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---

Notes by Aaron Hui based on Professor Mack’s lectures
...
27
...
7 psi
○ Standard atmospheric pressure (1 atm)
■ The pressure that support a column of mercury exactly 760 mm
high at 0℃ at sea level
...

■ Constant amount of gas
...

○ Ex: A sample of chlorine gas occupies a volume of 946 mL at a pressure
of 726 mmHg
...
46 * 103 mmHg

● Charle’s Law
○ V α T (K)
■ Fixed amount of gas
■ Constant pressure
○ T↑V↑
○

V1
T1

=

V2
T2

○ Temperature MUST be in Kelvin
■ T(K)= t(℃) +273
...
20 L at 125℃
...
54 L if the pressure
remains constant
...
15)K = 398 K

●

3
...
54 L
T2

3
...
54 L)
T 2 = 192 K
T 2 = (192 − 273
...
0℃
● Another form of Charles’ Law is Gay Lussac’s Law
○ P α T (K) - constant n, V
○

P1
T1

=

P2
T2

● Avogadro's Law - Relationship between Volume and Amount
○ V α number of moles (n) - constant T and P
○

V1
n1

=

V2
n2

○ Equal volumes of any 2 gases contain equal # of atoms/molecules and
equal # of moles
● Ideal Gases
○ Molecules do not interact with one another
...

○ Ideal gases occur at low pressures and high temperatures
...
414 L - ​standard molar
volume​
...
08206 mol
*K

3

● Ex: An aerosol spray deodorant can with a volume of 350
...
2 g of
propane gas (C​3​H​8​) as propellant
...
mL *

L
1000 mL

n = 3
...
350 L

mol C 3 H 8
44
...
0726 mol

T = (20
...
15)K = 293 K
nRT
V

P =

=

L*atm )(293 K)
(0
...
08206 mol
*K
0
...
0 atm

● Using Standard Molar Volume in Calculations
○ 22
...
40 gof NH​3​ ​at STP?
■

7
...
03 g N H 3

*

22
...
74 L N H 3

○ What volume (in Liters) is occupied by 49
...
8 g HCl *

mol HCl
36
...
414 HCl
mol HCl

= 30
...

○ Ex: A gas initially at 4
...
2 atm, and 66℃ undergoes a change so that
its final volume and temperature are 1
...
What is the final
pressure? Assume number of moles remains constant
...
15)K = 339 K
T 2 = (42 + 273
...
2 atm)(4
...
7 L)
315 K

P 2 = 2
...
A certain lightbulb containing argon at 1
...
What is the final pressure of argon in
the lightbulb (in atm)?
P 1V 1
n1 T 1

■

=

P 2V 2
n2 T 2
P1
T1

●

=

P2
T2

T 1 = (18 + 273
...
15)K = 359 K
1
...
48 atm
Lecture 5
...
17
● Density (d) Calculations
○

d=

m
v

=

PM
RT

m- mass , M- molar mass

○ Expressed in g/L not g/mL
● Molar mass of a gaseous substance
○ Can determine molar mass if formula is unknown
○

M=

dRT
P

d- density (g/L)

● What is the density (in g/L) of uranium hexafluoride (UF​6​) at 779 mmHg and
62℃?
○

D=

m
V

P V = nRT

T = (63 + 273
...
025 atm

Assume 1
...
00 mol)*(0
...
025 atm

mm (molar mass) U F 6 = 352
...
0 g
26
...
8 L

g
mol

= 13
...
10 L vessel contains 4
...
00 atm and 27
...
What is the
molar mass of the gas?
○

molar mass =

g
mol

P V = nRT
n=

PV
RT

=

T = (27
...
15)K = 300
...
00 atm)(2
...
15 K)
(0
...
65 g
0
...
0853 mol

= 54
...
414 L standard molar volume
...
She measures the
volume as 1
...
54 g at STP
...
15 L gas *

mol gas
22
...
0513 mol

1
...
0513 mol

= 30
...

