Properties of SolutionsChapter 17E-mail: benzene4president@gmail.comWeb-site: http://clas.sa.ucsb.edu/staff/terri/

Properties of Solutions – ch. 171. Predict the relative solubility of the following:

a. O2 in H20 vs. O2 in CCl4

b. CH3OH in H2O vs. CH3CH2CH2CH2OH in H2Oc. AgCl(s) in H2O at 25°C vs. AgCl(s) in H2O at 45°Cd. CO2(g) in C8H18 at 25°C vs. CO2(g) in C8H18 at 45°Ce. N2(g) in C6H6(g)at 1 atm vs. N2(g) in C6H6(g)at 8 atm

Properties of Solutions – ch. 17Solubility ⇒ the maximum amount of solute that can dissolve in

a given amount of solvent at any one T and P

Solubility Considerations1. Polarity of solute and solvent ⇒ “like dissolves like” 2. Temperature ⇒ Substances are more soluble at high T’s if

ΔSdiss > 0 (typical for solid solutes) and less soluble at high T’s if ΔSdiss < 0 (typical for gas solutes)

3. Pressure ⇒ gases are more soluble at high P’s (Le Chatlier’s principle)

Properties of Solutions – ch. 172. What is the molality of an aqueous solution that is 12.5% methanol

by weight (molar mass of CH3OH is 32.05 g/mol).

Molality ⇒ m =

Properties of Solutions – ch. 173. Calculate the molality and mole fraction of an aqueous solution that

is 8 M NaCl, the density of the solution is 1.18 g/ml.

Concentrations Variations

Molarity ⇒ M =

Molality ⇒ m =

Mole fraction ⇒ Xa =

Properties of Solutions – ch. 174. Calculate the heat of hydration for the following ionic solids.

a. KFb. RbFc. Compare the ion-dipole forces for K+ vs. Rb+ in H2O

Compound Lattice Energy Heat of Solution

KF -804 kJ/mol -15 kJ/mol

RbF -768 kJ/mol -24 kJ/mol

Properties of Solutions – ch. 175. A salt solution sits in an open beaker. Assuming constant temperature, the vapor pressure of the solution will…

a. increases over timeb. decreases over timec. stays the same over timed. We need to know which salt is in the solution to answer thise. We need to know the temperature and pressure to answer this

Properties of Solutions – ch. 17

Colligative properties ⇒ the properties of a solution with a non-volatile solute relative to a pure solvent

As the concentration of solute particles ↑vapor pressure ↓ “depression” ⇒ Pa = Xa Pa°

freezing point ↓ “depression” ⇒ ΔTf = - kf i m

boiling point ↑ “elevation” ⇒ ΔTb = kb i m

osmotic pressure ↑ ⇒ π = iMRT

i = vant Hoff factor =

Properties of Solutions – ch. 176. Rank the following aqueous solutions in order of increasing vapor

pressure:a. 0.1 M C6H12O6

b. 0.1 M KBrc. 0.05 M Na2SO4

d. 0.05 M CH3COOH

Properties of Solutions – ch. 177. Calculate the vapor pressure of a solution (in torr) made by

dissolving 159 g of ethylene glycol (HOCH2CH2OH – 62.08g/mol) in 500 g of water at 27 °C. At 27 °C the vapor pressure of pure water is 26.7 torr.

Properties of Solutions – ch. 17

8. Benzene (C6H6 – 78.12 g/mol) and toluene (C7H8 – 92.15 g/mol) form an ideal solution. What is the vapor pressure of a solution prepared by mixing 40 g of toluene with 15 g of benzene at 25 °C? At 25 °C the vapor pressures of pure toluene and pure benzene are 28 and 95 torr respectively. What is the mole fraction of the benzene in the vapor above the solution?

Properties of Solutions – ch. 179. Pentane and hexane form an ideal solution. What composition of a

pentane and hexane solution at 25 °C would give a vapor pressure of 350 torr? At 25 °C the vapor pressures of pure pentane and hexane are 511 torr and 150 torr respectively.

