KEY
Chemistry: Mixed Review
Solve each of the following problems, being sure to show your work and include all proper units.
1. Use the ideal gas law PV = nRT, to derive Boyle’s law and Charles’s law.
If n, R and T are constant:
PV = constant
P1V1 = P2V2
PV = nRT
If n, R and P are constant:
V/T = constant
(Boyle’s Law)
V1/T1 = V2/T2
V/T = nRP
(Charles’s Law)
2. A container holds 265 mL of chlorine gas, Cl2. Assuming that the gas sample is at STP, what is its mass?
V = 265 mL (0.265 L)
T = 273 K
P = 1 atm
n = ? mol
R = 0.0821 L.atm / mol.K
PV = nRT
n
PV
RT
n
(1 atm)(0.265 L)
(0.0821 L  atm/mol  K)(273 K )
n  0.0118 mol Cl 2
 71 g Cl 2 
  0.84 g Cl 2
x g Cl 2  0.0118 mol Cl 2 

 1mol Cl 2 
3. Suppose that 3.11 mol of carbon dioxide is at a pressure of 0.820 atm and a temperature of 39 oC. What is
the volume of the sample, in liters?
n = 3.11 mol CO2
P = 0.820 atm
T = 39oC + 273 = 312 K
V=?L
R = 0.0821 L.atm / mol.K
PV = nRT
V
nRT
P
V
(3.11 mol)(0.082 1L  atm/mol  K )(312 K)
0.820 atm
V  97.1 L
4. Compare the rates of diffusion of carbon monoxide, CO, and sulfur trioxide, SO3.
CO = carbon monoxide
m1 = 28 g
v1 = ?
SO3 = sulfur trioxide
m2 = 80 g
v2 = ?
v1
m2

v2
m1
v1
80 g

v2
28 g
CO diffuses 1.69x faster than SO3 (or SO3 is 0.59x slower)
v1 1.69

v2
1
5. A gas sample that has a mass of 0.993 g occupies 0.570 L. Given the temperature is 281 K and the
pressure is 1.44 atm, what is the molar mass of the gas?
P = 1.44 atm
PV = nRT
V = 0.570 L
n = ? mol
R = 0.0821 L.atm / mol.K
T = 281 K
n
PV
RT
2. 0.84 g Cl2
(1.44 atm)(0.570 L)
(0.0821 L  atm/mol  K)(281 K )
n  0.0356 mol
molar mass 
Answers: 1.
n
3. 97.1L
g
0.993 g

 28 g/mol
mol
0.0356 mol
4. CO is 1.69x faster
5. 28 g/mol
6. The density of a gas is 3.07 g/L at STP. Calculate the gas’s molar mass.
P = 1 atm
V=?L
T = 273 K
n = ? mol
D
mass
volume
3.07 g / L 
R = 0.0821
D = 3.07 g/L
Density is an intensive property.
Assume mass = 1.0 g
D
n
n
M
 M  68.8 g
22.4 L
(1 atm)(0.325 7 L)
(0.0821 L  atm/mol  K)(273 K )
n  0.01453 mol
M
M
 3.07 g/L 
V
V
V
PV
RT
molar mass 
1.0 g
 0.3257 L
3.07 g/L
g
1.0 g

mol
0.01453 mol
molar mass  68.8 g/mol
7. How many moles of helium gas would it take to fill a gas balloon with a volume of 1000. cm 3 when the
temperature is 32oC and the atmospheric pressure is 752 mm Hg?
 1 atm

  0.9895 atm
752 mm Hg 
P = 752 mm Hg
PV = nRT
760
mm
Hg


V=1L
n = ? mol
n
PV
RT
n
R = 0.0821 L.atm / mol.K
T = 32oC + 273 = 305 K
(0.9895 atm)(1 L)
(0.0821 L  atm/mol  K)(305 K )
n  0.0395 mol
8. A gas sample is collected at 16oC and 0.982 atm. If the sample has a mass of 7.40 g and a volume of 3.96
L, find the volume of the gas at STP and the molar mass.
(0.982 atm)(3.96 L) (1 atm)(V 2 )

289 K
273 K
P1V1 P2 V2

P1 = 0.982 atm
T1
T2
V1 = 3.96 L
o
T1 = 16 C + 273 = 289 K
P2 = 1 atm
PV
n
V2 = ? L
PV = nRT
RT
T2 = 273 K
n
molar mass 
(0.982 atm)(3.96 L)
(0.0821 L  atm/mol  K)(289 K )
g
7.4 g

mol
0.164 mol
V2  3.67 L
n  0.164 mol
molar mass  45 g/mol
9. An unknown gas effuses at 0.850 times the effusion rate of nitrogen dioxide, NO 2. Estimate the molar mass
of the unknown gas.
Unknown Gas
m1 = ? g
v1 = 0.850
Nitrogen dioxide, NO2
m2 = 46 g
v2 = 1
v1
m2

v2
m1
0.850

1
46 g
xg
V2  63.7 g
10. What will be the pressure if we open:
a. just valve A? 0.75 atm
b. just valve C?
c.
0.5 atm
any two valves? 0.5 atm
Use Boyle’s law:
(a) 1.5 atm of gas divided in 2 spheres = 0.75 atm
(b) 1.0 atm of gas divided in 2 spheres = 0.5 atm
(c) 1.5 atm of gas divided in 3 spheres = 0.5 atm
Answers: 6.
68.8 g/mol
7. 0.0395 mol
8. 3.67 L & 45 g/mol
9. 63.7 g/mol