Chemistry& 141
Clark College
Homework 5 SOLUTION
Follow the homework guidelines given in the syllabus! Show your work, use units and sig figs, where
appropriate.
1. A 1.5 L sample of gaseous CCl2F2, a refrigerant known as Freon-12, is held at 56 torr and 23.5 °C.
If the pressure is increased to 150 torr at the same temperature what will the new volume of gas be?
Since we are comparing two systems, we can leave the pressure units in torr – no need to convert to atm.
P1 = 56 torr
56 torr 1.5 L
PV
V1 = 1.5 L
P1 V1 = P2 V2 V2 = 1 1 =
= 0.56 L
P2 = 150 torr
P2
150 torr
V2 = ? L
Don’t forget to ask yourself if you answer makes sense with the relationship between P and V!
(
)(
)
2. In the system detailed in problem 1, how many grams of Freon-12 are in the system?
We can use either set of P and V data for the ideal gas law to find the number of moles, which is then
converted to grams. We will need to ensure that are units are proper for the ideal gas law!
P = 56 torr x
1 atm
= 0.074 atm
760 torr
PV = nRT n =
V = 1.5 L
n = ? mol
T = 23.5°C + 273.15 = 296.7 K
4.6 x 10-3 mol x
(
)(
)
0.074 atm 1.5L
PV
=
= 4.6 x 10-3 mol
RT
!
Liatm $
#" 0.08206 moliK &% 296.7 K
(
)
120.91 g
= 0.55 g CCl2 F2
1 mol
3. OMIT***A balloon is filled with air to an internal pressure of 1.25 atm at 21.2°C. That balloon is
carried to the bottom of a lake, where the temperature is now 10.9°C and the pressure is 5.95 atm.
a. What is the new volume of the balloon?
b. How many moles of air are in the balloon?
4. A sample of an organic compound weighs 0.495 g. This sample is a gas collected in a flask that has
an exact volume of 127 mL. At 98°C, the pressure of the gas is 754 mmHg in the flask. Determine
the number of moles for the compound, and use that to calculate the molecular weight of the organic
compound in the flask.
1 atm
= 0.992 atm
760 mmHg
1L
V = 127 mL x
= 0.127 L
1000 mL
n = ? mol
T = 98°C + 273.15 = 371 K
P = 754 mmHg x
Homework 5
PV = nRT
n=
(
)(
)
0.992 atm 0.127 L
PV
=
= 4.14 x 10-3 mol
RT
!
$
Liatm
#" 0.08206 moliK &% 371 K
(
0.495 g
-3
4.14 x 10 mol
Fall 2008
)
= 1.20 x 102 g/mol
Page 1 of 2
Chemistry& 141
Clark College
5. Octane burns in air according to the following unbalanced equation:
2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
a. If 75.0 g of octane react with excess oxygen, how many moles of carbon dioxide will be
generated?
75.0 g C8 H18 x
1 mol C8 H18
114.2 g C8 H18
x
16 mol CO 2
2 mol C8 H18
= 5.25 mol CO 2
b. How many liters of carbon dioxide will be generated if it is collected at 0.985 atm and 23.0°C?
P = 0.985 atm
V=?L
n = 5.25 mol
T = 23.0°C + 273.15 = 296.2 K
Homework 5
PV = nRT
!
V=
nRT
P
$
Liatm
(5.25 mol) #" 0.08206 moliK
&% ( 296.2 K )
=
= 1.30 x 10
(0.985 atm )
Fall 2008
2
L
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