Study Guide for Module 11B—Solutions II · PDF fileChemistry 1020, Module 11B Name ... *Although this number has been rounded, in the lab you would weigh as accurately as the balance

Chemistry 1020, Module 11B Name

Study Guide for Module 11B—Solutions II

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Reading Assignment: Sections 11.2 through 11.7 in Chemistry, 6th Edition by Zumdahl.

Guide for Your Lecturer:

1. Review of Preparing Solutions2. Solubility Rules3. The Solution Process and Analytical Concentrations4. Solutions Compared to Colloids and Suspensions; Review of Freezing Point, Boiling Point, Vapor

Pressure; Osmosis and Osmotic Pressure5. Qualitative Consideration of the Colligative Properties of Solutions

✔ 6. Calculation of Colligative Properties: Vapor Pressure Lowering✔ 7. Calculation of Colligative Properties: Osmotic Pressure✔ 8. Calculation of Colligative Properties: Freezing Point Depression✔ 9. Calculation of Colligative Properties: Boiling Point Elevation✔ 10. Molecular Weight of an Unknown from Freezing Point Depression or Boiling Point Elevation

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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 tobe 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•5H2O •Type cmpd: salt •Strength: strong

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Calculation |Steps to be used in preparing solution gfw of CuSO4•5H2O = 63.5 + 32.1 + 4*16.0 +5(2*1.01 + 16.0) = 249.7 g/mol

M = mol solute/L solnM = (g/mw)/LM*L = g/mwg = M*L*mw

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 CuSO4•5H2O **Although this number has been rounded, in the lab youwould weigh as accurately as the balance available allows.In General Chemistry, this would be to the nearest ±0.01 gso you would weight out a number close to 8.4 and recordits weight to two places past the decimal.

1) Weigh out approximately 8.4 gCuSO4•5H2O and then record the actualmass reading on the balance to the nearest0.01 g. 2) Add about 105/2 = 52 mL of H2O to thecontainer. 3) Add the 8.4 g CuSO4•5H2O to thewater in the container and mix until the solutedissolves. 4) Add enough additional H2O to get a totalof 105 mL of solution. 5) Mix the solution.

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)

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4)

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44 Xavier University of Louisiana


■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)

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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

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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

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■Review of Preparing Solutions (continued)1a) E. How would you prepare 93 milliliters of 0.24 M sodium carbonate solution from solid solute?


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 M1 = 3.4 M V1 = ?M2 = 0.87 MV2 = 112 mLM1V1 = M2V2V1 = M2V2/M1V1 = (0.87 M)(112 mL)/(3.4 M) = RV1 = 28.6588 mL = 29 mL of 3.4 M HBr**Although this number has been rounded, in the lab youwould measure volume as accurately as the equipmentavailable allows. In General Chemistry, this would be tothe nearest ±0.1 mL if you use a graduated cylinder or tothe nearest ±0.01 mL if you use a pipette. If you don’tknow what kind of equipment to use, assume a graduatedcylinder.

1) Measure out 28.7 mL of 3.4 M HBr

2) Put approximately (112 - 28.7)/2 = 42mL of H2O in a container. 3) Add the 28.7 mL of 3.4 M HBr to the container and mix. 4) Add additional water to the container until a total of 112 mL of solution is obtained. 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

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■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?


1)

2)

3)

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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

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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

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■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

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■Solubi l i t y Rules✔✔✔■Names and Formulas of Common Anions and Common Chemical Compounds2a) 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.

| || || || |

| || |



2b. Cations whose salts are generally soluble: Na+, K+, NH4+, and the other Group IA metal ions

Anions whose salts are all soluble: NO3-, CH3COO-, and ClO4-

Anions whose salts are generally soluble: Cl-, except for AgCl, Hg2Cl2*, and PbCl2Br-, except for AgBr, Hg2Br2*, PbBr2, and HgBr2I-, except for AgI, Hg2I2*, PbI2, and HgI2SO42- except for CaSO4, SrSO4, BaSO4, PbSO4, Hg2SO4*, &Ag2SO4

Anions whose salts are generally insoluble: S2-, except for those of IA and IIA metals and (NH4)2S.

CO32-, except for those of IA metals and (NH4)2CO3SO32-, except for those of IA metals and (NH4)2SO3PO43-, except for those of IA metals and (NH4)3PO4OH-, except for those of IA metals, Ba(OH)2, Sr(OH)2, and Ca(OH)2

*Hg22+ is mercury(I) ion (or mercurous ion). It is the only diatomic metallic cation youneed 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

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.



