1 Sample Exercise 13.1 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl 4 ) or in water: C 7 H 16, Na 2 SO 4, HCl, and I 2. Analyze We are given two solvents, one that is nonpolar (CCl 4 ) and the other that is polar (H 2 O), and asked to determine which will be the better solvent for each solute listed. Plan By examining the formulas of the solutes, we can predict whether they are ionic or molecular. For those that are molecular, we can predict whether they are polar or nonpolar. We can then apply the idea that the nonpolar solvent will be better for the nonpolar solutes, whereas the polar solvent will be better for the ionic and polar solutes. Solve C 7 H 16 is a hydrocarbon, so it is molecular and nonpolar. Na 2 SO 4, a compound containing a metal and nonmetals, is ionic. HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar. I 2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C 7 H 16 and I 2 (the nonpolar solutes) would be more soluble in the nonpolar CCl 4 than in polar H 2 O, whereas water would be the better solvent for Na 2 SO 4 and HCl (the ionic and polar covalent solutes).
2 Sample Exercise 13.1 Predicting Solubility Patterns Continued Arrange the following substances in order of increasing solubility in water: Answer: C 5 H 12 <C 5 H 11 Cl<C 5 H 11 OH<C 5 H 10 (OH) 2 (in order of increasing polarity and hydrogen-bonding ability)
3 Sample Exercise 13.2 A Henry s Law Calculation Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25 C. The Henry s law constant for CO 2 in water at this temperature is mol/l-atm. Analyze We are given the partial pressure of CO 2, P CO2, and the Henry s law constant, k, and asked to calculate the concentration of CO 2 in the solution. Plan With the information given, we can use Henry s law, Equation 13.4, to calculate the solubility, S CO2. Solve S CO2 = kp CO2 = ( mol/l-atm)(4.0 atm) = 0.14 mol/l = 0.14 M Check The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO 2 and the value of Henry s constant. Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25 C under a CO 2 partial pressure of atm. Answer: M
4 Sample Exercise 13.3 Converting Units of Temperature (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn 2+. What is the concentration of Zn 2+ in parts per million? (a) Analyze We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute. Plan We can calculate the mass percentage by using Equation The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve Comment The mass percentage of water in this solution is ( )% = 88.1%.
5 Sample Exercise 13.3 Converting Units of Temperature Continued (b) Analyze In this case we are given the number of micrograms of solute. Because 1 g is g, 5.4 g = g Plan We calculate the parts per million using Equation Solve (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution? Answer: (a) 2.91%, (b) 90.5 g of NaOCl
6 Sample Exercise 13.4 Calculation of Molarity A solution is made by dissolving 4.35 g glucose (C 6 H 12 O 6 ) in 25.0 ml of water at 25 C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g/ml. Analyze We are asked to calculate a molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water). Plan We use the molar mass of C 6 H 12 O 6 to convert grams to moles. We use the density of water to convert milliliters to kilograms. The molality equals the number of moles of solute divided by the number of kilograms of solvent (Equation 13.9). Solve Use the molar mass of glucose, g/mol, to convert grams to moles: Because water has a density of 1.00 g/ml, the mass of the solvent is (25.0 ml)(1.00 g/ml) = 25.0 g = kg Finally, use Equation 13.9 to obtain the molality: What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10 H 8 ) in 425 g of toluene (C 7 H 8 )? Answers: m
7 Sample Exercise 13.5 Calculation of Mole Fraction and Molarity An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution. Analyze We are asked to calculate the concentration of the solute, HCl, in two related concentration units, given only the percentage by mass of the solute in the solution. Plan In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let s assume that there is exactly 100 g of solution. Because the solution is 36% HCl, it contains 36 g of HCl and (100 36) g = 64 g of H 2 0. We must convert grams of solute (HCl) to moles to calculate either mole fraction or molality. We must convert grams of solvent (H 2 O) to moles to calculate mole fractions and to kilograms to calculate molality. Solve (a) To calculate the mole fraction of HCl, we convert the masses of HCl and H 2 O to moles and then use Equation 13.7:
