Download Study Guide for Module 11B—Solutions II

Chemistry 1020, Module 11B
Name
Study Guide for Module 11B—Solutions II
■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■
Reading Assignment:
Sections 11.2 through 11.7 in Chemistry, 6th Edition by Zumdahl.
Guide for Your Lecturer:
1.
2.
3.
4.
✔
✔
✔
✔
✔
5.
6.
7.
8.
9.
10.
Review of Preparing Solutions
Solubility Rules
The Solution Process and Analytical Concentrations
Solutions Compared to Colloids and Suspensions; Review of Freezing Point, Boiling Point, Vapor
Pressure; Osmosis and Osmotic Pressure
Qualitative Consideration of the Colligative Properties of Solutions
Calculation of Colligative Properties: Vapor Pressure Lowering
Calculation of Colligative Properties: Osmotic Pressure
Calculation of Colligative Properties: Freezing Point Depression
Calculation of Colligative Properties: Boiling Point Elevation
Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point Elevation
■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■
Homework
Note:
✔ indicates problems to be stressed on drill quizzes and hour exams.
Xavier University of Louisiana
43
Chemistry 1020, Module 11B
■Review of Preparing Solutions (From Module 4)
Note: The procedure used to prepare the solution is the same as that taught in Chemistry 1010 and the
accompanying laboratory. It is:
Step 1: Measure out ___ g of the solute (or ___ ml of the original solution).
Step 2: Put approximately one-half of the water in the "total volume" into a container.
(If solution is being made from solid, "total volume" = volume of solution to be made.
If solution is being made from another solution, "total volume" = volume of solution to
be made - volume of solution to be diluted.)
Step 3: Add the substance from Step 1 to the water in Step 2 and mix until all is dissolved.
Step 4: Add water until the total volume to be made is obtained.
Step 5: Mix the solution. (Note: If the solution is made in a volumetric flask, cap and invert to mix.)
1a) S. How would you prepare 105 milliliters of 0.32 M copper sulfate penta hydrate solution from pure solute?
•Formula: CuSO4 •5H 2 O
•Type cmpd: salt
•Strength: strong
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
gfw of CuSO 4 •5H 2 O = 63.5 + 32.1 + 4*16.0 + 1) Weigh out approximately 8.4 g
5(2*1.01 + 16.0) = 249.7 g/mol
CuSO4 •5H 2 O and then record the actual
mass reading on the balance to the nearest
M = mol solute/L soln
0.01 g.
M = (g/mw)/L
M*L = g/mw
2) Add about 105/2 = 52 mL of H2 O to the
g = M*L*mw
container.
L soln = 105 mL * (1 L/1000 mL) = 0.105 L
g = (0.32 mol/L)(0.105 L)(249.7 g/mol) =
R
= 8.39 g = 8.4 g CuSO 4 •5H 2 O *
3) Add the 8.4 g CuSO4 •5H 2 O to the
water in the container and mix until the solute
dissolves.
4) Add enough additional H2 O to get a total
*Although this number has been rounded, in the lab you
would weigh as accurately as the balance available allows. of 105 mL of solution.
In General Chemistry, this would be to the nearest ±0.01 g
so you would weight out a number close to 8.4 and record
5) Mix the solution.
its weight to two places past the decimal.
A. How would you prepare 112 milliliters of 0.23 M sodium sulfate solution from sodium sulfate hexahydrate?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
44
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■Review of Preparing Solutions (continued)
1a) B. How would you prepare 123 milliliters of 0.45 M cesium chlorate solution from cesium chlorate dihydrate?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
C. How would you prepare 34 milliliters of 0.78 M glucose (gfw=180.0) from solid solute?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
D. How would you prepare 81 milliliters of 0.13 M urea (gfw=60.0) solution from solid solute?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
Xavier University of Louisiana
45
Chemistry 1020, Module 11B
■Review of Preparing Solutions (continued)
1a) E. How would you prepare 93 milliliters of 0.24 M sodium carbonate solution from solid solute?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
b) S. How would you prepare 112 milliliters of 0.87 M hydrobromic acid from 3.4 M hydrobromic acid?
•Formula: HBr
•Type cmpd: acid
•Strength: strong
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
M 1 = 3.4 M
1) Measure out 28.7 mL of 3.4 M HBr
V1 = ?
M 2 = 0.87 M
V 2 = 112 mL
2) Put approximately (112 - 28.7)/2 = 42
M 1 V 1 = M2 V 2
mL of H2 O in a container.
V 1 = M2 V 2 /M 1
V 1 = (0.87 M)(112 mL)/(3.4 M) =
3) Add the 28.7 mL of 3.4 M HBr to the
R
container and mix.
V 1 = 28.6588 mL = 29 mL of 3.4 M HBr*
*Although this number has been rounded, in the lab you
4) Add additional water to the container until
would measure volume as accurately as the equipment
a total of 112 mL of solution is obtained.
available allows. In General Chemistry, this would be to
the nearest ±0.1 mL if you use a graduated cylinder or to
the nearest ±0.01 mL if you use a pipette. If you don’t
know what kind of equipment to use, assume a graduated
cylinder.
5) Mix the solution.
