ADVANCED HONORS CHEMISTRY - CHAPTER 14 NAME:
THE BEHAVIOR OF GASES
DATE:
GRAHAM'S LAW WORKSHEET - ANSWERS - V8
PAGE:
1. How fast would a molecule of sulfur dioxide travel if an atom of krypton (aarrgghh!) travels an average of
750 m/s at 200. oC?
DATA TABLE
Gas 1
Gas 2
Name
Sulfur Dioxide
Krypton
Formula
SO2
Kr
64 g
mol
⎛ m⎞
?⎜
⎝ sec ⎟⎠
84 g
mol
Molar Mass
Velocity
VelocitySulfur Dioxide
Velocity Krypton
( Velocity
Krypton
)
750 m
sec
Molar Mass Krypton
=
Molar MassSulfur Dioxide
⎛ VelocitySulfur Dioxide ⎞
⎜
⎟ = Velocity Krypton
⎝ Velocity Krypton ⎠
(
VelocitySulfur Dioxide =
( Velocity
VelocitySulfur Dioxide =
Krypton
)(
)
⎛
Molar Mass Krypton ⎞
⎜
⎟
⎜⎝ Molar MassSulfur Dioxide ⎟⎠
Molar Mass Krypton
Molar MassSulfur Dioxide
⎛ 750 m ⎞ ⎛ 84 g ⎞
⎜⎝ sec ⎟⎠ ⎜ mol ⎟
⎝
⎠
64 g
mol
VelocitySulfur Dioxide =
1
860 meters
sec
)
2. A fluorine gas molecule travels at about 300. m/s at room temperature. How fast would a molecule of
methanol, CH3OH, travel at the same temperature?
DATA TABLE
Gas 1
Gas 2
Name
Fluorine
Methanol
Formula
F2
CH3OH
Molar Mass
38.0 g
mol
Velocity
300. m
sec
32.0 g
mol
⎛ m⎞
? ⎜
⎝ sec ⎟⎠
Velocity Fluorine
=
Velocity Methanol
Molar Mass Methanol
Molar Mass Fluorine
Cross Multiply To Get The Variable in the Numerator.
( Velocity
( Velocity
)(
Molar Mass Fluorine = Velocity Methanol
)(
Molar Mass Fluorine
Fluorine
Fluorine
)
Molar Mass Methanol
Velocity Methanol =
) = (Velocity
( Velocity
Velocity Methanol =
(
Fluorine
)(
Methanol
)(
)(
Molar Mass Methanol
)
Molar Mass
)
Molar Mass
Methanol
Methanol
Molar Mass Fluorine
)
Molar Mass Methanol
⎛ 300. m ⎞ ⎛ 38.0 g ⎞
⎟
⎜⎝ sec ⎟⎠ ⎜
mol ⎠
⎝
Velocity Methanol =
32.0 g
mol
327 meters
sec
2 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8
3. Which gas diffuses faster at the same temperature, carbon dioxide or xenon? To three significant figures,
how much faster does it diffuse?
We are solving for the ratio of the rates of diffusion. In this type of problem it is generally understood
that, unless told otherwise, we are comparing the faster moving gas, the lighter gas,
to the slower moving gas, the heavier gas.
Thus, since carbon dioxide is the lighter gas, it will diffuse faster.
Therefore, we will make gas 1 carbon dioxide.
DATA TABLE
Gas 1 (Lighter Gas)
Gas 2 (Heavier Gas)
Name
Carbon Dioxide
Xenon
Formula
CO2
Xe
Molar Mass
44.0 g
mol
131 g
mol
Velocity
-
-
RateCarbon Dioxide
=
Rate Xenon
Molar Mass Xenon
Molar MassCarbon Dioxide
Since we are solving for the ratio of the rates of diffusion,
the left side of the equation can be expressed as one variable:
RateCarbon Dioxide
=x
Rate Xenon
RateCarbon Dioxide
=
Rate Xenon
x=
Molar Mass Xenon
Molar MassCarbon Dioxide
Molar Mass Xenon
Molar MassCarbon Dioxide
x=
131 g
mol = 1.73
44.0 g
mol
Carbon dioxide will diffuse 1.73 times faster than xenon.