■The Solution Process and Analytical Concentrations3a) 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: NaCH3COO •Type cmpd: salt •Strength: strong

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• NaCH3COO(s) ➙ Na+(aq) + CH3COO-(aq)

-----------------------------------------------------------------------------------------------------------------------------------------------A. ammonium sulfide

•Formula: •Type cmpd: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------B. copper(II) nitrate


-----------------------------------------------------------------------------------------------------------------------------------------------C. zinc(II) sulfate


-----------------------------------------------------------------------------------------------------------------------------------------------D. calcium chloride


-----------------------------------------------------------------------------------------------------------------------------------------------E. rubidium sulfite


-----------------------------------------------------------------------------------------------------------------------------------------------



■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 Na2SO4 •Name: sodium sulfate •Type cmpd: salt •Strength: strong •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Na2SO4(s) ➙ 2 Na+(aq) + SO42-(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, [SO42-] = 0.30 M

-----------------------------------------------------------------------------------------------------------------------------------------------A. 0.62 M CaCl2

•Name •Type cmpd: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------B. 0.25 M HNO3

•Name: •Type cmpd: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------C. 0.78 M KOH


-----------------------------------------------------------------------------------------------------------------------------------------------D. 0.35 M Na3PO4


-----------------------------------------------------------------------------------------------------------------------------------------------E. 0.13 M K2SO4


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■Solutions Compared to Colloids and Suspensions; Selected Properties of Solutions4a) 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 ishomogeneous all the way down to the molecular level. In a suspension the solute particles are stillsmall enough to stay suspended for a while but are large enough to settle out {precipitate} if themixture is left to stand for a while. In a colloid, the solute particles are intermediate in size of thesolute particles in a solution and the size in a suspension. As a result, they are small enough to staysuspended indefinitely {i.e. don't settle out as in the suspension} but are too large to consider thesolution 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 lightthrough it. If the light beam passes through without being reflected, it is a solution. If the light beamis reflected to the side {i.e. can be seen passing through the mixture in the way a light beam can beseen passing through a cloud of dust), the mixture is a colloid. The solute particles in a solution arenot 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) Define osmosis. (p. 535, A37)

-----------------------------------------------------------------------------------------------------------------------------------------------f) Briefly compare what happens to solute and solvent during diffusion (question 4c ) with what happens during

osmosis (question 4e).

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■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)

0.5 m 0.03 m urea urea

1-In which direction does the water flow as the system goesto equilibrium? From the more dilute 0.03 m solution tothe more concentrated 0.5 m solution.2-In which direction does the solute flow as the system goesto equilibrium? Solute does not flow. The membrane issemipermeable which means that only H2O can flow.3-What happens to the concentration of the 0.03 m side ofthe 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 thesystem as it goes to equilibrium?

It decreases as water flows in.5-What happens to the volume of solution in the 0.03 mside 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 sideof 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 ina beaker.

1-In which direction does the water flow as the system goesto equilibrium?

2-In which direction does the solute flow as the system goesto equilibrium?

3-What happens to the concentration of the solution insidethe balloon as the system goes to equilibrium?

4-What happens to the concentration of the solution in thebeaker as the system goes to equilibrium?

5-What happens to the volume of solution in the balloon asthe system goes to equilibrium?

6-What happens to the volume of solution outside theballoon as the system goes to equilibrium?

-----------------------------------------------------------------------------------------------------------------------------------------------



■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.

The solution inside the upside funnelabove is 0.03 m in urea. That outsideis 0.1 m in urea. There is asemipermable membrane across thetop of the funnel (on the bottom ofthe diagram) separating the two.



3-What happens to the concentration outside the funnel asthe system goes to equilibrium?

4-What happens to the concentration inside the funnel as thesystem goes to equilibrium?

5-What happens to the volume of solution outside the funnelas the system goes to equilibrium?

6-What happens to the volume of solution inside the funnelas 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 semipermeablemembrane separating the two halves of the tube..

0.4 m 0.1 m glucose glucose



3-What happens to the concentration of the solution on theleft as the system goes to equilibrium?

4-What happens to the concentration of the solution on theright as the system goes to equilibrium?

5-What happens to the volume of solution on the left as thesystem goes to equilibrium?