A. How would you prepare 34 milliliters of 0.98 M sulfuric acid from a 5.2 M solution of sulfuric acid?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
46
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■Review of Preparing Solutions (continued)
1b) B. How would you prepare 123 milliliters of 1.2 M hydrochloric acid from 2.0 M hydrochloric acid?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
C. How would you prepare 565 milliliters of 0.17 M nitric acid from 1.3 M nitric acid?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
D. How would you prepare 81 milliliters of 1.5 M acetic acid from a 2.8 M solution of acetic acid?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
Xavier University of Louisiana
47
Chemistry 1020, Module 11B
■Review of Preparing Solutions (continued)
1b) E. How would you prepare 78 milliliters of 0.57 M phosphoric acid from a 4.5 M solution of phosphoric
acid?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Calculation
|Steps to be used in preparing solution
1)
2)
3)
4)
5)
48
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■S o l u b i l i t y R u l e s
✔✔✔■Names and Formulas of Common Anions and Common Chemical Compounds
2a) List the following by name and formula. (NOTE: This comes directly from Chemistry 1010, Module 0.)
A.
Four polyatomic anions which contain S.
___________________________________________
Name
|Formula
||Name
|Formula
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
B.
Two polyatomic anions which contain only N and O.
|
||
|
|
||
|
|
||
|
C.
Four polyatomic anions which contain Cl.
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
D.
Four polyatomic anions which contain P.
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
E.
Four polyatomic anions which contain C.
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
F.
Two polyatomic anions which contain Cr.
|
||
|
|
||
|
G.
List four other important polyatomic anions.
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
H.
Four Group VIIA monoatomic anions.
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
|
||
|
I.
Two Group VIA monoatomic anions.
|
||
|
|
||
|
|
||
|
Xavier University of Louisiana
49
Chemistry 1020, Module 11B
Na+, K +, NH4+, and the other Group IA metal ions
NO3-, CH3COO-, and ClO4Cl -, except for AgCl, Hg2Cl 2*, and PbCl2
Br-, except for AgBr, Hg2Br2*, PbBr2, and HgBr2
I-, except for AgI, Hg2I2*, PbI 2, and HgI2
SO 42- except for CaSO 4, SrSO 4, BaSO 4, PbSO4, Hg 2SO 4*, &Ag 2SO 4
Anions whose salts are generally insoluble: S 2-, except for those of IA and IIA metals and (NH 4)2S.
CO32-, except for those of IA metals and (NH4)2CO3
SO 32-, except for those of IA metals and (NH4)2SO 3
PO43-, except for those of IA metals and (NH4)3PO4
OH-, except for those of IA metals, Ba(OH)2, Sr(OH)2, and Ca(OH)2
*Hg 2 2 + is mercury(I) ion (or mercurous ion). It is the only diatomic metallic cation you
need to know.
A. List the names and formulas of seven cations almost all of whose salts are water soluble. Then list exceptions.
Formula
Name
Exceptions
2b. Cations whose salts are generally soluble:
Anions whose salts are all soluble:
Anions whose salts are generally soluble:
B. List the names and formulas of three anions almost all of whose salts are water soluble. Then list exceptions.
C. List the names and formulas of four anions most of whose salts are water soluble. Then list exceptions.
D. List the names and formulas of five anions most of whose salts are not soluble in water. Then list exceptions.
50
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■The Solution Process and Analytical Concentrations
3a) Draw a sketch showing solid sodium chloride dissolving in water to form solvated sodium ion and chloride ions.
Write the balanced chemical equation for the solution of NaCl in water.
----------------------------------------------------------------------------------------------------------------------------------------------b) List three steps which together might be used to explain how the solution process occurs. (See pp. 516)
Step 1:
----------------------------------------------------------------------------------------------------------------------------------------------Step 2:
----------------------------------------------------------------------------------------------------------------------------------------------Step 3:
c)
----------------------------------------------------------------------------------------------------------------------------------------------Write a balanced chemical equation showing the given substance dissolving in water.
S. sodium acetate
•Formula: NaCH3 COO
•Type cmpd: salt
•Strength: strong
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
NaCH 3 COO(s) ➙ Na +(aq) + CH 3 COO -(aq)
----------------------------------------------------------------------------------------------------------------------------------------------A. ammonium sulfide
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------B. copper(II) nitrate
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------C. zinc(II) sulfate
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------D. calcium chloride
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------E. rubidium sulfite
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
-----------------------------------------------------------------------------------------------------------------------------------------------
Xavier University of Louisiana
51
Chemistry 1020, Module 11B
■The Solution Process and Analytical Concentrations (continued)
3d) Use an INITIAL/CHANGE/FINAL chart to determine the “real” concentrations of the species in the solution.