Properties of Solutions – ch. 1710. After substance A is mixed with substance B the solution feels hotter than before

they were mixed. What deviation from Raoult’s Law (if any) would be expected for this solution?a. no deviationb. positive deviation – A and B form stronger forces than pure A and Bc. positive deviation – A and B form weaker forces than pure A and Bd. negative deviation – A and B form stronger forces than pure A and Be. negative deviation – A and B form weaker forces than pure A and B

Properties of Solutions – ch. 17Ideal vs. Non-Ideal Solutionsa. ΔHsol’n ~ 0 Ideal solution – the actual vapor pressures will agree with ⇒Raoult’s lawb. ΔHsol’n > 0 Non-Ideal solution – the forces in the solution are ⇒ weaker than in the pure substances resulting in higher VPs than expected from Raoult’s Lawc. ΔHsol’n < 0 Non-Ideal solution – the forces in the solution are ⇒ stronger than in the pure substances resulting in lower VPs than expected from Raoult’s Law

Properties of Solutions – ch. 1711. Predict if the following solutions will be ideal or non-ideal. What type of deviation from Raoult’s Law would you expect for the non-ideal solutions?

a. carbon tetrachloride (CCl4) and dichloromethane (CH2Cl2)b. water and acetone (CH3COCH3)c. carbon tetrachloride and benzene (C6H6)

Properties of Solutions – ch. 1712. The vapor pressures of several solutions of water and butanol were

determined at various compositions and the data is given below:

a. Are the solutions of water and butanol ideal?b. Which of the above solutions would have the lowest boiling point?

XH2O VP (torr)

0.00 47.0

0.27 52.5

0.58 58.3

0.72 46.9

1.00 40.2

Properties of Solutions – ch. 1713. Rank the following aqueous solutions by their boiling points, and

freezing points.a. 0.1 M CH2Ob. 0.1 M LiFc. 0.05 M (NH4)2SO4

Properties of Solutions – ch. 17

Properties of Solutions – ch. 1714. Calculate the boiling point and freezing point of a solution made by

dissolving 110 g of K3PO4 (212.3g/mol) in 800 mL of water at 1 atm. For water kb = 0.51 °Ckg/mol and kf = 1.86 °Ckg/mol.

Properties of Solutions – ch. 1715. Calculate the osmotic pressure of a solution made by dissolving 83

g of glucose (C6H12O6) in 100 mL of water at 30 °C.

Properties of Solutions – ch. 17Osmosis the diffusion of water thru a semipermeable membrane⇒Osmotic pressure () the minimum pressure that must be applied to keep osmosis from ⇒occurring = iMRT as the concentration gradient ↑ ⇒ ⇒Reverse osmosis increasing the conc gradient by the movement of water thru a ⇒semipermeable membrane by applying a pressure that is > the osmotic pressure

Properties of Solutions – ch. 1716. A solution contains 3.75 g of a nonvolatile hydrocarbon in 95 g of

acetone. The boiling points of pure acetone and the solution are 55.9 °C and 56.5 °C respectively. What is the molar mass of the hydrocarbon? For acetone the Kb = 1.71 °CKg/mol.

Properties of Solutions – ch. 1717. A solution that contains 29 g of non-volatile/non-ionizing solute in

126 g of water has a vapor pressure of 723.4 torr at 100 °C. What is the molar mass of the solute?

Properties of Solutions – ch. 17 – Feldwinn18. A solid mixture contains MgCl2 (molar mass= 95.218 ) and 𝑔 𝑚𝑜𝑙

NaCl (molar mass = 58.443 ). When 0.500 g of this 𝑔 𝑚𝑜𝑙solid is dissolved in enough water to form 1.000 L of solution, the

osmotic pressure at 25.0˚C is observed to be 0.3950 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.)

Properties of Solutions – Answers1. Predict the relative solubility of the following:

a. O2 in H20 vs. O2 in CCl4

b. CH3OH in H2O vs. CH3CH2CH2CH2OH in H2Oc. AgCl(s) in H2O at 25°C vs. AgCl(s) in H2O at 45°Cd. CO2(g) in C8H18 at 25°C vs. CO2(g) in C8H18 at 45°Ce. N2(g) in C6H6(g)at 1 atm vs. N2(g) in C6H6(g)at 8 atm

2. What is the molality of an aqueous solution that is 12.5% methanol by weight (molar mass of CH3OH is 32.05 g/mol).

molality = 12.% if there’s 100 g of solution 12.5 g of⇒ CH3OH and 87.5 g of watermole of CH3OH = = 0.39 mol ⇒ m= = 4.46m