44
...
414 L

= 1
...
14% B and 21
...
At 27℃, 74
...
12 atm
...
0934 g, what is the
molecular formula?
○

78
...
81 g B

21
...
008 g H

= 7
...
69 mol H

B 7
...
69 = B H 3
7
...
228

P V = nRT
n=

PV
RT

=

T = (27 + 273
...
15 K

(1
...
0743 L)
L*atm )(300
...
08206 mol
*K

= 0
...
6 g
13
...
0934 g
0
...
6 g/mol

=2

= (BH 3 )2
= B2H 6
● What is the volume of CO​2​ produced at 37
...
00 atm when 5
...
60 g C 6 H 12 O6 *

mol C 6 H 12 O6
180
...
186 mol CO2

T = (37
...
15)K = 310
...
15 K)
(0
...
08206 mol
*K
1
...
73 L CO2

● If 34
...
mmHg, how many grams of NaN​3
were used in the reaction:
2N aN 3 (s) → 2N a (s) + 3N 2 (g)
L N 2 ⇒ (P V = nRT ) ⇒ mol N 2 ⇒ mol N aN 3 ⇒ g N aN 3
n=

PV
RT

=

(1
...
6 L)
L*atm )(75+273
...
0826 mol
*K

1
...
21 mol N 2

65
...
4 g N aN 3

● Review: At STP (0℃ and 1 atm), 1 mole of any ideal gas occupies 22
...

○ What is the volume of N​2​ produced at STP when 52
...
3 g N aN 3 ⇒ mol N aN 3 ⇒ mol N 2 ⇒ L N 2
mol N aN 3
65
...
3 g N aN 3 *

*

3 mol N 2
2 mol N aN 3

*

22
...
0 L N 2

● At any constant T and P, coefficients relate the volume as well as mol (only for
gases)
○ Calculate the volume of O​2​ in L for the complete combustion of 14
...

2C 4 H 10 (g) + 13O2 (g) → 8CO2 (g) + 10H 2 O (g)
14
...
9 L O2

○ Calculate the liters of CO​2​ produced during the complete combustion of
27
...

2C 8 H 18 (g) + 25O2 (g) → 16CO2 (g) + 18H 2 O (g)
27
...
14 L sample of HCl gas at 2
...
Calculate the molarity of the acid
solution assuming no change in volume
...
61 atm)(2
...
15 K)
(0
...
226 mol HCl
0
...
15)K = 301
...
226 mol HCl

= 0
...
4 x 10​5​ L and a
pressure of 7
...
A solution of LiOH with negligible volume
was introduced
...
2 x 10​-3​ atm
...
9*10

62
...
2*10 atm)(2
...
08206 mol
*K

mol Li2 CO3
mol CO2

*

73
...
8 mol CO2
= 4
...
0 L of C​4​H​10​ and 25
...
25 atm?
2C 4 H 10 (g) + 13O2 (g) → 8CO2 (g) + 10H 2 O (g)
○ Method 1:
5
...
5 L O2 needed

O2 is the limiting reagent because need 32
...
0 L O2 *

8 L CO2
13 L O2

= 15
...
0 L C 4 H 10 *
25
...
L CO2

= 15
...
5 g of C​4​H​10​ and 14
...
5 g C 4 H 10 *
14
...
12 g C 4 H

mol O2
32
...
172 mol CO2

= 0
...
172 mol CO2 *

22
...
9 L CO2

● Mixtures of Gases
○ Each particle of a gas acts independently
9

○ Partial pressure​ - the pressure exerted by a gas in a mixture
○ Air

P​T​ = 1
...
78(1
...
78 atm
P O2 = 0
...
00 atm) = 0
...
01(1
...
01 atm
P total = P N 2 + P O2 + P other = 1
...
How much oxygen (in g) is produced?
2M gO (s) → 2M g (s) + O2 (g)
P O2 = P T − P H 2 O
P O2 = (757 − 23
...
964 atm
T = (25 + 273
...
964 atm)(0
...
08206 mol
*K

1
...
00 g O2
mol O2

= 1
...
045 g O2

Lecture 5
...
17
● Mixture of A and B
○

PA =

nA RT
V

PB =

nB RT
V

10

(nA +nB )RT
V
n
( X i) = n i
T

PT = PA + PB =
○ Mole fraction

XA =

nA
nA +nB

XA + XB = 1

P i = X iP T
○ A sample of natural gas contains 8
...
421 moles of C​2​H​6
and 0
...
If the total pressure of the gases is 1
...
116 mol
(8
...
42+0
...
0132