8 Sample Exercise 13.5 Calculation of Mole Fraction and Molarity Continued (b) To calculate the molality of HCl in the solution, we use Equation We calculated the number of moles of HCl in part (a), and the mass of solvent is 64 g = kg: A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution. Answer: (a) , (b) m
9 Sample Exercise 13.6 Calculation of Molarity Using the Density of the A solution with a density of g/ml contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene. Calculate the molarity of the solution. Analyze Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the solution (0.876 g/ml). Plan The molarity of a solution is the number of moles of solute divided by the number of liters of solution (Equation 13.8). The number of moles of solute (C 7 H 8 ) is calculated from the number of grams of solute and its molar mass. The volume of the solution is obtained from the mass of the solution (mass of solute + mass of solvent = 5.0 g g = 230 g) and its density. Solve The number of moles of solute is The density of the solution is used to convert the mass of the solution to its volume: Molarity is moles of solute per liter of solution:
10 Sample Exercise 13.6 Calculation of Molarity Using the Density of the Continued Check The magnitude of our answer is reasonable. Rounding moles to 0.05 and liters to 0.25 gives a molarity of (0.05 mol)/(0.25 L) = 0.2 M The units for our answer (mol/l) are correct, and the answer, 0.21, has two significant figures, corresponding to the number of significant figures in the mass of solute (2). Comment Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263) are similar in magnitude, the molarity and molality are also similar in magnitude: (0.054 mol C 7 H 8 )/(0.225 kg solvent) = 0.24 m A solution containing equal masses of glycerol (C 3 H 8 O 3 ) and water has a density of 1.10 g/ml. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution. Answer: (a) 10.9 m, (b) XC 3 H 8 O 3 = 0.163, (c) 5.97 M
11 Sample Exercise 13.7 Calculation of Vapor Pressure of a Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/ml at 25 C. Calculate the vapor pressure at 25 C of a solution made by adding 50.0 ml of glycerin to ml of water. The vapor pressure of pure water at 25 C is 23.8 torr (Appendix B), and its density is 1.00 g/ml. Analyze Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute. Plan We can use Raoult s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, X solvent, is the ratio of the number of moles of solvent (H 2 O) to total moles of solution (moles C 3 H 8 O 3 + moles H 2 O). Solve To calculate the mole fraction of water in the solution, we must determine the number of moles of C 3 H 8 O 3 and H 2 O: We now use Raoult s law to calculate the vapor pressure of water for the solution:
12 Sample Exercise 13.7 Calculation of Vapor Pressure of a Continued Comment The vapor pressure of the solution has been lowered by 23.8 torr 23.2 torr = 0.6 torr relative to that of pure water. The vapor-pressure lowering can be calculated directly using Equation together with the mole fraction of the solute, C 3 H 8 O 3 : = (0.024)(23.8 torr) = 0.57 torr. Notice that the use of Equation gives one more significant figure than the number obtained by subtracting the vapor pressure of the solution from that of the pure solvent. The vapor pressure of pure water at 110 C is 1070 torr. A solution of ethylene glycol and water has a vapor Pressure of 1.00 atm at 110 C. Assuming that Raoult s law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290
13 Sample Exercise 13.8 Calculating Of Boiling-Point Elevation and Freezing-Point Depression Automotive antifreeze consists of ethylene glycol, CH 2 (OH)CH 2 (OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water. Analyze We are given that a solution contains 25.0 mass % of a nonvolatile, nonelectrolyte solute and asked to calculate the boiling and freezing points of the solution. To do this, we need to calculate the boiling-point elevation and freezing-point depression. Plan To calculate the boiling-point elevation and the freezing-point depression using Equations and 13.13, we must express the concentration of the solution as molality. Let s assume for convenience that we have 1000 g of solution. Because the solution is 25.0 mass % ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. Using these quantities, we can calculate the molality of the solution, which we use with the molal boiling-point-elevation and freezing-point-depression constants (Table 13.3) to calculate T b and T f.we add T b to the boiling point and subtract T f from the freezing point of the solvent to obtain the boiling point and freezing point of the solution.