(pp. 142-3)
S. 0.30 M Na 2SO4
•Name: sodium sulfate
•Type cmpd: salt
•Strength: strong
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Na 2 SO 4 (s) ➙ 2 Na+(aq) + SO4 2- (aq)
initial
0.30
0
0
_
∆
-0.30
+0.60
+ 0.30
final
0
0.60
_ 0.30
Therefore, “real” concentrations of species in solution are: [Na+] = 0.60 M, [SO4 2- ] = 0.30 M
----------------------------------------------------------------------------------------------------------------------------------------------A. 0.62 M CaCl2
•Name
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------B. 0.25 M HNO 3
•Name:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------C. 0.78 M KOH
•Name:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------D. 0.35 M Na3PO4
•Name:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------E. 0.13 M K2SO4
•Name:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
-----------------------------------------------------------------------------------------------------------------------------------------------
52
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions
4a) Distinguish among solutions, colloids, and suspensions. (All three are mixtures. They differ in terms of
the size of "solute" particles. In a solution, the solute particles are small and the mixture is
homogeneous all the way down to the molecular level. In a suspension the solute particles are still
small enough to stay suspended for a while but are large enough to settle out {precipitate} if the
mixture is left to stand for a while. In a colloid, the solute particles are intermediate in size of the
solute particles in a solution and the size in a suspension. As a result, they are small enough to stay
suspended indefinitely {i.e. don't settle out as in the suspension} but are too large to consider the
solution as truly homogeneous.)
----------------------------------------------------------------------------------------------------------------------------------------------b) Outline a procedure you could use to determine if a given mixture were a solution, colloid, or a suspension.
(Let the mixture sit for a while. If a precipitate settles out, it is a suspension. If not, shine light
through it. If the light beam passes through without being reflected, it is a solution. If the light beam
is reflected to the side {i.e. can be seen passing through the mixture in the way a light beam can be
seen passing through a cloud of dust), the mixture is a colloid. The solute particles in a solution are
not large enough to reflect the light, those in the colloid are.)
c)
----------------------------------------------------------------------------------------------------------------------------------------------Describe what happens (overall) to solute and solvent as two solutions of differing concentrations diffuse
together.
d)
----------------------------------------------------------------------------------------------------------------------------------------------Sketch the overall movement of urea and water if a 0.5 M solution of urea is placed in contact with a 0.2 M
solution without mixing.
e)
f)
----------------------------------------------------------------------------------------------------------------------------------------------Define osmosis. (p. 535, A37)
----------------------------------------------------------------------------------------------------------------------------------------------Briefly compare what happens to solute and solvent during diffusion (question 4c ) with what happens during
osmosis (question 4e).
-----------------------------------------------------------------------------------------------------------------------------------------------
Xavier University of Louisiana
53
Chemistry 1020, Module 11B
■Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions
(continued)
4g) S. Answer the following questions concerning the diagram below which represents a cell in which two
solutions are separated by a semipermeable membrane. (pp. 535-6)
1-In which direction does the water flow as the system goes
to equilibrium? From the more dilute 0.03 m solution to
the more concentrated 0.5 m solution.
2-In which direction does the solute flow as the system goes
to equilibrium? Solute does not flow. The membrane is
semipermeable which means that only H2 O can flow.
3-What happens to the concentration of the 0.03 m side of
the system as it goes to equilibrium?
It increases as water flows out.
4-What happens to the concentration of the 0.5 m side of the
0.5 m
0.03 m
system as it goes to equilibrium?
urea
urea
It decreases as water flows in.
5-What happens to the volume of solution in the 0.03 m
side of the system as it goes to equilibrium?
It decreases as water flows out.
6-What happens to the volume of solution in the 0.5 m side
of the system as it goes to equilibrium?
It increases as water flows in.
----------------------------------------------------------------------------------------------------------------------------------------------A. Answer the following questions concerning the diagram below which represents a balloon made from a
semipermeable membrane filled with a 0.5 m glucose solution and immersed in a 0.2 m glucose solution in
a beaker.
1-In which direction does the water flow as the system goes
to equilibrium?
2-In which direction does the solute flow as the system goes
to equilibrium?
3-What happens to the concentration of the solution inside
the balloon as the system goes to equilibrium?
4-What happens to the concentration of the solution in the
beaker as the system goes to equilibrium?
5-What happens to the volume of solution in the balloon as
the system goes to equilibrium?
6-What happens to the volume of solution outside the
balloon as the system goes to equilibrium?
-----------------------------------------------------------------------------------------------------------------------------------------------
54
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions
(continued)
4g) B. Answer the following questions concerning the diagram below which represents a cell in which two
solutions are separated by a semipermeable membrane.
1-In which direction does the water flow as the system goes
to equilibrium?
2-In which direction does the solute flow as the system goes
to equilibrium?
3-What happens to the concentration outside the funnel as
the system goes to equilibrium?
4-What happens to the concentration inside the funnel as the
system goes to equilibrium?
The solution inside the upside funnel
above is 0.03 m in urea. That outside
is 0.1 m in urea. There is a
semipermable membrane across the
top of the funnel (on the bottom of
the diagram) separating the two.
5-What happens to the volume of solution outside the funnel
as the system goes to equilibrium?
6-What happens to the volume of solution inside the funnel
as the system goes to equilibrium?
----------------------------------------------------------------------------------------------------------------------------------------------C. Answer the following questions concerning the diagram below which represents a “U-tube” with a 0.4 m
glucose solution on the left (below) and a 0.1 m glucose solution on the right. There is a semipermeable
membrane separating the two halves of the tube..
1-In which direction does the water flow as the system goes
to equilibrium?
2-In which direction does the solute flow as the system goes
to equilibrium?
3-What happens to the concentration of the solution on the
left as the system goes to equilibrium?