6-What happens to the volume of solution on the right asthe system goes to equilibrium?

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■Qualitat ive Consideration of the Coll igative Properties of Solutions5a) 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 explainwhat 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 Covered1)

2)

3)

4)

i) Define "particle concentration". (particle concentration = concentration of solute * number of particleseach 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)



■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

ConcentrationsWhich solution has theindicated property?

Which solution conductselectrity better?

S1. 0.16 m urea nonelectrolyte 0.16 m Higher fp? The twohave the same parti-

Neither conduct since noions are in either.

0.16 m glucose nonelectrolyte 0.16 m cle concentrations sothey have the same fp.

S2. 0.23 m NaCl strongelectrolyte

0.46 m Lower vapor pressure?0.23 m NaCl since

0.23 m NaCl since ithas higher concentration

0.13 m NaCl strongelectrolyte

0.26 m its particle conc. ishigher.

of ions.

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 mNa2SO4

Higher vapor pressure?

0.37 m glucose

Solutions Strength ParticleConcentrations

Which solution has theindicated property 1?

Which solution conductselectrity better?

D. 0.16 m urea Higher bp?

0.16 m NaNO3

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 NaNO3 Lower freezing point?

0.17 m KBr



✔✔✔■Calculation of Colligative Properties: Vapor Pressure Lowering6a) 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 fractionof water in the solution. )

-----------------------------------------------------------------------------------------------------------------------------------------------c) Write the equation for the statement in b) . (Psoln = Xwater in soln*Ppure water)

-----------------------------------------------------------------------------------------------------------------------------------------------d) 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 knowthe mole fraction of solute in the solution rather than mole fraction of water. Use Xsolute + Xpure water = 1 toobtain an equation which relates change in vapor pressure as the solution forms (i.e. the difference between thevapor pressure of the solution from that of pure water) to the mole fraction of solute. Note: This is theequation usually used when considering colligative properties of water.( VP = Ppure water - Psoln = Xsolute*Ppure water )

-----------------------------------------------------------------------------------------------------------------------------------------------e) When can the equation in d above be used? (i.e. What are its limitations?)

-----------------------------------------------------------------------------------------------------------------------------------------------f) 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: C6H12O6 •Type cmpd: not a, b, or s •Strength: nonelectrolyte

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Psoln = Pwater*Xwater

Pwater = 17.54 torr (from data sheet)Xwater = (mol water)/(total moles)mol water = 251 mL H2O *

* 1 g H2O

1 mL H2O * 1 mol H2O

18.02 g H2O =

= 13.93 mol H2Omol glucose = 42.0 g glucose *

* 1 mol glucose

180.0 g glucose = 0.2333 mol glucose

Xwater = 13.93 mol H2O

13.93 + 0.2333 total moles =

= 0.9838

Psoln = Pwater*Xwater = (17.54 torr)(0.9838) = 17.26 torr *

= 17.3 torr∆P = 17.54 - 17.3 torr = 0.24 torr difference which rounds to 0.2 torrTherefore, it is 0.2 torr lower.

*Reminder: Carry one more place in intermediate steps which involve multiplication or division. Then roundround 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: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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 Coll igative Properties: Osmotic Pressure7a) 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 thesolution would cause the osmotic pressure to double. Therefore, the osmotic pressure of thesolution 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?



✔✔✔■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(NH2)2 •Type cmpd: not a, b, or s •Strength: nonelectrolyte

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Osmotic pressure = ∏ = MRT and M = (mol urea)/(L soln)

mol urea = 12 g urea * 1 mol urea

60.0 g urea = 0.200 mol urea

L soln = (12 g urea + 93 g H2O) * 1.00 mL1.01 g *

1 L1000 mL = 0.104 L soln

∏ = MRT = 0.200 mol urea

0.104 L * 0.0821L*atmmol*K * (273 + 25 K) =

R= 46.9 atm = 47 atm

-----------------------------------------------------------------------------------------------------------------------------------------------A. Calculate the osmotic pressure at 50oC 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 75oC 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: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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 75oC 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 conductelectricity 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: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••



✔✔✔■Calculation of Colligative Properties: Freezing Point Depression8a) 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 concentrationof the solution would cause the freezing point to decrease twice as much. That is, ∆T for nextsolution 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 -0.48oC. 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)



-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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: C6H12O6 •Type cmpd: not a, b, or s •Strength: nonelectrolyte

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••∆Tf = Kf*mm = moles solute/kg solvent = moles glucose/kg water

moles solute = 22 g glucose * 1 mol glucose180 g glucose = 0.122 mol glucose

kg solvent = 61 g H2O * 1 kg H2O

1000 g H2O = 0.061 kg H2O

∆Tf = 1.86oCm *

0.122 mol glucose0.061 kg H2O = 3.72oC which rounds to 3.7oC*

Tf = freezing point - ∆Tf = 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 resulthas the same number of decimal places as the least precise measurement used in the calculation. Thus, when3.7oC is subtracted from 0.00oC, you get -3.7 oC.