14 Sample Exercise 13.8 Calculating Of Boiling-Point Elevation and Freezing-Point Depression Continued Solve The molality of the solution is calculated as follows: We can now use Equations and to calculate the changes in the boiling and freezing points: T b = K b m = (0.51 C/m)(5.37 m) = 2.7 C T f = K f m = (1.86 C/m)(5.37 m) = 10.0 C Hence, the boiling and freezing points of the solution are Boiling point = (normal bp of solvent) + T b = C C = C Freezing point = (normal fp of solvent) Tf = 0.0 C 10.0 C = 10.0 C Comment Notice that the solution is a liquid over a larger temperature range than the pure solvent. Calculate the freezing point of a solution containing kg of CHCl 3 and 42.0 g of eucalyptol (C 10 H 18 O), a fragrant substance found in the leaves of eucalyptus trees. (See Table 13.3.) Answer: 65.6 C
15 Sample Exercise 13.9 Freezing-Point Depression in Aqueous s List the following aqueous solutions in order of their expected freezing point: m CaCl 2, 0.15 m NaCl, 0.10 m HCl, m CH 3 COOH, 0.10 m C 12 H 22 O 11. Analyze We must order five aqueous solutions according to expected freezing points, based on molalities and the solute formulas. Plan The lowest freezing point will correspond to the solution with the greatest concentration of solute particles. To determine the total concentration of solute particles in each case, we must determine whether the substance is a nonelectrolyte or an electrolyte and consider the number of ions formed when an electrolyte ionizes. Solve CaCl 2, NaCl, and HCl are strong electrolytes, CH 3 COOH (acetic acid) is a weak electrolyte, and C 12 H 22 O 11 is a nonelectrolyte. The molality of each solution in total particles is as follows: Because the freezing points depend on the total molality of particles in solution, the expected ordering is 0.15 m NaCl (lowest freezing point), 0.10 m HCl, m CaCl 2, 0.10 m C 12 H 22 O 11, and m CH 3 COOH (highest freezing point).
16 Sample Exercise 13.9 Freezing-Point Depression in Aqueous s Continued Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(NO 3 ) 2, 2 mol of KCl, 3 mol of ethylene glycol (C 2 H 6 O 2 )? Answer: 2 mol of KCl because it contains the highest concentration of particles, 2 m K + and 2 m Cl, giving 4 m in all
17 Sample Exercise Calculating Involving Osmotic Pressure The average osmotic pressure of blood is 7.7 atm at 25 C. What molarity of glucose (C 6 H 12 O 6 ) will be isotonic with blood? Analyze We are asked to calculate the concentration of glucose in water that would be isotonic with blood, given that the osmotic pressure of blood at 25 C is 7.7 atm. Plan Because we are given the osmotic pressure and temperature, we can solve for the concentration, using Equation Solve Comment In clinical situations the concentrations of solutions are generally expressed as mass percentages. The mass percentage of a 0.31 M solution of glucose is 5.3%. The concentration of NaCl that is isotonic with blood is 0.16 M, because NaCl ionizes to form two particles, Na + and Cl (a M solution of NaCl is M in particles). A 0.16 M solution of NaCl is 0.9 mass % in NaCl. This kind of solution is known as a physiological saline solution.
18 Sample Exercise Calculating Involving Osmotic Pressure Continued What is the osmotic pressure at 20 C of a M sucrose (C 12 H 22 O 11 ) solution? Answer: atm, or 37 torr
19 Sample Exercise Molar Mass from Freezing-Point Depression A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving g of the substance in 40.0 g of CCl 4. The boiling point of the resultant solution was C higher than that of the pure solvent. Calculate the molar mass of the solute. Analyze Our goal is to calculate the molar mass of a solute based on knowledge of the boiling-point elevation of its solution in CCl 4, T b = C, and the masses of solute and solvent. Table 13.3 gives K b for the solvent (CCl 4 ), K b = 5.02 C/m. Plan We can use Equation 13.12, T b = K b m, to calculate the Molality of the solution. Then we can use molality and the quantity of solvent (40.0 g CCl 4 ) to calculate the number of moles of solute. Finally, the molar mass of the solute equals the number of grams per mole, so we divide the number of grams of solute (0.250 g) by the number of moles we have just calculated.