0.4 m
glucose
0.1 m
glucose
4-What happens to the concentration of the solution on the
right as the system goes to equilibrium?
5-What happens to the volume of solution on the left as the
system goes to equilibrium?
6-What happens to the volume of solution on the right as
the system goes to equilibrium?
-----------------------------------------------------------------------------------------------------------------------------------------------
Xavier University of Louisiana
55
Chemistry 1020, Module 11B
■Qualitative Consideration of the Colligative Properties of Solutions
5a) Define colligative property (p. 531,A31)
----------------------------------------------------------------------------------------------------------------------------------------------b) Define vapor pressure. (p. 484, A41)
c)
----------------------------------------------------------------------------------------------------------------------------------------------If liquid water is placed in a flask and the flask is closed, some of the water evaporates and gives rise to vapor
pressure. Draw a stoppered flask showing molecules of water in both the liquid and vapor states. Then explain
what causes the vapor pressure. (p. 484)
d)
----------------------------------------------------------------------------------------------------------------------------------------------Define melting point. (p. 491)
e)
----------------------------------------------------------------------------------------------------------------------------------------------Define boiling point. (p. 491)
f)
----------------------------------------------------------------------------------------------------------------------------------------------Define osmotic pressure. (p. 535)
----------------------------------------------------------------------------------------------------------------------------------------------g) What happens to each of the following when the concentration of a solution increases?
-vapor pressure of water above the solution (pp. 524-5)
-----------------------------------------------------------------------------------------------------------------------------------------------osmotic pressure of the solution (p. 536-7)
-----------------------------------------------------------------------------------------------------------------------------------------------freezing point of the solution (p. 533)
-----------------------------------------------------------------------------------------------------------------------------------------------boiling point of the solution (p. 531)
----------------------------------------------------------------------------------------------------------------------------------------------h) List four colligative properties of solutions AND indicate where you will study them in the remainder of this module.
Colligative Property
Learning Goal Where Covered
1)
2)
3)
4)
i)
Define "particle concentration". (particle concentration = concentration of solute * number of particles
each solute molecule breaks into when in solution)
j)
What are the two factors which determine the particle concentration of a solution? (1) concentration of solute,
2) the extent of dissociation of the solute)
56
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■Qualitative Consideration of the Colligative Properties of Solutions (continued)
5k) For each of the given pairs of solutions, indicate the strength of the solutes and the particle concentrations.
Then answer the questions in the 4th and 5th columns. INCLUDE EXPLANATIONS.
Solutions
Strength
Particle
Which solution has the
Which solution conducts
Concentrations indicated property?
electrity better?
nonelectrolyte
0.16 m
Neither conduct since no
S1. 0.16 m urea
Higher fp? The two
ions are in either.
have the same parti0.16 m
cle concentrations so
0.16 m glucose nonelectrolyte
they have the same fp.
strong
0.46
m
0.23 m NaCl since it
S2. 0.23 m NaCl
Lower vapor pressure?
electrolyte
0.23 m NaCl since
has higher concentration
strong
0.26 m
its particle conc. is
of ions.
0.13 m NaCl
electrolyte
higher.
A. 0.62 m urea
Higher osmotic press.?
0.16 m glucose
B.
0.86 m glucose
Lower boiling point?
0.29 m glucose
C.
0.15 m
Na2SO4
0.37 m glucose
Solutions
D.
Higher vapor pressure?
Strength
0.16 m urea
Particle
Concentrations
Which solution has the
indicated property 1?
Higher bp?
Which solution conducts
electrity better?
0.16 m NaNO 3
E.
0.13 m NaCl
Higher bp?
0.44 m glucose
F.
0.75 m urea
Lower vapor press.?
0.23 m glucose
G.
0.34 m NaBr
Higher freezing point?
0.21 m NaBr
H.
0.17 m urea
Lower vapor pressure?
0.38 m urea
I.
0.17 m NaNO 3
Lower freezing point?
0.17 m KBr
Xavier University of Louisiana
57
Chemistry 1020, Module 11B
✔✔✔■Calculation of Colligative Properties: Vapor Pressure Lowering
6a) In general how does vapor pressure of a solution change as its concentration changes? (p. 524-5)
----------------------------------------------------------------------------------------------------------------------------------------------b) In words, what is the quantitative relationship between vapor pressure of a solution a nonelectrolyte and the
concentration of that solution? (The vapor pressure of the solution is directly proportional to the mole fraction
of water in the solution. )
c)
d)
e)
f)
----------------------------------------------------------------------------------------------------------------------------------------------Write the equation for the statement in b) . (Psoln = Xwater in soln*Ppure water)
----------------------------------------------------------------------------------------------------------------------------------------------In real life, chemists are generally more interested in the how much the vapor pressure of a solution differs from
the vapor pressure of pure water than in the vapor pressure of the solution itself. In addition, they usually know
the mole fraction of solute in the solution rather than mole fraction of water. Use Xsolute + Xpure water = 1 to
obtain an equation which relates change in vapor pressure as the solution forms (i.e. the difference between the
vapor pressure of the solution from that of pure water) to the mole fraction of solute. Note: This is the
equation usually used when considering colligative properties of water.
( VP = Ppure water - P soln = Xsolute*Ppure water )
----------------------------------------------------------------------------------------------------------------------------------------------When can the equation in d above be used? (i.e. What are its limitations?)
