1. Properties of Solutions

2.  Molarity = mol/L  Mass percent = mass of solute/mass of solution x 100  Mole fraction of component A = X A = n A n A + n B Molality = moles of solute ÷ kg of water

3.  A solution is prepared by mixing 1.00 g ethanol (C 2 H 5 OH) with 100.0 g water to give a final volume of 101.0 mL. Calculate the molarity, mass percent, mole fraction and molality of this solution.  0.215M  0.990%  0.00389  0.217m

4.  Normality is defined as number of equivalents per liter of solution. For acid-base reaction, the equivalent is the mass of acid or base that can furnish or accept exactly 1 mole of protons (H + ) or OH -. Acid or baseMolar massEquivalent mass M vs N HCl36.5 1M=1N H 2 SO 4 9898/2 = 491M=2N NaOH40 1M=1N Ca(OH) 2 7474/2 = 37 1M=2N

5.  For oxidation-reduction reactions, normality is defined as the equivalent of oxidizing or reducing agent that can accept or furnish 1 mole of electrons. Thus 1 equivalent of reducing agent will react with exactly 1 equivalent of oxidizing agent. The equivalent mass of an oxidizing or reducing agent can be calculated from the number of elecrons in its half-reactions.  MnO 4 - + 5e - + 8H +  Mn 2+ + 4H 2 O  Since MnO 4 - ion present in 1 mole of KMnO 4 consumes 5 moles of electrons, the equivalent mass is the molar mass divided by 5. 158g/5 = 31.6 g

6.  The electrolyte in automobile lead storage batteries is a 3.75M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality and normality of the solution.  1.230 g/mL = mass/1000 mL = 1.230 x 10 3 g  3.75 mol x 98g/mol = 368 g  1230. g solution – 368 g H 2 SO 4 = 862 g H 2 O  368 g H 2 SO 4 ÷ 1230. g solution x 100 = 29.9% H 2 SO 4  3.75 mol H 2 SO 4 ÷ 862 g H 2 O x 1 kg/1000g = 4.35 m  H 2 SO 4 has 2 equivalents. So 2 x 3.75 = 7.50 N

7.  Steps to forming a solution (like dissolves like):  Solute separates into individual components.  Intermolecular forces in the solvent are overcome to make room for the solute.  The solute and solvent interact to form a solution.  Enthalpy (heat) of solution:  Δ H soln = Δ H 1 + Δ H 2 + Δ H 3  Where Δ H soln can be + or – depending on whether its exothermic or endothermic

8.  Consider the following:  NaCl (s)  Na + (g) + Cl - (g) Δ H 1 = 786 kJ/mol  H 2 O (l) + Na + (g) + Cl - (g)  Na + aq) + Cl - (aq)  Δ H hyd = Δ H 2 + Δ H 3 = -783 kJ/mol  First step Δ H 1 is large to break ionic bonds.  Δ H 2 is large because hydrogen bonds must be broken to make room for the solute.  Δ H 3 is large and negative to account for the attractions between ions and water molecules.  The heat of solution = 786 kJ/mol – 783 kJ/mol = 3 kJ/mol

9.  The energy it took to dissolve sodium chloride was little and it turns out that processes that require large amounts of energy tend not to occur.  Decide whether liquid hexane (C 6 H 14 ) or liquid methanol (CH 3 OH) is the more appropriate solvent for the substances grease (C 20 H 42 ) and potassium iodide (KI). ΔH1ΔH1 ΔH2ΔH2 ΔH3ΔH3 Δ H soln Outcome Polar solute, polar solventLarge Large, NegativeSmall Solution Forms Nonpolar solute, Polar solventSmallLargeSmall Large, positive No solution forms Nonpolar solute, Nonpolar solventSmall Solution Forms Polar solute, Nonpolar solventLargeSmall Large, positive No solution forms

10.  Pressure has little effect on solids or liquids but it does have significant effect on the solubility of gases in liquids. Carbonated beverages are bottled under high pressure conditions.  As the amount of pressure is increased, the number of gas molecules per unit volume increases, and the gas enters the solution at a higher rate than it leaves. As the concentration of dissolved gas increases, the rate of the escape of the gas also increases until a new equilibrium is reached where the solution contains more dissolved gas than before.

11.  C is concentration of dissolved gas  k is a constant characteristic of a particular solution.  P is partial pressure of the gaseous solute above the solution.  A certain soft drink is bottled so that a bottle at 25 o C contains CO 2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO 2 in the atmosphere is 4.0 x 10 -4 atm, calculate the equilibrium concentrations of CO 2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO 2 in aqueous solution is 3.1 x 10 -2 mol/L· atm at 25 o C.