20 Sample Exercise Molar Mass from Freezing-Point Depression Continued Solve From Equation we have Thus, the solution contains mol of solute per kilogram of solvent. The solution was prepared using of solvent (CCl 4 ). The number of moles of solute in the solution is therefore The molar mass of the solute is the number of grams per mole of the substance: Camphor (C 10 H 16 O) melts at C, and it has a particularly large freezing-point-depression constant, K f = 40.0 C/m. When g of an organic substance of unknown molar mass is dissolved in g of liquid camphor, the freezing point of the mixture is found to be C. What is the molar mass of the solute? Answer: 110 g/mol
21 Sample Exercise Molar Mass from Osmotic Pressure The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 ml of solution. The osmotic pressure of the solution at 25 C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass. Analyze Our goal is to calculate the molar mass of a high-molecular-mass protein, based on its osmotic pressure and a knowledge of the mass of protein and solution volume. Plan The temperature (T = 25 C) and osmotic pressure ( = 1.54 torr) are given, and we know the value of R so we can use Equation to calculate the molarity of the solution, M. In doing so, we must convert temperature from C to K and the osmotic pressure from torr to atm. We then use the molarity and the volume of the solution (5.00 ml) to determine the number of moles of solute. Finally, we obtain the molar mass by dividing the mass of the solute (3.50 mg) by the number of moles of solute. Solve Solving Equation for molarity gives
22 Sample Exercise Molar Mass from Osmotic Pressure Continued Because the volume of the solution is 5.00 ml = L, the number of moles of protein must be The molar mass is the number of grams per mole of the substance. The sample has a mass of 3.50 mg = g. The molar mass is the number of grams divided by the number of moles: Comment Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way to determine the molar masses of large molecules. A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form L of solution. The osmotic pressure of this solution was found to be 1.21 kpa at 25 C. Calculate the molar mass of the polystyrene. Answer: g/mol
Sample Exercise 13.1 (p. 534) By the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Na 2 SO 4(s) + 10 H 2
13.4 Ways of Expressing Concentration All methods involve quantifying the amount of solute per amount of solvent (or solution). Concentration may be expressed qualitatively or quantitatively. The terms
Lecture Presentation Chapter 13 Properties of Yonsei University homogeneous mixtures of two or more pure substances: may be gases, liquids, or solids In a solution, the solute is dispersed uniformly throughout
Colligative properties 1 1. What does the phrase like dissolves like mean. 2. Why does the solubility of alcohols decrease with increased carbon chain length? Alcohol in water (mol/100g water) Methanol
Role of Disorder in Solutions Disorder (Entropy) is a factor Solutions mix to form maximum disorder Two Ways to Form Solutions 1. Physical Dissolving (Solvation) NaCl(s) Na + (aq) + Cl - (aq) C 12 H 22
Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 13 Properties of are homogeneous mixtures of two or more pure substances. In a solution,
Chem 112 Solutions From the book (10, 12, 16, 18, 22, 24 52, 54, 56, 58, 62, 64, 66, 68, 74, 76, 78, 80, 82, 86, 88, 90, 92, 106 and 116) 1. Which of the following compounds are nonelectrolytes? A. NaF
COLLIGATIVE PROPERTIES: A colligative property is a property that depends only on the number of solute particles present, not their identity. The properties we will look at are: lowering of vapor pressure;
13 Sample Test 1 SAMPLE TEST 1. CHAPTER 12 1. The molality of a solution is defined as a. moles of solute per liter of solution. b. grams of solute per liter of solution. c. moles of solute per kilogram
Colligative Properties Vapor pressures have been defined as the pressure over a liquid in dynamic equilibrium between the liquid and gas phase in a closed system. The vapor pressure of a solution is different
16 SOLUTIONS SECTION 16.1 PROPERTIES OF SOLUTIONS (pages 471 477) This section identifies the factors that affect the solubility of a substance and determine the rate at which a solute dissolves. Solution
Name: Class: Date: Unit 13 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The dissolution of water in octane (C 8 H 18 ) is prevented by.