----------------------------------------------------------------------------------------------------------------------------------------------S. A certain solution is prepared at 20oC by adding 42.0 grams of glucose (gfw = 180.0) to 251 ml of water.
How does the vapor pressure of a solution differ from that of pure water at the same temperature? (pp. 526-8)
•Formula: C 6 H12 O6
•Type cmpd: not a, b, or s
•Strength: nonelectrolyte
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
P soln = P water *X water
13.93 mol H2 O
P water = 17.54 torr (from data sheet)
X water = 13.93 + 0.2333 total moles =
X water = (mol water)/(total moles)
= 0.9838
mol water = 251 mL H2 O *
1 g H2O
1 mol H2 O
* 1 mL H O * 18.02 g H O =
P soln = P water *X water
2
2
= (17.54 torr)(0.9838) = 17.26 torr *
= 13.93 mol H2 O
= 17.3 torr
mol glucose = 42.0 g glucose *
∆P = 17.54 - 17.3 torr = 0.24 torr difference
1 mol glucose
* 180.0 g glucose = 0.2333 mol glucose
which rounds to 0.2 torr
Therefore, it is 0.2 torr lower.
*Reminder: Carry one more place in intermediate steps which involve multiplication or division. Then round
round to the number you want when you finish multiplication/division before you do addition or subtraction.
----------------------------------------------------------------------------------------------------------------------------------------------A. How much does the vapor pressure of pure water at 60oC change if the water is used to prepare a solution
by adding 21.3 grams of urea (CH4N2O) to 302 ml of H20?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------58
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
✔✔✔■Calculation of Colligative Properties: Vapor Pressure Lowering (continued)
6f) B. What is the vapor pressure of a solution prepared by adding 45.3 grams of glucose (C6H12O6) to 153 ml of
water at 50oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------C. How much is the vapor pressure of a solution lowered by adding 25.9 grams of nonelectrolyte with a
molecular weight of 120.3 to 151 ml of water at 90oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------D. How much is the vapor pressure lowered by adding 55.7 grams of glucose (C6H12O6) to 125 ml of water
at 60oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------E. What is the vapor pressure at 80oC of a solution prepared by adding 42.4 grams of glucose (gfw = 180.0) to
125 mL of water?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------Xavier University of Louisiana
59
Chemistry 1020, Module 11B
✔✔✔■Calculation of Colligative Properties: Vapor Pressure Lowering (continued)
6f) F. How many grams of urea (gfw = 60.0) must be added to 212 mL of water to make a solution whose vapor
pressure at 70oC is 213 torr?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------✔✔✔■Calculation of Colligative Properties: Osmotic Pressure
7a) In general how does osmotic pressure of a solution change as its concentration changes? (p. 536-7)
----------------------------------------------------------------------------------------------------------------------------------------------b) State a quantitative relationship between osmotic pressure and concentration of a solution in words. (p. 537)
c)
----------------------------------------------------------------------------------------------------------------------------------------------S. A solution of urea is found to have an osmotic pressure of 4.2 atm. What would be the osmotic pressure of
a solution with twice the concentration?
Since osmotic pressure is directly proportional to concentration, doubling the concentration of the
solution would cause the osmotic pressure to double. Therefore, the osmotic pressure of the
solution twice the concentration would be 2*4.2 atm = 8.4 atm.
----------------------------------------------------------------------------------------------------------------------------------------------A. A solution of urea is found to have an osmotic pressure of 8.4 atm. What would be the osmotic pressure of
a solution with three times the concentration?
----------------------------------------------------------------------------------------------------------------------------------------------B A solution of glucose is found to have an osmotic pressure of 5.6 atm. What would be the osmotic
pressure of a solution with one-half the concentration?
----------------------------------------------------------------------------------------------------------------------------------------------C. A solution of urea is found to have an osmotic pressure of 2.4 atm. What would be the osmotic pressure of
a solution with one-fourth the concentration?
----------------------------------------------------------------------------------------------------------------------------------------------D. A solution of glucose is found to have an osmotic pressure of 9.6 atm. What would be the osmotic
pressure of a solution with one-third the concentration?
60
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
✔✔✔■Calculation of Colligative Properties: Osmotic Pressure (continued)
7d) Write the equation showing how osmotic pressure of solution is related to concentration. (p. 537)
e)
----------------------------------------------------------------------------------------------------------------------------------------------What is the constant in the equation in d above called?
f)
----------------------------------------------------------------------------------------------------------------------------------------------When can the equation in d above be used? (i.e. What are its limitations?)