Properties of Solutions – Answers3. Calculate the molality and mole fraction of an aqueous solution that

is 8 M NaCl, the density of the solution is 1.18 g/ml. (molar mass of NaCl is 58.44 g/mol)

molality = and mole fraction = 8 M NaCl 8 moles of NaCl per 1 L of solution⇒(8 mol NaCl)(58.44 g/mol) = 467.5 g NaCl(1 L of sol’n)(1000mL/L)(1.18 g/L) = 1180 g of sol’n – 467.5g = 712.5 g of water = 39.5 mol watermolality = = 11.2 mmole fraction= = 0.168

Properties of Solutions – Answers4. Calculate the heat of hydration for the following ionic solids.

a. KFb. RbFc. Compare the ion-dipole forces for K+ vs. Rb+ in H2O

ΔHhydration = ΔHLE + ΔHsol’n

KF ⇒ ΔHhydration = (-804kJ/mol) + (-15kJ/mol) = -819kJ/molRbF ⇒ ΔHhydration = (-768kJ/mol) + (-24kJ/mol) = -792kJ/molSince both species have fluoride we can compare the ion-dipole forces of K+ vs. Rb+ the more negative the heat of hydration the stronger ⇒the force K⇒ + has the stronger ion-dipole force this makes sense ⇒since K+ is smaller and can form shorter/stronger bonds with water

Properties of Solutions – Answers5. A salt solution sits in an open beaker. Assuming constant temperature, the vapor pressure of the solution will…

b. decreases over timeas water evaporates away the concentration of the salt increases

causing the VP to decrease6. Rank the following aqueous solutions in order of increasing vapor pressure:

a. 0.1 M C6H12O6 i=1⇒b. 0.1 M KBr i=2⇒c. 0.05 M Na2SO4 i=3⇒d. 0.05 M CH3COOH i=1⇒

as (conc)x(i) ↑ VP ↓ b < c < a < d

Properties of Solutions – Answers7. Calculate the vapor pressure of a solution (in torr) made by

dissolving 159 g of ethylene glycol (HOCH2CH2OH – 62.08g/mol) in 500 g of water at 27 °C. At 27 °C the vapor pressure of pure water is 26.7 torr.

(159g)/(62.08g/mol) = 2.6 mol HOCH2CH2OH(500g)/(18.02g/mol) = 27.7 mol waterPwater = (27.7mol/30.3mol)(26.7torr) = 24.4torr

Relative FP or VP => b < c < a < dRelative BP or OP => b > c > a > d

Properties of Solutions – Answers8. Toluene and benzene form an ideal solution. What is the vapor

pressure of a solution prepared by mixing 40 g of toluene with 15 g of benzene at 25 °C? At 25 °C the vapor pressures of pure toluene and pure benzene are 28 and 95 torr respectively.(40g)/(92.15g/mol) = 0.43 mol toluene(15g)/(78.12g/mol) = 0.19 mol benzenePtoluene = (0.43 mol/0.62 mol)(28 torr) = 19.6 torrPbenzene = (0.19 mol/0.62 mol)(95 torr) = 29.1 torrPsolution = 48.7 torr

Properties of Solutions – Answers9. Pentane and hexane form an ideal solution. What composition of a

pentane and hexane solution at 25 °C would give a vapor pressure of 350 torr? At 25 °C the vapor pressures of pure pentane and hexane are 511 torr and 150 torr respectively.Ptotal = Ppentane + Phexane

Ptotal = XpPp° + XhPh°Since Xp + Xh = 1 => Xp = 1 – Xh Ptotal = (1 – Xh) Pp° + XhPh°350 torr = (1 – Xh)(511 torr) + Xh(150 torr)Xh = 0.45 => Xp = 0.55

Properties of Solutions – Answers

10. After substance A is mixed with substance B the solution feels hotter than before they were mixed. What deviation from Raoult’s Law (if any) would be expected for this solution?d. negative deviation – A and B form stronger forces than pure A and B

A hotter solution means that the heat of solution was exothermic – the heat given off is due to forming strong forces

Properties of Solutions – Answers11. Predict if the following solutions will be ideal or non-ideal. What type of deviation from Raoult’s Law would you expect for the non-ideal solutions?