48 Practice Problems for Ch. 17 - Chem 1C - Joseph 1. Which of the following concentration measures will change in value as the temperature of a solution changes? A) mass percent B) mole fraction C) molality
Chapter 14 Solutions 1 14.1 General properties of solutions solution a system in which one or more substances are homogeneously mixed or dissolved in another substance two components in a solution: solute
CHAPTER 14 Solutions The Dissolution Process 1. Effect of Temperature on Solubility 2. Molality and Mole Fraction Colligative Properties of Solutions 3. Lowering of Vapor Pressure and Raoult s Law 4. Fractional
12.1 Thirsty Solutions: Why You Shouldn t Drink Seawater 544 12.2 Types of Solutions and Solubility 546 12.3 Energetics of Solution Formation 551 12.4 Solution Equilibrium and Factors Affecting Solubility
Chapter 14 Mixtures Warm Up What is the difference between a heterogeneous and homogeneous mixture? Give 1 example of a heterogeneous mixture and 1 example of a homogeneous mixture. Today s Agenda QOTD:
12.3 Colligative Properties Changes in solvent properties due to impurities Colloidal suspensions or dispersions scatter light, a phenomenon known as the Tyndall effect. (a) Dust in the air scatters the
Problems - Chapter 13 (with solutions) 1) The following question concerns mixing of liquids. a) Consider mixing two liquids where mixing is exothermic (Hsoln < 0). Would you expect a solution to form (yes/maybe/no)?
118 ChemQuest 39 Name: Date: Hour: Information: Molarity Concentration is a term that describes the amount of solute that is dissolved in a solution. Concentrated solutions contain a lot of dissolved solute,
Prelab Assignment Before coming to lab: Lab 9. Colligative Properties an Online Lab Activity Chemistry 162 - K. Marr Revised Winter 2014 This lab exercise does not require a report in your lab notebook.
CHEMISTRY The Central Science Properties of Solutions The Solution Process Solutions: Air; brass; body fluids; sea water When a solution forms some questions we can ask are: What happens on a molecular
Colligative Properties of Solutions Freezing Point Depression Objectives: To investigate the colligative property of freezing point depression; to examine the relationship between freezing point depression
Solutions 1 Solutions Concentration Solution concentration = how much solute dissolved in solvent Coffee crystal = solute Water = solvent Liquid Coffee = solution so a solute is dissolved in solvent to
Purpose: Colligative Properties: Freezing Point Depression and Molecular Weight The first purpose of this lab is to experimentally determine the van't Hoff (i) factor for two different substances, sucrose
Colligative Properties Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable version of this book, as well as other interactive content, visit www.ck12.org
Chapter 13 Properties of Solutions Learning goals and key skills: Describe how enthalpy and entropy changes affect solution formation Describe the relationship between intermolecular forces and solubility,
Chapter Thirteen Physical Properties Of Solutions 1 Solvent: Solute: Solution: Solubility: Types of Solutions Larger portion of a solution Smaller portion of a solution A homogeneous mixture of 2 or more
John W. Moore Conrad L. Stanitski Peter C. Jurs Solubility & Intermolecular Forces Solution = homogeneous mixture of substances. It consists of: http://academic.cengage.com/chemistry/moore solvent - component
Colligative Properties Discussion Chem. 1A The material covered today is found in sections Chapter 12.5 12.7 This material will not be covered in lecture, you will have homework assigned. Chem. 1A Colligative
Chemistry 100 Bettelheim, Brown, Campbell & Farrell Ninth Edition Introduction to General, Organic and Biochemistry Chapter 6 Solutions & Colloids Solutions Components of a Solution Solvent: The substance
A.P. Chemistry Practice Test: Ch. 11, Solutions Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is
Chapter 13 Properties of Solutions 13.1 The Solution Process - Solutions are homogeneous mixtures of two or more pure substances. - In a solution, the solute is dispersed uniformly throughout the solvent.