----------------------------------------------------------------------------------------------------------------------------------------------g) S. Calculate the osmotic pressure at room temperature (25oC) of a solution prepared by adding 12 grams of
urea (gfw=60.0) to 93 grams of water. The density of the solution is 1.01 g/ml. (Section 11.6 in text)
•Formula: CO(NH 2 )2
•Type cmpd: not a, b, or s
•Strength: nonelectrolyte
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Osmotic pressure = ∏ = MRT and M = (mol urea)/(L soln)
1 mol urea
mol urea = 12 g urea * 60.0 g urea = 0.200 mol urea
1.00 mL
1L
L soln = (12 g urea + 93 g H2 O) * 1.01 g * 1000 mL = 0.104 L soln
0.200 mol urea
L*atm
∏ = MRT =
* 0.0821 mol*K * (273 + 25 K) =
0.104 L
R
= 46.9 atm = 47 atm
----------------------------------------------------------------------------------------------------------------------------------------------A. Calculate the osmotic pressure at 50 oC of a solution prepared by adding 23 g of urea (gfw = 60.0) to 112 g
of H2O. The density of the solution is 1.08 g/ml.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------B. Calculate the osmotic pressure at 75 oC of a solution prepared by adding 43 grams of glucose (gfw = 180.0)
to 234 grams of water. The density of the solution is 1.07 g/ml.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
-----------------------------------------------------------------------------------------------------------------------------------------------
Xavier University of Louisiana
61
Chemistry 1020, Module 11B
✔✔✔■Calculation of Colligative Properties: Osmotic Pressure (continued)
7g) C. How many grams of glucose (gfw = 180.0) must be added to enough water to make 183 mL of a solution
with an osmotic pressure of 1.3 atm at 45oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------D. How many grams of urea (gfw = 60.0) must be added to enough water to obtain 115 mL of a solution with
an osmotic pressure of 2.7 atm at 35oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------E. Calculate the osmotic pressure at 75 oC of a solution prepared by adding 52 g of a substance with gfw =
140.3 to 512 g of water. The density of the solution is 1.03 g/mL. The solution does not conduct
electricity when a low voltage is applied.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------F. How many grams of urea (gfw = 60.0) must be added to enough water to make 255 mL of a solution with
an osmotic pressure of 3.2 atm at 37oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
62
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
✔✔✔■Calculation of Colligative Properties: Freezing Point Depression
8a) In general how does freezing point of a solution change as its concentration changes? (p. 533)
----------------------------------------------------------------------------------------------------------------------------------------------b) State a quantitative relationship between change in freezing point and concentration of a solution in words. (p. 534)
c)
----------------------------------------------------------------------------------------------------------------------------------------------S. A solution of urea is found to freeze at -0.23oC. At what temperature would a solution with twice the
concentration freeze?
Since change in freezing point is directly proportional to concentration, doubling the concentration
of the solution would cause the freezing point to decrease twice as much. That is, ∆T for next
solution is 2*0.23oC = 0.46oC. Therefore, the solution freezes at -0.46oC.
----------------------------------------------------------------------------------------------------------------------------------------------A. A solution of urea is found to freeze at -.016oC. At what temperature would a solution with three times
the concentration freeze?
----------------------------------------------------------------------------------------------------------------------------------------------B A solution of glucose is found to freeze at -0.48 oC. At what temperature would a solution with one-fourth
the concentration freeze?
----------------------------------------------------------------------------------------------------------------------------------------------C. A solution of urea is found to freeze at -0.32oC. At what temperature would a solution with one-half the
concentration freeze?
----------------------------------------------------------------------------------------------------------------------------------------------D. A solution of urea is found to freeze at -0.58oC. At what temperature would a solution with four times the
concentration freeze?
d)
----------------------------------------------------------------------------------------------------------------------------------------------Write the equation showing how change in freezing point of solution is related to concentration. (p. 534)
e)
----------------------------------------------------------------------------------------------------------------------------------------------What is the constant in the equation in d above called?
f)
----------------------------------------------------------------------------------------------------------------------------------------------When can the equation in d above be used? (i.e. What are its limitations?)
-----------------------------------------------------------------------------------------------------------------------------------------------
Xavier University of Louisiana
63
Chemistry 1020, Module 11B
✔✔✔■Calculation of Colligative Properties: Freezing Point Depression (continued)
8g) S. Calculate the freezing point of a solution prepared by dissolving 22 grams of glucose, gfw = 180.0, in 61
grams of water. (pp. 534)
•Formula: C 6 H12 O6
•Type cmpd: not a, b, or s •Strength: nonelectrolyte
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
∆T f = K f*m
m = moles solute/kg solvent = moles glucose/kg water
1 mol glucose
moles solute = 22 g glucose * 180 g glucose = 0.122 mol glucose
1 kg H 2 O
kg solvent = 61 g H 2 O * 1000 g H O = 0.061 kg H 2 O
2
oC 0.122 mol glucose
∆T f = 1.86 m * 0.061 kg H O = 3.72oC which rounds to 3.7oC*
2
Tf = freezing point - ∆T f = 0.00oC - 3.7oC = -3.7oC*
*Note: The rule for rounding after multiplying or dividing is different from that after adding or subtrracting.
Therefore, use significant figures to round after multiplying, then add or subtract. Finally, round so the result
has the same number of decimal places as the least precise measurement used in the calculation. Thus, when
3.7 o C is subtracted from 0.00o C, you get -3.7 o C.