12.  C = (3.1 x 10 -2 mol/L atm)(5.0 atm) =.16 mol/L  C = (3.1 x 10 -2 mol/L atm)(4.0 x 10 -4 atm) = 1.2 x 10 -5 mol/L  Notice how much smaller the concentration of CO 2 is in the second scenario, this explains why sodas go “flat”.  Solutions Crash Course (8:05) Solutions Crash Course (8:05)

13.  Most solids dissolve more readily in higher temperatures while gases do the opposite.  Thermal pollution is created from using bodies of water to cool down industrial processes such as nuclear power plants. The increased temperatures decrease the amount of dissolved oxygen and tend to form a less dense layer of water that “floats” on the colder water which inhibits the normal absorption of oxygen.  This is particularly damaging to in deep lakes.

14.  Aqueous solutions and pure solvents have different vapor pressures. The diagram on the left shows that eventually the water will be absorbed by the other solution to achieve equilibrium because the initial pressure of the pure solvent (water) is greater than the aqueous solution.  Thus as the water emits vapor to attempt to reach equilibrium, the aqueous solution absorbs vapor to try to lower the vapor pressure towards its equilibrium value.  Equilibrium vapor pressure can only be reached when all the water is absorbed into the aqueous solution.

15.  The presence of a nonvolatile solute lowers the vapor pressure of a solvent.  P soln is the observed vapor pressure of the solution.  X solvent is the mole fraction of the solvent.  P 0 solvent is the vapor pressure of the pure solvent. The presence of a nonvolatile solute inhibits the escape of solvent molecules from the liquid and so lowers the vapor pressure of the solvent.

16.  Raoult’s law is a linear equation y=mx+b, where  y = P soln  x = x solvent  m = P 0 solvent  b = 0  Calculate the expected vapor pressure at 25 o C for a solution prepared by dissolving 158.0 g of sucrose, (342.30 g/mol) in 643.5 cm 3 of water. At 25 o C the density of water is 0.997 g/cm 3 and the vapor pressure is 23.76 torr.

17.  P soln = (X H20 )(P H20 )  X = mol H20 ÷ (mol H20 + mol sucrose )  Find moles of water.  Find moles of sucrose.  158g sucrose x 1 mol/342.3 =.4616 mol sucrose  643.5 cm 3 H 2 O x.997g/cm 3 = 641.6 g H 2 O  641.6 g H 2 O x 1 mol/18.02 g H 2 O = 35.60 mol H 2 O  X = 35.63 mol H 2 O ÷ (35.63 mol +.4616 mol) =.9873  P soln = (.9873)(23.76 torr) = 23.46 torr  Was the pressure less with the addition of sugar?

18.  When calculating the vapor pressure for solutions with ionic solutes, you will find that adding 1 mole of NaCl will lower the vapor pressure twice as much as a non-ionic solute. Why?  Predict the vapor pressure of a solution prepared by mixing 35.0 g of solid Na 2 SO 4 (142 g/mol) with 175 g water at 25 o C. The vapor pressure of pure water at 25 o C is 23.76 torr.

19.  n H20 = 175 g H 2 O x 1 mol H 2 O/18.02 g H 2 O = 9.72 mol H 2 O  n Na2SO4 = 35.0 gNa 2 SO 4 x 1 mol Na 2 SO 4 /142. gNa 2 SO 4 = 0.246 mol Na 2 SO 4  n solute = 3 (0.246 mol) = 0.738 mol Na 2 SO 4  X H2O = 9.72 mol H 2 O ÷ (9.72 mol H 2 O + 0.738 mol Na 2 SO 4 ) =  0.929  P soln = (0.929)(23.76 torr) = 22.1 torr

20.  If the solute is volatile (contributes to the vapor pressure), Raoult’s law is modified to:  P total = X A P 0 A + X B P 0 B  When there is a strong attraction between solute and solvent, the reaction is exothermic and there will be a negative deviation from Raoult’s law. Ex: polar acetone and polar water.

21.  If the two liquids mix endothermically, it indicates that the solute-solvent interactions are weaker than the interactions among the molecules in the pure liquids and the deviation from Raoult’s law will be positive.  Ex: hexane and ethanol

22.  When the solution is made up of very similar liquids, the enthalpy of solution is very close to zero, and thus the solution closely obeys Raoult’s law (ideal behavior).  Ex: benzene and toluene

23.  A solution is prepared by mixing 5.81 g acetone (C 3 H 6 O, molar mass 58.1 g/mol) and 11.9 g chloroform (HCCl 3 molar mass 119.4 g/mol) At 35 o C, this solution has a total vapor pressure of 260. torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35 o C are 345 and 293 torr respectively.  To see if this solution behaves ideally, calculate the expected vapor pressure using Raoult’s law: P total = X A P 0 A + X B P 0 B

24.  X acetone = 5.81 g acetone x 1 mol/58.1g =.100 mol acetone  X chloroform = 11.9 g chloroform x 1mol/119.4 g =.100 mol acetone  Since the solution contains equal numbers of moles of acetone and chloroform, we will find the average pressures. (.100 mol ÷ (.100 mol +.100 mol) =.500mol  X acetone = 0.500molX chloroform = 0.500 mol  P total (0.500 mol)(345 torr) + (0.500 mol)(293 torr) = 319 torr  The expected value is higher than the observed pressure of 260 torr. This shows that the solution does not behave ideally. The strong attraction between the polar molecules lowers the tendency of these molecules to escape from solution.