= (0
...
37 atm) = 0
...
0% oxygen, 78
...
9% argon by volume has a total pressure of 790
...
What is the
partial pressure of each gas?
■

P O2 = 0
...
mmHg) = 166 mmHg
P N 2 = 0
...
mmHg) = 616 mmHg
P Ar = 0
...
mmHg) = 7 mmHg

● Hydrogen gas is generated when Ca reacts with water
...
The volume of gas
collected was 641 mL
...
82 mmHg at 30℃
...
82)mmHg = 956 mmHg
956 mmHg *

1 atm
760 mmHg

= 1
...
15)K = 303 K
11

nH 2 =

PV
RT

=

(1
...
641 L)
L*atm )(303 K)
(0
...
0325 mol H 2 *

2
...
0325 mol H 2 *

mol Ca
mol H 2

= 0
...
0655 g H 2

Part 2:

●

*

40
...
30 g Ca

Unequal Pressures
○ Density Hg = 13
...
00 g/mL
○ 1 mmHg = 13
...
7 mmHg?

P gas = P atm − P H 2 O − P height dif f erence
P h = 154 mmH 2 O *

1 mmHg
13
...
3 mmHg

P gas = (743
...
5 − 11
...
9 mmHg
● Kinetic molecular theory of gases - explains the behavior of gases that act
ideally
...
V gas particle << (much less than) space between particles (particles have
negligible volume)
2
...
Collisions are perfectly elastic
(the total kinetic energy remains constant)
3
...

4
...
K E = 12 m(↑)v 2 (↓) α T
Example: 1 mol He and 1 mol Ne at 273 K
→ same KE
12

→ He is faster
● Maxwell Speed Distribution Curves
○ At higher temperatures, more molecules are moving faster
●

Diffusion - the gradual mixing of molecules of one gas with molecules of another

by virtue of their kinetic properties
...
5 rHCl or

■

rHCl =

2
3

3
2

= 1
...
5
rHCl

rN H 3

● Graham’s law of effusion
○ Effusion - gas under pressure escapes into another area by passing
through a small opening
...
37 = 1
...
4 rO2
● If CO gas effuses at a rate that is 1
...
48 rX
rCO
rX

√

= (1
...
48)2 =

MX 2
)
M CO

MX
28
...
4 g/mol
13

● Real Gases
○ 1 mol of ideal gas
■ PV = nRT
n=

PV
RT

= 1
...

● Modified Ideal Gas Equation
○ Van der Waals equation - nonideal gas
■

(P +

an2
)(V
V2

− nb) = nRT

● Corrected P due to attractive forces
● Corrected Volume
● (equation wont be tested)
●

Note that on SA-6 #13 only Zn reacts with HCl not Cu

INTERMOLECULAR FORCES AND LIQUIDS AND
SOLIDS
Lecture 5
...
17
● For Molecular Compounds:
○ Intermolecular ​forces - attractive forces ​between​ molecules
■ Determines boiling point, melting point, state at room temp, etc
...
Covalent
bonds)
●

For Ionic Compounds:
○ “Intermolecular forces” are the ionic bonds (ion-ion forces)

●

Intermolecular vs Intramolecular
○ 41 kJ to vaporize 1 mole of water (inter)
14

○ 930 kJ to break all O-H bonds in 1 mol of water (intra)
○ Generally, ​inter​molecular forces are much weaker than ​intra​molecular
forces
...


■
■ Constant motion of electrons within molecules can create
instantaneous dipole​
...


●

C H 3 OCH 3 no, H is not attached to FON
...


○

Al3+ + H 2 O
18

■ Dispersion, Ion-Dipole force
●

Summary of Intermolecular Forces

○
● Due Tuesday
○ SA7 and Exp 9-Voicethread

Lecture 5
...
17

● Surface Tension - the resistance of a liquid to spread out and increase its surface
are
...

●

Cohesion - the intermolecular attraction between like molecules
...