----------------------------------------------------------------------------------------------------------------------------------------------A. Calculate the freezing point of a solution prepared by dissolving 1.8 grams of urea, gfw = 60.0 in 301 g H2O.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------B. How many grams of glucose (gfw=180.0) must be added to 152 grams of water in order to obtain a solution
with a freezing point of -0.012 oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
-----------------------------------------------------------------------------------------------------------------------------------------------
64
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
✔✔✔■Calculation of Colligative Properties: Freezing Point Depression (continued)
8g) C. Calculate the freezing point of a solution prepared by dissolving 32 grams of glucose, gfw = 180.0, in 112
grams of water.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------D. Calculate the freezing point of a solution prepared by dissolving 15 grams of urea, gfw = 60.0, in 251
grams of water.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------E. Calculate the freezing point of a solution prepared by dissolving 25 grams of glucose (gfw = 180.0) in 123
g of water.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------F. How many grams of a substance with gfw = 75.0 must be added to 412 grams of water to obtain a solution
with a freezing point of -1.35 oC? The solution does not conduct electricity.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------Xavier University of Louisiana
65
Chemistry 1020, Module 11B
✔✔✔■Calculation of Colligative Properties: Boiling Point Elevation
9a) In general how does boiling point of a solution change as its concentration changes? (p. 531-2)
----------------------------------------------------------------------------------------------------------------------------------------------b) State a quantitative relationship between change in boiling point and concentration of a solution in words. (p. 532)
c)
----------------------------------------------------------------------------------------------------------------------------------------------S. A solution of urea is found to boil at 101.12oC. At what temperature would a solution with twice the
concentration boil if pressure is kept constant?
Since change in boiling point is directly proportional to concentration, doubling the concentration of
the solution would cause the boiling point to increase twice as much. That is, ∆T for new solution
is 2*1.12oC = 2.24oC. Therefore, the solution boils at 102.24oC. (Note: You have to add the
change to the boiling point of water to get the boiling point of the solution.)
----------------------------------------------------------------------------------------------------------------------------------------------A. A solution of urea is found to boil at 102.36oC. At what temperature would a solution with three times
the concentration boil if pressure is kept constant?
----------------------------------------------------------------------------------------------------------------------------------------------B A solution of glucose is found to boil at 103.22oC. At what temperature would a solution with one-fourth
the concentration boil if pressure is kept constant?
----------------------------------------------------------------------------------------------------------------------------------------------C. A solution of urea is found to boil at 100.98oC. At what temperature would a solution with one-half the
concentration boil if pressure is kept constant?
----------------------------------------------------------------------------------------------------------------------------------------------D. A solution of urea is found to boil at 102.48oC. At what temperature would a solution with four times the
concentration boil if pressure is kept constant?
d)
----------------------------------------------------------------------------------------------------------------------------------------------Write the equation showing how change in boiling point of solution is related to concentration. (p. 532)
e)
----------------------------------------------------------------------------------------------------------------------------------------------What is the constant in the equation in d above called?
f)
----------------------------------------------------------------------------------------------------------------------------------------------When can the equation in d above be used? (i.e. What are its limitations?)
-----------------------------------------------------------------------------------------------------------------------------------------------
66
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
✔✔✔■Calculation of Colligative Properties: Boiling Point Elevation (continued)
9g)
S. Calculate the boiling point of a solution prepared by dissolving 15 grams of urea (gfw=60.0) in 105
grams of water.
•Formula: CO(NH2 )2
•Type cmpd: not a, b, or s •Strength: nonelectrolyte
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
∆Tb = K b*m
m = moles solute/kg solvent = moles urea/kg water
1 mol urea
moles solute = 15 g urea * 60.0 g urea = 0.250 mol urea
1 kg H 2 O
kg solvent = 105 g H 2 O * 1000 g H O = 0.105 kg H 2 O
2
oC 0.250 mol urea
∆Tb = 0.51 m * 0.105 kg H O = 1.21oC which rounds to 1.2oC*
2
Tb = boiling point + ∆Tb = 100.00 oC + 1.2oC = 101.2oC*
*Note: The rule for rounding after multiplying or dividing is different from that after adding or subtrracting.
Therefore, use significant figures to round after multiplying, then add or subtract. Finally, round so the result
has the same number of decimal places as the least precise measurement used in the calculation. Thus, when
1.2 o C is added to 100.00o C, you get 100.00 o C + 1.2 o C rounds to 101.2 o C.
----------------------------------------------------------------------------------------------------------------------------------------------A. What is the boiling point of a solution prepared by adding 57 grams of glucose (gfw=180.0) to 125 grams
of water?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------B. How many grams of glucose (gfw=180.0) must be added to 205 grams of water in order to obtain a solution
which has a boiling point of 100.023oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
-----------------------------------------------------------------------------------------------------------------------------------------------
Xavier University of Louisiana
67
Chemistry 1020, Module 11B
✔✔✔■Calculation of Colligative Properties: Boiling Point Elevation (continued)
9g) C. What is the boiling point of solution prepared by adding 63 grams of a substance with gfw = 122.1 to
142 grams of water? The solution does not conduct electricity when low voltage is applied.
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------D. How many grams of water must be added to 12 grams of urea (gfw=60.0) in order to obtain a solution with
a boiling point of 100.16oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------E. What is the boiling point of a solution prepared by adding 32 grams of a urea (gfw = 60.0) to 152 mL of
water?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------F. How many grams of urea (gfw = 60.0) must be added to 455 mL of water to obtain a solution with a
boiling point of 101.63oC?