16.3 Colligative of Solutions > 16.3 Colligative of Solutions > CHEMISTRY & YOU Chapter 16 Solutions 16.1 of Solutions 16.2 Concentrations of Solutions 16.3 Colligative of Solutions 16.4 Calculations Involving
Concentration of Solutions and Molarity The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. A dilute solution is one that contains a small
Name: Honors Unit 10 Notes Solutions [Chapter 10] Objectives: 1. Students will be able to calculate solution concentration using molarity, molality, and mass percent. 2. Students will be able to interpret
Guide to Chapter 11. Solutions and their properties We will spend three lecture days on this chapter. You may want to start by reviewing the concepts of heterogeneous solutions (Chapter 2) Read the introductory
Colligative Properties of Nonvolatile Solutes 01 Colligative Properties of Nonvolatile Solutes 02 Colligative Properties: Depend on the amount not on the identity There are four main colligative properties:
version: master Exam 1 - VDB/LaB/Spk This MC portion of the exam should have 19 questions. The point values are given with each question. Bubble in your answer choices on the bubblehseet provided. Your
DETERMINATION OF MOLAR MASS BY FREEZING-POINT DEPRESSION 141 Determination of Molar Mass by Freezing-Point Depression OBJECTIVES: Gain familiarity with colligative properties of nonelectrolyte solutions
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PRACTICE PROBLEMS FOR TEST 2 (March 11, 2009) 1) What is the overall order of the following reaction, given the rate law? A) 1st order B) 2nd order C) 3rd order D) 4th order E) 0th order 2NO(g) + H 2(g)
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1. SOLUTIONS Solutions to Exercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure
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Colligative properties CH102 General Chemistry, Spring 2014, Boston University here are four colligative properties. vapor-pressure lowering boiling-point elevation freezing-point depression osmotic pressure
CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show
Practice Exercises 1.1 C H S = k H P H S P H S = 1.0 atm C H S = 0.11 mol HS 4.08g HS L 1 mol H S =.7 g L 1.7 g L 1 H S = k H (1.0 atm H s S) k H =.7 g L 1 Hydrogen sulfide is more soluble in water than
Chapter 14 Solutes and Solvents A solution is a homogeneous mixture of two or more substances. The relative abundance of the substances in a solution determines which is the solute and which is the solvent.
1. Classify the following reaction as precipitation, acid-base or oxidation-reduction: Ce4+(aq) + Fe2+(aq) Ce3+(aq) + Fe3+(aq) Ans. oxidation-reduction 2. Classify the following reaction as precipitation,
Experiment 5: Molecular Weight Determination From Freezing Point Depression PURPOSE To become familiar with colligative properties and to use them to determine the molar mass of a substance APPARATUS AND
CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 13 The Properties of Mixtures: Solutions and Colloids Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction
Appendix C Molarity C.1 MOLARITY AND THE MOLE The molar mass is the mass of a mole of a pure substance while the molarity, M, is the number of moles of a pure substance contained in a liter of a solution.
Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer s Law Aqueous Solutions In Chemistry, many reactions take place in water. This is also true for Biological processes. Reactions that take
, the van t Hoff Factor, and Molar Mass Objectives To understand colligative properties. To find the freezing point depression of a solution. To determine the van't Hoff factor for acetic acid dissolved
Purpose To observe freezing point depression of different solutions and calculate the molar mass of commercial antifreeze. Introduction In winter, trucks dispense dirt and salt onto the roads and antifreeze
1 Chapter 7 Solutions Solutions: Solute and Solvent Solutions are homogeneous mixtures of two or more substances form when there is sufficient attraction between solute and solvent molecules have two components:
Lecture Presentation Chapter 12 Solutions Sherril Soman, Grand Valley State University Thirsty Seawater Drinking seawater can cause dehydration. Seawater Is a homogeneous mixture of salts with water Contains
Determination of Molar Mass by Boiling Point Elevation of Urea Solution CHRISTIAN E. MADU, PhD AND BASSAM ATTILI, PhD COLLIN COLLEGE CHEMISTRY DEPARTMENT Purpose of the Experiment Determine the boiling
Molarity Practice Worksheet Find the molarity of the following solutions: 1) 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution. 2) 0.5 grams of sodium chloride is dissolved to make
Colligative Properties Vapour pressure Boiling point Freezing point Osmotic pressure Learning objectives Describe meaning of colligative property Use Raoult s law to determine vapor pressure of solutions
Chapter 13 part 4: Colligative Properties Read: BLB 13.5-13.6 HW: BLB 13:9, 58, 61, 67, 69, 75 Packet 13:13-18 Know:, Colloids!vapor pressure lowering Raoult s Law: P A = X A P A!boiling point elevation
12A. A Molar Mass from Freezing-Point Depression Time: 2 hours Required chemicals and solutions: Reagent Requirement/5 Pairs Preparation of 1 L p C 6 H 4 Cl 2 2.0 g Cyclohexane 100 ml Other required materials:
Physical Chemistry Lecture 34 Colligative Properties of Solutions Colligative properties Solution properties that reflect the number of particles in solution Vapor-pressure lowering Freezing-point depression
Name: Class: Date: Chapter 13 & 14 Practice Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Acids generally release H 2 gas when they react with a.
Experiment 34 EVALUATION OF DEICER AND ANTIFREEZE PERFORMANCE FV 1/11 MATERIALS: PURPOSE: beakers: 400 ml; 150 ml; 100 ml (3); 50 ml (2); 10 ml and 25 ml graduated cylinders; thermometer; 25 x 200 mm test
Chapter 3 Stoichiometry of Formulas and Equations Chapter 3 Outline: Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing
Calculations involving concentrations, stoichiometry MUDr. Jan Pláteník, PhD Mole Unit of amount of substance the amount of substance containing as many particles (atoms, ions, molecules, etc.) as present
Solutions Properties of Solutions... the components of a mixture are uniformly intermingled (the mixture is homogeneous). Solution Composition 1. Molarity (M) = 4. Molality (m) = moles of solute liters
TYPES OF SOLUTIONS A solution is a homogeneous mixture of two substances: a solute and a solvent. Solute: substance being dissolved; present in lesser amount. Solvent: substance doing the dissolving; present
AP* Chemistry PROPERTIES OF SOLUTIONS IMPORTANT TERMS Solution a homogeneous mixture of two or more substances in a single phase. Does not have to involve liquids. Air is a solution of nitrogen, oxygen,
Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass
Osmolality Explained What is osmolality? Simply put, osmolality is a measurement of the total number of solutes in a liquid solution expressed in osmoles of solute particles per kilogram of solvent. When
Chapter 8 1 Chapter 8 Solutions Solutions to In-Chapter Problems 8.1 A heterogeneous miture does not have a uniform composition throughout a sample. A solution is a homogeneous miture that contains small
PRACTICING SKILLS Concentration Chapter 14 1. For 2.56 g of succinic acid in 500. ml of water: The molality of the solution: Molality = #mole solute/kg solvent: With a density of water of 1.00 g/cm3, 500.
CH302 LaBrake and Vanden Bout Colligative Properties PROBLEM #1: Give the molecular formula, the van t hoff factor for the following Ionic Compounds as well as guess the solubility of the compounds. If
Major chemistry laws. Mole and Avogadro s number. Calculating concentrations. Major chemistry laws Avogadro's Law Equal volumes of gases under identical temperature and pressure conditions will contain
Chemistry 9 SOLUTIONS Y ou know that when sugar or salt is added to water, it dissolves. The resulting mixture is called a solution. Solutions play an important role in our life. In industry, solutions