○
●

Properties of Liquids - Viscosity
○ Viscosity - a measure of a fluid’s resistance to flow
■ Strong intermolecular forces → high viscosity
19

●

Water is a unique substance
○ Ice is less dense than water

●

Phase Changes from liquid to gas

○
●

Evaporation (Vaporization)
○ Liquids are constantly evaporating even when the temperature is less than
the boiling point
...

●

Boiling point - vapor pressure = atm pressure

●

To figure which substance has stronger IMF,
○ Compare vapor pressures (at the same temperature)
○ Compare boiling points (at the same vapor pressure)

20

○
●

Volatile liquid ○ Easily forms gas molecules due to weak intermolecular forces
...
Why?
■ A higher vapor pressure has weaker IMF
● HF - ​Hydrogen bonds, dipole dipole, dispersion
● HCl - ​dipole dipole, dispersion
■ HCl is weaker, so it has higher vapor pressure
...

■ All dispersion forces (if only C and H, the comp is nonpolar so only
disp)
●
C 3H 8
Low VP
Strong IMF

< C 2H 6

< CH 4
→

High VP
Weaker IMF
21

Higher molar
Mass

Lowest MM

● Since they are all nonpolar and dispersion forces, compare
molar masses
...

●

Boiling
○ Boiling point- temperature at which the (equilibrium) vapor pressure of a
liquid is = to the external pressure
...


●
○ Higher molar weight = higher boiling point b/c stronger IMFs
●

Molar heat of vaporization ( ΔH vap )
○ Energy required to vaporize 1 mol of liquid

●

Dispersion Forces and Boiling Point
○ Which substance will have a larger boiling point:
■ F​2​ (38
...
46 g/mol)
● F has dispersion, HCl has dipole dipole AND dispersion
○ HCl is stronger IMF so higher BP
■ Xe or Ar
● Both dispersion,​ but Xe has higher MM so it is stronger
IMF with higher BP
■ NaCl or H​2​O
● NaCl has ion-ion
22

● Water has H-bond, d-d, disp
○ NaCl is a solid (with a higher melting point) and
ion-ion is significantly stronger than h-bonds
● Critical Temperature and Pressure
○ Critical Temperature ( T c ) - the temperature above which the gas cannot
liquefy, no matter what the pressure
...

○ Stronger IMF → High T c
●

Phase Changes for Solids to Liquids

○
○ Melting point​ of a solid or ​freezing point​ of a liquid - the temperature at
which the solid and liquid phases coexist in equilibrium
○ Normal Melting point - ​the melting point at 1 atm
...

○

ΔH f us < ΔH vap always

●

Supercooling - a liquid is temporarily cooled to below its freezing point
...

● Calculate the amount of energy (in kJ) released when 346 g of steam at 182℃ is
converted to liquid water at 0℃
...
Mix 87
...
40 M ammonium sulfate + 225
...
20 M potassium
hydroxide
...
5 mL)(0
...
0 mL)(0
...


C hange
E end
E (M )

25
312
...


− 45

+ 45

0

0

25
...
11M


...
080M

0

45
312
...
14M

25

V olumef = 87
...
0 mL = 312
...
18
...


■
■ ↑P - condensation
■ ↓T - gas to solid or deposition
■ For water, since the slope of the line between solid and liquid is
negative, then the density of the solid is less than the liquid
...


○
●

Types of Solids
○ Crystalline solid - ordered
26

○ Amorphous solid - no order
●

X-ray diffraction by crystals
○ X-ray diffraction - scattering of x-rays by the units of a crystalline solid
○ Max von Laue - crystals diffract x-rays since the λ of x-rays is similar to the
distance between lattice points
...
154 nm are diffracted from a crystal at an
angle of 14
...
Assuming that n=1, what is the distance (in pm) between
tlayers in the crystal?
■ 1000 pm = 1 nm
■

2 d sinθ = nλ
●

●

d=

nλ
2sinθ

103 pm

=

(1)(0
...
17)

= 315 pm

Types of Crystals
○ Ionic Crystals
■ Basic units: cations and anions
■ Forces: ionic (ion-ion)
■ High melting point
■ Poor conductor of heat and electricity unless melted or dissolved in
water
■ Hard, brittle
○ Covalent Network
27