a. carbon tetrachloride (CCl4) and dichloromethane (CH2Cl2)non-ideal with positive deviations - CCl4 has LDF whereas CH2Cl2 is polar and has dipole-dipole in addition to the LDF – upon mixing the CH2Cl2 will be unable to have as many dipole-dipole forces and therefore weakerb. water and acetone (CH3COCH3) non-ideal with negative deviations – water is polar and can H-Bond however acetone is polar but can’t H-Bond with itself but can H-Bond with water – therefore the forces in solution are strongerc. carbon tetrachloride and benzene (C6H6) ideal - both are non-polar and will have LDFs when pure and in solution

Properties of Solutions – Answers12. The vapor pressures of several solutions of water and butanol were

determined at various compositions and the data is given below:

a. are the solutions of water and butanol ideal? No – since the VP’s of some of the solutions are outside of the range of pure water and pure butanol – since they are outside on the high side the solution will have positive deviations

b. which of the above solutions would have the lowest boiling point? Highest VP will result in the lowest BP => XH2O = 0.58

XH2O VP (torr)

0.00 47.0

0.27 52.5

0.58 58.3

0.72 46.9

1.00 40.2

Properties of Solutions – Answers13. Rank the following aqueous solutions by their boiling points, and

freezing points.a. 0.1 M CH2Ob. 0.1 M LiFc. 0.05 M (NH4)2SO4

as (conc)x(i) ↑ FP ↓ and BP ↑Relative FP => b < c < a Relative BP => a < c < b

Properties of Solutions – Answers14. Calculate the boiling point and freezing point of a solution made by

dissolving 110 g of K3PO4 (212.3g/mol) in 800 mL of water at 1 atm. For water kb = 0.51 °Ckg/mol and kf = 1.86 °Ckg/mol.

(110 g K3PO4)/(212.3g/mol) = 0.52 mol K3PO4

m = 0.52 mol/0.8 kg = 0.65m ΔTf = -(0.65 mol/kg)(4)(1.86 °Ckg/mol) = -4.8°C ΔTb = (0.65 mol/kg)(4)(0.51°Ckg/mol) = 1.3°CSince pure water has a FP = 0°C and BP = 100°C then the solution will have a FP = -4.8°C and BP = 101.3°C

Properties of Solutions – Answers15. A solution contains 3.75 g of a nonvolatile hydrocarbon in 95 g of

acetone. The boiling points of pure acetone and the solution are 55.9 °C and 56.5 °C respectively. What is the molar mass of the hydrocarbon? For acetone the Kb = 1.71 °CKg/mol.

molar mass = g/mol => given grams and you can get moles fromΔTb = kb i m => mol = ΔTb kgsolvent/ kb i mol = (56.5 °C – 55.9 °C)(0.095kg)/(1.71°CKg/mol)(1)mol = 0.056molar mass = 3.75g/0.056mol = 66.8g/mol

Properties of Solutions – Answers16. Calculate the osmotic pressure of a solution made by dissolving 83

g of glucose (C6H12O6) in 100 mL of water at 30 °C.

π = iMRTπ = (1)((83g/180g/mol)/(01.L))(0.08206atmL/molK)(303K)π = 115 atm

Properties of Solutions – Answers17. A solution that contains 29 g of non-volatile/non-ionizing solute in

126 g of water has a vapor pressure of 723.4 torr at 100 °C. What is the molar mass of the solute?

molar mass = g/mol => given grams and you can get moles fromPwater = XwaterPwater° => nsolute = (nwPw°/Pw) – (nw) => nsolute = (126g/18g/mol)(760torr)/(723.4torr) – (126g/18g/mol) nsolute = 0.354 molesmolar mass = 29g/0.354mol = 82 g/mol

Properties of Solutions – Answers18. A solid mixture contains MgCl2 (molar mass= 95.218 ) and 𝑔 𝑚𝑜𝑙

NaCl (molar mass = 58.443 ). When 0.500 g of this solid 𝑔 𝑚𝑜𝑙is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25.0˚C is observed to be 0.3950 atm. What is the mass percent of MgCl2 in the solid?

If you have X g of MgCl2 then there’s 0.50 g – X g of NaClmoles of MgCl2 = X/95.218 = 0.0105Xmoles of NaCl = (0.5 – X)/58.443 = 0.00855 – 0.0171Xπ = iMRT = (iM (for MgCl2)+iM (for NaCl))RT 0.3950 = (3(0.0105X/1) + 2((0.00855 – 0.0171X)/1)(0.08206)(298)X = 0.352 g% MgCl2 = (0.352g/0.5g)(100) = 70.4%