Chapter 10 Molar Volume Review
Vapor Pressure of water at 28.0o is 3.78 kPa
Assume STP unless otherwise stated
1.
If 10.0 grams of a gas has a volume of 3.12 dm3 at STP, what is its molecular mass?
2.
How many grams are in 1,350 dm3 of CO2?
3.
How many dm3 are in 85.0 grams of CO2?
4.
How many moles are in 750.0 cm3 of PF3?
5.
If there are 5.00 dm3 of CO2 at 15.0o C and 100 kPa, how may moles of CO2 are present?
6.
If a gas has a mass of .620 grams and a volume of 212 cm3, what is its density in g/dm3?
7.
Find the mass of 8.50 m3 of CO2 at 101.3 kPa and 0.000 C.
8.
What is the molecular mass of a gas that has a mass of 2.10 grams and a volume of 150 dm3 at
18.0o C and 100.5 kPa?
9.
2.75 dm3 of N2 gas is collected over water at a temperature of 28.0o C and a pressure of 98.5 kPa.
What is the mass of the dry gas in grams?
Solutions
1.
PV =
M=
2.
m
RT
M
M =
or
mRT
PV
8.31 dm3.kPa
mol . K
10.0 g
1350 dm3
or
1 mol CO2
22.4 dm3
PVM
m=
RT
m=
101.3 kPa
1,350 dm3
273 K
101.3 kPa
44.0 g CO2
1 mol CO2
44.0 g CO2
1 mol CO2
3.12 dm
= 71.8 g/mol
3
= 2,650 g
mol . K
8.31 dm3.kPa
= 2,650 g
273 K
3.
4.
5.
85.0 g CO2
6.
V =
V=
85.0 g CO2
750.0 cm3 PF3
7.
n=
101.3 kPa
PV = nRT
m=
8.
PV =
M=
9.
PV =
m=
273 K
101.3 kPa
1 mol PF3
22.4 dm3
1 mol
44.0 g CO2
= 43.3 dm3
= .0335 mol
PV
RT
or
PV =
8.31 dm3.kPa
mol . K
1 dm3
1,000 cm3
n=
750.0 cm3 PF3
n=
or
1,000 cm3
1 dm3
m
RT
M
or
m
RT
M
2.10 g
m
RT
M
94.72 kPa
or
mol . K
8.31 dm3.kPa
= 2.92 g/dm3
1,000 dm3
1 m3
M =
8.31 dm3.kPa
mol . K
or
273 K
1 dm3
1,000 cm3
= .0335 mol
= .209 mol
288 K
This is the only problem that can’t
be done using the formula
PVM
RT
m=
8.50 m3
101.3 kPa
mol . K
8.31 dm3.kPa
PV
RT
5.00 dm3
100 kPa
.620 g
212 cm3
= 43.3 dm3
mRT
MP
or
n=
22.4 dm3
1 mol CO2
1 mol
44.0 g CO2
m=
2.75 dm3 N2
44.0 g CO2
1 mol CO2
mol . K
8.31 dm3.kPa
= 16,700 g
273 K
mRT
PV
291 K
100.5 kPa
150 dm3
= .337 g/mol
PVM
RT
28.0 g N2
1 mol N2
mol . K
8.31 dm3.kPa
= 2.92 g
301 K