■ Basic units: atoms
■ Forces: covalent bonds
■ No individual molecules
■ High melting point
■ Poor conductor of heat and electricity
■ Hard
■ Examples: C (diamond, graphite), SiO​2​ (Quartz)
○ Molecular Crystals
■ Basic units: molecules
■ Forces: intermolecular forces
■ Low melting point
■ Poor conductor of heat and electricity
■ Soft
■ Larger IMF = higher the melting point
○ Metallic Crystals
■ Basic units: metal atoms
■ Forces: metallic bonds
■ Soft to hard, range of melting points
■ Good conductors of heat and electricity
○ Atomic Solid
■ Solid containing atoms from a ​nonmetallic element​ (ex: solid form
of Ne and other noble gases)
■ Single atms held in place by weak dispersion forces
● Low melting points
○ Amorphous solids
■ Amorphous solid - no order
■ Glass
●
Type Solid

Basic Unit

IMF

MP

Example
28

Ionic

Ions

Ion ion

High

NaCl

Covalent
Network

Atoms

Covalent
bonds

High

C - (diamond/grahite)
SiO​2​ - (quartz)

Molecular

Molecules

H-bonds,
d-d,and/or
dispersion

Low

H​2​O, F​2

Metallic

Metal atoms

Metallic
bonding

High

Cu

Atomic

Nonmetal
atoms

dispersion

Low

Ne

●
●

Practice:
○ Classify the solid and what type of intermolecular forces are present:
■ SO​2​ - Molecular solid, dipole dipole, dispersion
■ KI - Ionic solids, ion ion
■ Zn - Metallic solid, metallic bonding
■ NaBr - Ionic solid, ion ion
■ CH​4 -​ Molecular solid, dipole dipole, dispersion
○ Arrange the following in order of increasing melting point
...


●

Hydration - an ion is surrounded by water molecules

●

“Like dissolves like”
○ Substances with similar ​INTER​molecular forces are soluble
■ Nonpolar dissolve in nonpolar
● CCl​4​ in C​6​H​6
■ Polar​ and ​ionic ​compounds dissolve in polar solvents
● C​2​H​5​OH​ in H​2​O
● NaCl ​in H​2​O or NH​3
○ Miscible​ - 2 liquids that are completely soluble in one another in all
proportions
...

● Concentration Units:
○ Percent by mass
■

% by mass =

mass of solute
mass of solution

* 100%

○ Mole Fraction (X)
■

XA =

mol A
total mol

○ Molarity
■

M=

mol solute
L soln

○ Molality
■

m=

mol solute
kg solvent

○ Parts per million (ppm)
■

1 ppm =

■

ppm =

1 g solute
106 g solution

g solute
g soln

6

* 10

■ Ex: 3
...
892 g of KCl is dissolved in 54
...
What is the
percent by mass of KCl in solution?
■

0
...
892 g +54
...
61% KCl

● Calculating molarity and molality
○ What is the ​molality​ of a solution containing 7
...
78 g urea *
0
...
203 kg H 2 O

mol urea
60
...
130 mol urea

= 0
...
36 g
of H​2​SO​4​ in water and diluting to a final volume of 50
...
00 g/mL)?
mol H 2 SO4
L soln

■ Molarity (M):

2
...
0241 mol H 2 SO4
0
...
08 g H 2 SO4

mol H 2 SO4
kg H 2 O

= 0
...
481 M H 2 SO4

50
...
00 g soln
mL soln

= 50
...
0 g − 2
...
64 g H 2 O
0
...
04764 kg H 2 O

= 0
...
73 m methanol (CH​3​OH) solution whose
density is 0
...
73 mol CH 3 OH
kg H 2 O

0
...
73 mol CH 3 OH → 1 kg of H 2 O
■

2
...
04 g CH 3 OH
mol CH 3 OH

= 87
...
5 g CH 3 OH = 1087
...
5 g soln *
2
...
11 L soln

mL soln
0
...
11 L soln

= 2
...
86 M ethanol (C​2​H​5​OH) solution whose density
is 0
...
927 g soln
mL soln