-----------------------------------------------------------------------------------------------------------------------------------------------A. Calculate the freezing point of a solution prepared by dissolving 1.8 grams of urea, gfw = 60.0 in 301 g H2O.


-----------------------------------------------------------------------------------------------------------------------------------------------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 -0.012oC?•Formula: •Type cmpd: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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 -1.35oC? The solution does not conduct electricity.•Formula: •Type cmpd: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■Calculation of Coll igative Propert ies: Boil ing Point Elevation9a) 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 ofthe solution would cause the boiling point to increase twice as much. That is, ∆T for new solutionis 2*1.12oC = 2.24oC. Therefore, the solution boils at 102.24oC. (Note: You have to add thechange 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)



-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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 = Kb*mm = moles solute/kg solvent = moles urea/kg water

moles solute = 15 g urea * 1 mol urea

60.0 g urea = 0.250 mol urea

kg solvent = 105 g H2O * 1 kg H2O

1000 g H2O = 0.105 kg H2O

∆Tb = 0.51oCm *

0.250 mol urea0.105 kg H2O = 1.21oC which rounds to 1.2oC*

Tb = boiling point + ∆Tb = 100.00oC + 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 resulthas the same number of decimal places as the least precise measurement used in the calculation. Thus, when1.2oC is added to 100.00oC, you get 100.00oC + 1.2oC rounds to 101.2oC.

-----------------------------------------------------------------------------------------------------------------------------------------------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: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

-----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■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: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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✔✔✔■Molecular Weight of an Unknown from Freezing Point Depression or Boiling PointElevation10. 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)

∆Tf = 0oC - Tf∆Tf = 0oC - (-0.53oC)

∆Tf = 0.53oC∆Tf = Kf*m =

= Kf * (g solute/mw solute)

kg solvent

∆Tf * (kg solvent) = Kf * g solute

mw solute∆Tf * (kg solvent) * (mw solute) = Kf

* (g solute)

mw solute = Kf * (g solute)

(kg solvent) * (∆Tf) =

mw solu= (1.86 oC/m) * (1.5 g)

(0.052 kg) * (0.53oC) R

= 101.2 g/mol = 1.0*102 g/mol

-----------------------------------------------------------------------------------------------------------------------------------------------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


-----------------------------------------------------------------------------------------------------------------------------------------------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?



✔✔✔■Molecular Weight of an Unknown from Freezing Point Depression or Boiling PointElevation (continued)10. E. 7.33 grams of an unknown nonelectrolyte is dissolved in 35.5 grams of water to produce a solution which


-----------------------------------------------------------------------------------------------------------------------------------------------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?

-----------------------------------------------------------------------------------------------------------------------------------------------

■Bonding/Model Act iv i ty to Improve Abi l i ty to Visual ize in 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).

----------------------------------------------------------------------------------------------------------------------------------------------



■Bonding/Model Activi ty to Improve Abil i ty 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/100grams of water. Use this information to sketch a curve illustrating how the vapor pressure of a solution ofNaCl 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 gsolute 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 thesolution 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 thesubstance is -0.240oC. The electrolyte has the formula AB and partly dissociates into A + and B- when itdissolves.

D. A 0.010 m solution of acetic acid, CH3COOH, has a freezing point of -0.0194oC. What is the percentdissociation 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 Kand 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 CS2 is 2.35oC/m and the boiling point is 46.23oC?

Note: All parts of this module as well as thefollowing Skills Module on Rounding Rules mustbe filled out when you arrive at drill. You will havequizzes on both.



Revised by JW Carmichael 5/9/2000; SJB 5/8/2001; SJB 2002 & 2003


Documents

Study Guide for Module 11B—Solutions II · PDF fileChemistry 1020, Module 11B Name ... *Although this number has been rounded, in the lab you would weigh as accurately as the balance