•Formula:
•Type cmpd:
•Strength:
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
----------------------------------------------------------------------------------------------------------------------------------------------68
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
✔✔✔■Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point
Elevation
10. S. 1.5 grams of an unknown nonelectrolyte is dissolved in 52 grams of water to produce a solution which has
a freezing point of -0.53oC. What is the molecular weight of the unknown. (pp. 533-5)
= 0 oC - Tf
= 0 oC - (-0.53oC)
= 0.53oC
= K f*m =
(g solute/mw solute)
= Kf *
kg solvent
∆Tf
∆Tf
∆Tf
∆Tf
o
g solute
∆Tf * (kg solvent) = Kf * mw solute mw solu= (1.86 C/m) * (1.5 g)
(0.052 kg) * (0.53 oC)
∆Tf * (kg solvent) * (mw solute) = Kf
R
* (g solute)
= 101.2 g/mol = 1.0*102 g/mol
K f * (g solute)
mw solute = (kg solvent) * (∆T ) =
f
----------------------------------------------------------------------------------------------------------------------------------------------A. 6.03 grams of an unknown nonelectrolyte is dissolved in 124 grams of water to produce a solution which
has a freezing point of -1.00oC. What is the molecular weight of the unknown?
----------------------------------------------------------------------------------------------------------------------------------------------B. 2.54 grams of an unknown nonelectrolyte is dissolved in 35.7 grams of water to produce a solution which
has a freezing point of -0.752oC. What is the molecular weight of the unknown?
----------------------------------------------------------------------------------------------------------------------------------------------C. 5.01 grams of an unknown nonelectrolyte is dissolved in 25.4 grams of water to produce a solution with a
boiling point of 101.21oC. What is the molecular weight of the unknown.
----------------------------------------------------------------------------------------------------------------------------------------------D. 7.03 grams of an unknown nonelectrolyte is dissolved in 46.6 grams of water to produce a solution with a
boiling point of 101.33oC. What is the molecular weight of the unknown?
Xavier University of Louisiana
69
Chemistry 1020, Module 11B
✔✔✔■Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point
Elevation (continued)
10. E. 7.33 grams of an unknown nonelectrolyte is dissolved in 35.5 grams of water to produce a solution which
has a freezing point of -0.75oC. What is the molecular weight of the unknown?
----------------------------------------------------------------------------------------------------------------------------------------------F. 8.21 grams of an unknown nonelectrolyte is dissolved in 75.2 grams of water to produce a solution which
as a boiling point of 100.65oC. What is the molecular weight of the unknown?
----------------------------------------------------------------------------------------------------------------------------------------------■B o n d i n g / M o d e l A c t i v i t y t o I m p r o v e A b i l i t y t o V i s u a l i z e i n 3 - D
A. a)
Draw the Lewis structure of CH3NO2 (nitromethane).
---------------------------------------------------------------------------------------------------------------------------------------------B. a) Draw the Lewis structure of CH3CH2Cl ( ethylchloride).
---------------------------------------------------------------------------------------------------------------------------------------------C. a) Draw the Lewis structure of CCl4 (carbon tetrachloride).
---------------------------------------------------------------------------------------------------------------------------------------------70
Xavier University of Louisiana
Chemistry 1020, Module 11B
Name
■Bonding/Model Activity to Improve Ability to Visualize in 3-D (continued)
D. a)
Draw the Lewis structure of C2H4 (ethylene).
---------------------------------------------------------------------------------------------------------------------------------------------A-D b) Then use your model set to assemble a model of the species.
■Challenge Questions
A. The vapor pressure of water at 60oC is 152 torr and the solubility of NaCl at that temperature is 51 grams/100
grams of water. Use this information to sketch a curve illustrating how the vapor pressure of a solution of
NaCl varies as 100.0 grams of NaCl is added to 100.0 ml of water 1 gram at a time. (i.e. Sketch vp versus g
solute added.)
B. 45 grams of NaCl is added to 100.0 grams of water and the water is boiled off. Sketch the boiling point of the
solution versus time if the solubility of NaCl in water at 100oC is 53 g/100ml.
C. What is the percent dissociation of a weak electrolyte if the freezing point of a 0.123 m solution of the
substance is -0.240 oC. The electrolyte has the formula AB and partly dissociates into A + and B- when it
dissolves.
D. A 0.010 m solution of acetic acid, CH3COOH, has a freezing point of -0.0194 oC. What is the percent
dissociation of acetic acid in the solution?
E. The boiling point of a solution prepared by adding 2.31 grams of an organic compound (93.8% C and 6.2% H)
to 86.7 grams of carbon tetrachloride was 350.83 K. The boiling point of pure carbon tetrachloride is 349.90 K
and the Kb of carbon tetrachloride is 4.48 K/m. What is the molecular formula of the organic compound?
F. What is the boiling point of a solution of 10.0 grams of P4 in 45.2 mL of carbon disulfide if the boiling point
elevation constant for CS 2 is 2.35oC/m and the boiling point is 46.23oC?
Note: All parts of this module as well as the
following Skills Module on Rounding Rules must
be filled out when you arrive at drill. You will have
quizzes on both.
Xavier University of Louisiana
71
Chemistry 1020, Module 11B
Revised by JW Carmichael 5/9/2000; SJB 5/8/2001; SJB 2002 & 2003
72
Xavier University of Louisiana