5
...
86 mol ethanol
■

1000 mL
L

1
...
86 mol ethanol *

*

0
...
07 g ethanol
mol ethanol

= 927 g soln

= 270
...
g ethanol = 657 g H 2 O = 0
...
86 mol ethanol
0
...
92 m ethanol

○ Calculate the molarity and molality of a 35
...
45 g/mL
...
4 g H 3 P O4
100 g soln

3
...
4 g H 3 P O4
100 g soln

*

⇒ (molarity) ⇒

mol H 3 P O4
L soln

mol H 3 P O4
kg H 2 O
mol H 3 P O4
97
...
45 g soln
mL soln

*

1000 mL
L

= 12
...
4 g H 3 P O4 → 100 g soln
■

35
...
99 g H 3 P O4

= 0
...
4 g = 64
...
0646 kg H 2 O
0
...
0646 kg H 2 O

= 5
...
6 percent (by mass) aqueous
solution of sodium chloride if the density is 2
...

■

44
...
6 g N aCl
100 g soln

2
...
44 g N aCl

*

*

2
...
7 M N aCl

■ Assume 44
...
6 g N aCl *

mol N aCl
58
...
763 mol N aCl

100 g − 44
...
4 g H 2 O = 0
...
763 mol N aCl
0
...
7 m N aCl

○ Calculate the ppm of a solution that is 5
...
23*10−4 g N O2
100 g soln

=

5
...
23 ppm N O 2
33

○ Calculate the % weight of a solution that is 32
...

10−4
10−4

■

32
...
21*10 g benzene
100 g soln

*

−3

g benzene
100 g soln

= 3
...

○ Solubility increases with increasing temp for most substances

○
● Temperature and Solubility of Gases
○ Solubility decreases with increasing temperature for ALL gases
■ Side note: solubility increases with increasing pressure for ALL
gases
...
25
...

○

c = kP

○ c - concentration ( M =

mol
L

) of dissolved gas

○ P - partial pressure (atm) of gas over solution
...
22 atm (Henry’s law constant for oxygen is 1
...

■

c = kP
34

c = (1
...
22 atm)
*atm
c = 2
...
9*10−4 mol O2
L soln

1
...
9 * 10−4 mol O2 → 1
...
00 g O2
mol O2

2
...
0 L soln *

1000 mL
L

*

= 0
...
00 g soln
mL soln

= 1000 g soln

1000 g − 0
...
9*10−4 mol O2
1
...
9 * 10−4 m O2

○ You have a sample of water at sea level and another on the top of Mt
...
Assuming that the temperature is the same, how would the
amount of O​2​ dissolved in water compare between the two samples?
■ Amount of oxygen would be lower on Mt
...

○ What would happen to the amount of dissolved oxygen if you took water in
a closed container at sea level and carried it up to the top of Mt
...
Everest
...

■ Relatively dilute solutions ≤ 0
...

○ Raoult’s Law
■

P 1 = X 1P 01
●

P 1 − v apor pressure SOLU T ION

●

P 0 1 − v apor pressure P U RE solvent

●

X 1 − mol f raction solvent

■ If the solution contains only one solute:
●

●

X1 + X2 = 1
○

X 1 − solvent

○

X 2 − solute

X1 = 1 − X2
○

ΔP = P 0 1 − P 1 = X 2 P 0 1
■

ΔP − v apor pressure depression

● Practice using Raoult’s Law
○ Calculate the vapor pressure of a solution made by dissolving 82
...
06 g/mol) in 212 mL of water at 35ºC
...
18
mmHg
...
4 g urea *

mol urea
60
...
00 g H 2 O
mL H 2 O

11
...
8+1
...
37 mol urea
*

mol H 2 O
18
...
8 mol H 2 O

= 0
...
896)(42
...
8 mmHg
■

ΔP = X 2 P 0 1 OR ΔP = P 0 1 − P 1
●

ΔP = X 2 P 0 1
36

○

ΔP = (1 − 0
...
18 mmHg)
ΔP = 4
...
18 mmHg − 37
...
4 mmHg

● Boiling Point Elevation
○ Lower vapor pressure → higher boiling point
○ More solute particles → greater increase in boiling point
○

ΔT b = T b − T 0 b
■ Always positive
...

