463 mL of acetic acid?
(Assume the density of acetic acid is 1.08 g/mL)
Step 1)
 1 mol 
x mol NaCl = 117 g NaCl 

 58.5 g 
x = 2 mol NaCl
Step 2)
 1.08 g   1 kg 
x kg = 463 mL acetic acid 
 = 0.5 kg

 1 mL   1000 g 
Step 3)
m=
mol solute
kg solvent
ΔTb = K b  m  i
Step 4)

2 mol
0.5 kg

 4 molal

ΔTb = 3.07 o C/molal  4  (2)
ΔTb = 14.46 C
o
117.9o C
Step 5) normal boiling point = + 24.6o C
142.5o C
ANSWERS:
"expected" boiling point
5.
??
6. 5.4 molal
7. 142.5oC
Colligative Properties
Useful Equations:
Tf = Kf  m
Solvent
acetic acid
acetone
benzene
camphor
carbon tetrachloride
chloroform
ether
ethyl alcohol
ethylene glycol
methyl alcohol
formic acid
napthalene
phenol
toluene
water
Tb = Kb  m
Normal
Freezing Point
(oC)
16.6
- 95.4
5.5
178.8
- 23.0
- 63.5
- 116.3
- 117.3
- 13.5
- 97.8
8.3
80.2
40.9
- 94.5
0.0
molality =
Molal Freezing
Point Constant
Kf (oC/molal)
- 3.90
- 4.90
- 39.7
- 1.79
- 2.77
- 6.8
- 7.40
- 1.86
mol of solute
kg of solvent
Normal Boiling
Point (oC)
117.9
56.2
80.1
207.4
76.7
61.2
34.6
78..3
197.2
64.5
100.8
217.7
181.8
110.7
100.0
MM =
grams
moles
Molal Boiling
Point Constant
Kb (oC/molal)
3.07
1.71
2.53
5.61
5.03
3.63
2.02
1.22
0.83
3.56
3.56
3.33
0.512
................................................................................................................................................................. C U T H E R E ..................................................................................................................................................................
Colligative Properties
Useful Equations:
Tf = Kf  m
Solvent
acetic acid
acetone
benzene
camphor
carbon tetrachloride
chloroform
ether
ethyl alcohol
ethylene glycol
methyl alcohol
formic acid
napthalene
phenol
toluene
water
Tb = Kb  m
Normal
Freezing Point
(oC)
16.6
- 95.4
5.5
178.8
- 23.0
- 63.5
- 116.3
- 117.3
- 13.5
- 97.8
8.3
80.2
40.9
- 94.5
0.0
molality =
Molal Freezing
Point Constant
Kf (oC/molal)
- 3.90
- 4.90
- 39.7
- 1.79
- 2.77
- 6.8
- 7.40
- 1.86
mol of solute
kg of solvent
Normal Boiling
Point (oC)
117.9
56.2
80.1
207.4
76.7
61.2
34.6
78..3
197.2
64.5
100.8
217.7
181.8
110.7
100.0
MM =
grams
moles
Molal Boiling
Point Constant
Kb (oC/molal)
3.07
1.71
2.53
5.61
5.03
3.63
2.02
1.22
0.83
3.56
3.56
3.33
0.512
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