KEY
Practice Problems: IMFs and Phases
CHEM 1A
Temperature 
1. How much energy (in kJ) is required to completely vaporize 200.0 g of 25.00ºC liquid water?
Properties of Water
Specific Heats (C):
gas = 1.84 J/gºC
liquid = 4.184 J/gºC
solid = 2.09 J/gºC
Heat of Vaporization:
Hvap = 40.7 kJ/mol
Heat of Fusion:
Hfus = 6.01 kJ/mol
Gas
E
boiling pt.
F
D
B
melting pt.
C
Liquid
Solid
A
Heat added at a constant rate 
Heating 200.0 g of liquid water to the boiling point (segment DE):
q1 = m Cliq T = m Cliq (Tf – Ti) 200.0 g
62760 J
1 kJ
103 J
4.184 J
100.00ºC – 25.00ºC
= 62.760 kJ
Vaporizing 200.0 g of liquid water (segment EF):
200.0 g H2O
1 mol H2O
18.016 g H2O
Sum of the energies:
qtotal = q1 + q2
= 11.10124334 mol H2O
q2 = n Hvap = 11.10124334 mol H2O
515 kJ
Answer: ________________
40.7 kJ
1 mol H2O
= 62760 J
g ºC
= 451.8206039 kJ
62. 760
kJ (q1)
+ 451. 8206039 kJ (q2)
514. 5806039 kJ (qtotal)
2. How much energy (in kJ) is required to melt 150.0 g of –18.00 ºC ice, and bring the resulting liquid
Temperature 
water up to 25.00ºC?
Properties of Water
Specific Heats (C):
gas = 1.84 J/gºC
liquid = 4.184 J/gºC
solid = 2.09 J/gºC
Heat of Vaporization:
Hvap = 40.7 kJ/mol
Heat of Fusion:
Hfus = 6.01 kJ/mol
Gas
F
boiling pt.
G
E
C
melting pt.
D
Liquid
B
Solid
A
Heat added at a constant rate 
Heating 150.0 g of ice to the melting point (segment BC):
q1 = m Csolid T = m Csolid (Tf – Ti) 150.0 g
5643 J
1 kJ
103 J
2.09 J
0.00ºC – –18.00ºC
= 5643 J
g ºC
= 5.643 kJ
Melting 150.0 g of ice (segment CD):
150.0 g H2O
1 mol H2O
18.016 g H2O
= 8.325932504 mol H2O
q2 = n Hfus = 8.325932504 mol H2O
6.01 kJ
1 mol H2O
= 50.03885435 kJ
Heating 150.0 g of liquid water (segment DE):
q3 = m Cliq T = m Cliq (Tf – Ti) 150.0 g
15690 J
71.4 kJ
Answer: ________________
1 kJ
103 J
4.184 J
25.00ºC – 0.00ºC
= 15690 J
g ºC
= 15.690 kJ
Sum of the energies:
qtotal = q1 + q2 + q3
5.6 43
kJ (q1)
+ 50.0 3885435 kJ (q2)
+ 15.6 90
kJ (q3)
71.3 7185435 kJ (qtotal)
3. Iridium (Ir, atomic # 77) has a face centered cubic unit cell with an edge length of 383.3 pm.
Calculate the density (in g/cm3) of solid iridium.
Face Centered Cubic:
4 atoms Ir
8 corners ( 1/8
1 mol Ir
192.2 g Ir
23
6.022 x 10 atoms Ir
383.3 pm
atom
atom
) + 6 faces ( 1/2
) = 4 atoms Ir
corner
face
10−12 m
1 pm
= 1.27665227 x 10−21 g Ir
1 mol Ir
1 cm
= 3.833 x 10−8 cm
−2
10 m
Vcube = (edge length)3 = (3.833 x 10−8 cm)3 = 5.63140105 x 10−23 cm3
density =
mass
volume
=
22.67 g/cm3
Answer: ________________
1.27665227 x 10−21 g
5.63140105 x 10
−23
3
cm
= 22.6702425 g/cm3