■

T 0 b − boiling point P U RE solvent

■

T b − boiling point SOLU T ION

■

T b > T 0b

■

ΔT b = K b m

ΔT b > 0

● m - molality
● K​b​ - constant (℃/m) for solvent
● Freezing Point Depression
○ Freezing - disordered → ordered, remove energy
○ Solution is more disordered so need to remove more energy (freeze at a
lower temperature)
○ The more solute particles → more disorder → greater the decrease in
freezing point
...

■

T 0 f − f reezing point P U RE solvent

■

T f − f reezing point SOLU T ION

■

T 0f > T f

■

ΔT f = K f m

ΔT f > 0

37

● m - molality
● K​f​ - constant (℃/m) for solvent
...
01 g and the K​f​ for water equals 1
...

■

478 g antif reeze *

mol
62
...
202 kg H 2 O

= 2
...
86℃
)(2
...
48℃
m
0℃ − 4
...
48 (freezing point of soln)
● Molar Mass determination of solute
○ A 7
...
05ºC
...
12 ºC/m
...
05℃ = ( 5
...
205 m =

0
...
301 kg benzene *
7
...
0617 mol

n=

0
...
0617 mol (C 5 H 4 )n

= 127 g/mol

127 g/mol
64
...
85 g of an organic compound in 100
...
16ºC
...
50ºC
...
12 ºC/m
...
50℃ − 5
...
12℃
)x
m
38

x(or molality) = 0
...
1000 kg benzene *
0
...
0066 mol

0
...

kg benzene

= 0
...


= 128 g/mol = 1
...

■ Vapor-pressure lowering P 1 = X 1 P 0 1
■ Boiling-point elevation​ ΔT b = K b m
■ Freezing-point depression ΔT f = K f m
● Colligative Properties - ELECTROLYTES
○

N aCl (aq) → N a+ (aq) + C l− (aq)
0
...
1 m N a+ + 0
...
2 m ions

○ van’t Hoff factor (i)​ =

actual # particles
# f ormula units

i ​ should be
nonelectrolytes

1

NaCl

2

CaCl​2

3

○ Boiling-point elevation​ - ΔT b = iK b m
○ Freezing-point depression​ - ΔT f = iK f m
● Example problem:
○ Which solution would have the lowest freezing point?
(Lowest fp = most particles)
■

0
...
20 = 0
...
20 m HC 2 H 3 O2
● i=139

●

0
...
40 m
○

■

closer to 0
...
10 m HCl
● i=1
●

■

0
...
20 m N a3 P O4
● i=4
●

0
...

● SA9 Tips (NAG SAG)
○ Always Soluble
■ N itrates
■ A cetates
■ G roup 1
○ Soluble with some exceptions (but exceptions not tested)
■ S uflates
■ A mmonium
■ G roup 7
○ Insolubles
■

Ag + , CO3 2− , P O4 3− , C 2 O4 2− , S 2− , O2− , OH −
● Silver is always insoluble, but all else can be soluble if paired
with a NAG SAG
...


40



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Title: General Chemistry Exam 3 Notes - Everything you need!
Description: I used to be a chemistry instructor for students struggling in General Chemistry. If you are struggling and need a quick crash course or refresher for your THIRD Gen Chem exam, this is THE PERFECT study guide/crash course packet for you. Here are 40 pages of extremely detailed notes covering everything you would expect to be on the THIRD exam of your General Chemistry course. The notes are all in large font with all equations and examples completed AND typed out for your convenience. All examples are worked out STEP BY STEP, including EXPLANATIONS every step of the way. TOPICS INCLUDE: 1. Gases – Properties and Gas Laws 2. Density, Molar Mass & Gas Stoichiometry 3. Kinetic Molecular Theory & Real Gases 4. Liquids, Intermolecular Forces, Properties 5. Phase Changes, Phase Diagrams 6. Solids, Intro to Solutions 7. Solutions, Solubility & Concentration units 8. Colligative Properties 9. Organic Chemistry

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