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Lecture

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Department
Chemistry
Course Code
CHM135H1
Professor
Kris Quinlan

Page:
of 2
LECTURE 6 – Solutions
Slide No. Notes
1-dilute – not much solute
-concentrated – lots of solute
-saturated – maximum amount of solute dissolved
-unsaturated – less than maximum amount of solute dissolved
-supersaturated – produced by heating the solution, dissolving more
solute, then letting it cool
2- molality is independent of temperature
3- 1.45g (mol/342.3g) / (30.0ml)(1.00g/mol)(kg/1000g) = 0.141 mol/kg
4- miscible = can be mixed together in any proportions
5-network solids – ex. diamonds – insoluble in all solvents
-metals – insoluble in liquid solvents; can dissolve in other metals
-ex. brass = zinc and copper
-ionic solids – will dissolve if solvent has dipole-ion interactions
strong enough
-molecular solids – will dissolve in solvent with same type of
intermolecular forces
-ex. glucose can H-bond with water
6-Ans1: C
-Ans2: B – depends on substance
-Ans3 – gas solubility decreases
7-pushing down forces gas molecules into liquid phase
-higher rate of gas dissolving means more liquid than changing into
gas phase
-rate dissolving = rate into gas phase
8- Torr = mmHg
9- (a) to (b), vapour pressure is lower
12 -P of solution = X of water times P of water
-100.0mL of water = (1.00g/mL)(mol/18g) = 5.56 mol water
-100.0mL of ethylene glycol = (1.15g/mL)(mol/62.0g) = 1.86 mol
ethylene glycol
-X of water = 5.56 mol / (5.56 mol + 1.86 mol) = 0.749
-P of solution = (0.749)(525.8mmHg) = 394 mmHg
13 -higher vapour pressure if separated – B because less concentrated so
more of B evaporates
-Ans: A
-Because in the closed box there is no vapour pressure until A and B
starts evaporating
-Both will be happy if they have equal concentrations – thus reaching
equilibrium
14 - adding more sucrose increases BP
15 -Tb = kb times m = (2.53 degrees C kg/ mol)(0.10 mol/kg) = 0.253
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degrees C – this is the increment by which BP increased
-Tb = 80.10 degrees C + 0.253 degrees C = 80.35 degrees C
-Ans: B
-A -> Na+ and Cl- therefore 0.200M ions in total
-B -> Ca2+ and 2Cl- therefore 0.300M ions in total
16 - FP depression is decreasing therefore negative sign
17 -Tf = -kf times m
-(2.70 – 6.50 degrees C) = -20.2 degrees C/mol times m
-m = 0.188 mol/kg
-0.188 mol/kg (75.0 g)(kg/1000g) = 0.0141mol
-1.50 g/ 0.0141mol = 106 g/mol
18 - solvent moves from lower concentration to higher concentration
19 -purpose = solute
-M represents total molarity of solute
20 - = MRT = 1.0 mol solute / kg of solvent
-***Be careful that the questions gives molality, but wants molarity
-molarity is mols of solute over L of solution
-assume 1 kg of solvent
-1.0 mol sucrose (342.3g/mol) = 342.3 g
-solution: 1000g + 342.3 g = 1342.3 g solution
-1342.3 g (mL / 1.1g) = 1220.3 mL solution
- = 1.0mol/1220.3mL (1000mL/L)(0.08206Latm/Kmol)(298K) = 20
atm
-Pressure on the right (sucrose) side is greater by 20 atm than the
pressure exerted on the left side
-No net movement of solvent
22 -Ans1 – A
-Intermolecular forces – boiling – molecules uncharged
-A has dipole-dipole; B and C both have dispersion forces
-Ans2 – B
-***note that all A to D are in water
-A won’t dissociate
-B will dissociate – 3mol
-C will not dissociate – 1 mol
-D – 2 mol
-A or C which has the higher BP? Same
-In order of increasing BP: B, D, (A and C)
NOTE -amount of solute – need to look at what’s in solution
-FP, BP, vapour pressure – first see if pure substance or solution?
-Pure substance – intermolecular forces – overcoming these forces
dispersion, dipole-dipole, H-bonding – strongest is H-bonding
-Solution – concentration of solute (what is in the solution)
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