25.  By adding a nonvolatile solute to a solvent, the freezing point of the solution will be lowered (depressed) and the boiling point will be elevated.  The change in boiling point can be represented by the equation: Δ T = K b m solute  Where Δ T is the boiling point elevation, or difference between the boiling point of the solution and that of the pure solvent.  K b is a constant that is characteristic of the solvent- called the molal boiling-point elevation constant.  m solute is the molality of the solute in the solution.

26.  A solution was prepared by dissolving 18.00 g glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34 o C. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution.  K b for water is 0.51 o C · kg/mol  Boiling point of water is 100.00 o C so:  Δ T = 100.34 o C – 100.00 o C = 0.34 o C  m solute = 0.34 o C ÷ 0.51 o C · kg/mol = 0.67 kg/mol  The solution was prepared using 0.1500 kg water  m solute = 0.67 mol/kg = mol solute/kg solvent = n glucose /.1500 kg  n glucose = (0.67 mol/kg)(.1500 kg)= 0.10 mol glucose  0.10 mol glucose has a mass of 18.00 g and 1 mol = 180.00 g

27.  What mass of ethylene glycol (C 2 H 6 O 2 molar mass = 62.1 g/mol) the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at -10.0 o F (-23.3 o C)? Assume the density of water is exactly 1.00 g/mL. K f for water is 1.86 o C · kg/mol  Δ T = 0 o C – (-23.3 o C) = 23.3 o C  m solute = 23.3 o C ÷ 1.86 o C · kg/mol = 12.5 mol/kg  (12.5 mol/kg)(10.0 kg) = 1.25 x 10 2 mol  (1.25 x 10 2 mol)(62.1 g/mol) = 7.76 x 10 3 g

28.  A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 0.546g was dissolved in 15.0 g benzene, and the freezing point depression was determined to be 0.240 o C. Calculate the molar mass of the hormone.  K f for benzene is 5.12 o C · kg/mol  m hormone =.240 o C ÷ 5.12 o C · kg/mol =.0469 mol/kg  (.0469 mol/kg)(.0150 kg) = 7.04 x 10 -4 mol  0.546 g ÷ 7.04 x 10 -4 mol = 776 g/mol

29.  A solution a pure solvent separated by a semipermeable membrane will only let the solvent through.  Eventually, the solution will have a higher volume than the solvent—this is called osmosis.  There is greater hydrostatic pressure on the solution than the on the pure solvent. This excess pressure is called osmotic pressure.  To calculate osmotic pressure: π = MRT  Where π = osmotic pressure  M = molarity of the solution  R = gas law constant  T = Kelvin temperature

30.  To determine the molar mass of a certain protein, 1.00 x 10 -3 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 o C. Calculate the molar mass of the protein.  1.12 torr x 1atm/760 torr = 1.47 x 10 -3 atm  M = 1.47 x 10 -3 atm ÷ (0.08206 L · atm/K· mol)(298K) = 6.01 x 10 -5 mol/L  (1L/6.01 x 10 -5 mol)(1.00 x 10 -3 g/.001L) = 1.66 x 10 4 g/mol

31.  What concentration of sodium chloride in water is needed to produce an aqueous solution isotonic with blood ( π = 7.70 atm at 25 o C) Π = MRT  M = 7.70 atm ÷ (0.08206 L · atm/K· mol)(298K) = 0.315 mol/L  This represents the total molarity of solute particles but remember that NaCl ionizes into Na + and Cl - so the concentration needed of solute is 0.315 M ÷ 2 = 0.158 M.

32.  If a solution in contact with pure solvent across a semipermeable membrane is subjected to an external pressure larger than its osmotic pressure, reverse osmosis occurs.  The pressure will cause a net flow of solvent from the solution to the solvent. In reverse osmosis, the semipermeable membrane acts as a molecular filter to remove solute particles.  This is used in desalination processes.

33.  Colligative properties of solutions depend on number of particles. If a 0.1m solution of glucose has a K f of 0.186 o C, then 0.1m NaCl would have a K f twice that because it breaks up into 2 ions.  The relationship between moles of solute dissolved and moles of particles in solution is called the van’t Hoff factor, i.  i = moles of particles in solution moles of solute dissolved So: π = i MRT

34.  The observed osmotic pressure for a 0.10 M solution of Fe(NH 4 ) 2 (SO 4 ) 2 at 25 o C is 10.8 atm. Compare the expected and experimental values for i.  The ionic solid dissociates in water to produce 5 ions so the expected value for i is 5.  Calculate i using the above equation.  i = π ÷ MRT  i = 10.8 atm ÷ (0.10M x 0.08206 L · atm/mol · K x 298K) = 4.4  A possible explanation for the difference could be ion pairing.

35.  Colloids are solutions that have suspended particles that will scatter light called the Tyndall effect. This is the difference between a colloid and a suspension.  Colloidal particles have a center particle surrounded by a layer of positive ions and then negative ions. They repel one another in a solution and thus stay suspended rather than settling out.