Chem 1C – Chapter 17 – CLAS
1. Predict the relative solubility of the following:
a. O2 in H20 vs.O2 in CCl4
b. CH3OH in H2Ovs. CH3CH2CH2CH2OH in H2O
c. AgCl(s) in H2O at 25°C vs.AgCl(s) in H2O at 45°C
d. CO2(g) in C8H18 at 25°Cvs. CO2(g) in C8H18 at 45°C
e. N2(g) in C6H6(g)at 1 atmvs.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).
If you have 100 g of solution there will be 12.5 g of methanol and 87.5 g of water ⇒ molality =
12.5 g CH3OHx = 0.272 mole CH3OH ⇒ molality = = 3.1mol/kg
3. 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. If you have 1 L of solution there are 8 mol of NaCl⇒ since the density of the solution is 1.18 g/mL then 1L or 1000 mL of solution weighs 1180 g ⇒ some of this is water and some is NaCl⇒
8 mol of NaCl x x = 467.5 g NaCl ⇒ so 1180g – 467.5 g = 721.5 g of water ⇒
molality= = 11.1mol/kg
mol fraction of NaCl = ⇒ 721.5 g water= 40.1 moles of water
mol fraction of NaCl = = 0.17
4. Calculate the heat of hydration for the following ionic solids.
Heat of hydration = Lattice energy + Heat of solution
a. KF
Heat of hydration for KF =( -804 kJ/mol) + (-15 kJ/mol) = -815 kJ/mol
b. RbF
Heat of hydration for RbF=( -786 kJ/mol) + (-24 kJ/mol) = -810 kJ/mol
c. Compare the ion-dipole forces for K+vs.Rb+ in H2O
The more negative the heat of hydration the stronger the ion-dipole forces – K+ has the stronger forces – this should also make sense since K+ is a smaller ion and therefore can form shorter/stronger bonds
5. A salt solution sits in an open beaker. Assuming constant temperature, the vapor pressure of the solution will…
a. increases over time
b. decreases over time
c. stays the same over time
d. We need to know which salt is in the solution to answer this
e. We need to know the temperature and pressure to answer this
6. Rank the following aqueous solutions in order of increasing vapor pressure:
a. 0.1 M C6H12O6
b. 0.1 M KBr
c. 0.05 M Na2SO4
d. 0.05 M CH3COOH
7. 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.
For non-volatile soluted⇒Psol’n = XsolventP°solvent
159 g HOCH2CH2OH x = 2.56 mole HOCH2CH2OH and 500 g H2O x = 27.8 mol H2O
XH2O = ⇒ XH2O = = 0.916
Psol’n = (0.916)(26.7 torr) = 24.4 torr
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? For solutions with 2 volatile species ⇒Psol’n = XaPa°+ XbPb°
40 g toluene x = 0.434 mole toluene
15 g benzene x = 0.192 mole benzene
Xtoluene = = 0.693
Xbenzene = = 0.307
Psol’n = (0.693)(28 torr) + (0.307)(95 torr) = 48.6 torr
mole fraction of a gas or vapor = = 0.6
9. 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.For solutions with 2 volatile species ⇒Psol’n = XaPa°+ XbPb°
sinceXhexane + Xpentane = 1 ⇒Psol’n = XhexanePhexane°+ (1- Xhexane )Ppentane°
350 torr = (Xhexane)(150 torr) + (1- Xhexane )(511 torr) ⇒ Xhexane = 0.446 ⇒ Xpentane = 0.554
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?
a. no deviation
b. positive deviation – A and B form stronger forces than pure A and B
c. positive deviation – A and B form weaker forces than pure A and B
d. negative deviation – A and B form stronger forces than pure A and B
e. negative deviation – A and B form weaker forces than pure A and B
11. 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 polar and non-polar give positive deviations
b. waterand acetone (CH3COCH3)non-ideal polar H-bonding with polar with O give negative deviations
c. carbon tetrachloride and benzene (C6H6) non-polar and non-polar are ideal
12. 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 – data is illustrating positive deviations
b. Which of the above solutions would have the lowest boiling point?Highest VP is lowest BP
13. Rank the following aqueous solutions by their boiling points, and freezing points.
a. 0.1 M CH2Oas concentration x i ↑ FP ↓ BP ↑
b. 0.1 M LiFFP ⇒ b < c < a
c. 0.05 M (NH4)2SO4BP ⇒ a < c < b
14. 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.ΔTf = – kfim andΔTb = kb im
110 g K3PO4 x = 0.518 mole K3PO4 ⇒ m = = 0.648 mole/kg
ΔTf = –(1.86 °Ckg/mol)(4)( 0.648 mole/kg) = 4.8 °C ⇒ since the FP of pure water is 0 °C the FP of this solution is – 4 °C
ΔTb = kb im = (0.51 °Ckg/mol)(4)(0.648 mol/kg) = 1.32 °C ⇒ since the BP of pure water is 100 °C the BP of this solution is 101.32 °C
15. 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 from
ΔTb = kb i m => mol =
mol =
mol = 0.033
molar mass = 3.75g/0.056mol = 66.8g/mol
16. Calculate the osmotic pressure of a solution made by dissolving 83 g of glucose (C6H12O6) in 100 mLof water at 30 °C. π = iMRT
π = ( ) (0.08206atmL/molK)(303K)
π = 115 atm
17. 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 from Pwater = XwaterPwater° => nsolute = (nwPw°/Pw) – (nw) =>
nsolute = (126g/18g/mol)(760torr)/(723.4torr) – (126g/18g/mol)
nsolute = 0.354 moles
molar mass = 29g/0.354mol = 82 g/mol
18. A solid mixture contains MgCl2 (molar mass= 95.218 ????) and NaCl (molar mass = 58.443 ????). When 0.500 g of this solid 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.)
If you have X g of MgCl2 then there’s 0.50 g – X g of NaCl
moles of MgCl2 = X/95.218 = 0.0105X
moles of NaCl = (0.5 – X)/58.443 = 0.00855 – 0.0171X
π = 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%