3 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8
4. Which gas diffuses faster at the same temperature, nitrogen or ammonia? To three significant figures, how
much faster does it diffuse?
We are solving for the ratio of the rates of diffusion. In this type of problem it is generally understood
that, unless told otherwise, we are comparing the faster moving gas, the lighter gas,
to the slower moving gas, the heavier gas.
Thus, since ammonia is the lighter gas, it will diffuse faster.
Therefore, we will make gas 1 ammonia.
DATA TABLE
Gas 1 (Lighter Gas)
Gas 2 (Heavier Gas)
Name
Ammonia
Nitrogen
Formula
NH3
N2
Molar Mass
17.0 g
mol
28.0 g
mol
Velocity
-
-
Rate Ammonia
=
Rate Nitrogen
Molar Mass Nitrogen
Molar Mass Ammonia
Since we are solving for the ratio of the rates of diffusion,
the left side of the equation can be expressed as one variable:
Rate Ammonia
=x
Rate Nitrogen
x=
x=
Molar Mass Nitrogen
Molar Mass Ammonia
28.0 g
mol = 1.28
17.0 g
mol
Ammonia will diffuse 1.28 times faster than nitrogen.
4 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8
5. A helium atom travels an average of 1,000. m/s at 250. oC. How fast would an atom of radon travel at the
same temperature?
DATA TABLE
Gas 1
Gas 2
Name
Helium
Radon
Formula
He
Rn
Molar Mass
4.00 g
mol
Velocity
1,000. m
sec
222 g
mol
⎛ m⎞
? ⎜
⎝ sec ⎟⎠
Velocity He
=
Velocity Rn
Molar Mass Rn
Molar Mass He
Cross Multiply To Get The Variable in the Numerator.
( Velocity )(
Molar Mass He = Velocity Rn
( Velocity )(
Molar Mass He
He
He
)
Molar Mass Rn
Velocity Rn =
Velocity Rn =
(
)(
) = (Velocity )(
Rn
Molar Mass Rn
)
Molar Mass
)
Molar Mass
( Velocity )(
He
Molar Mass He
Rn
Rn
)
Molar Mass Rn
⎛ 1,000. m ⎞ ⎛ 4.00 grams ⎞
⎟
⎜⎝ sec ⎟⎠ ⎜
mole
⎝
⎠
222 grams
mole
Velocity Rn =
134 m
sec
5 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8
6. A nitrogen gas molecule travels at about 500. m/s at room temperature. How fast would a molecule of
ethanol, C2H5OH, travel at the same temperature?
DATA TABLE
Gas 1
Gas 2
Name
Nitrogen
Ethanol
Formula
N2
C2H5OH
Molar Mass
28.0 g
mol
Velocity
500. m
sec
46.1 g
mol
⎛ m⎞
? ⎜
⎝ sec ⎟⎠
Velocity N
VelocityC
=
2
2 H 5 OH
Molar MassC H OH
2
5
Molar Mass N
2
Cross Multiply To Get The Variable in the Numerator.
( Velocity )(
N2
( Velocity )(
N2
Molar Mass N
Molar Mass N
Molar MassC H OH
2
VelocityC
2
) = (Velocity
)(
Molar MassC H OH
) = (Velocity
)(
Molar Mass C2 H5OH
2
C 2 H5 OH
C 2 H5 OH
5
Velocity ) (
(
=
N2
2 H 5 OH
5
Molar Mass C2 H5OH
Molar Mass N
2
)
Molar MassC H OH
2
VelocityC
2
2 H 5 OH
=
5
⎛ 500. m ⎞ ⎛ 28.0 grams ⎞
⎟
⎜⎝ sec ⎟⎠ ⎜
mole
⎝
⎠
VelocityC
46.1 grams
mole
2 H 5 OH
=
390. m
sec
6 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8
)
)
7. Helium gas effuses through an opening at a rate 2.000 times faster than that of unknown gas. What is the
molecular mass of this unknown gas?
DATA TABLE
Gas 1
Gas 2
Name
Helium
Unknown
Formula
He
?
Molar Mass
4.003 g
mol
⎛ g ⎞
? ⎜
⎝ mol ⎟⎠
Velocity
-
-
Molar MassUnknown
Rate of Effusion He
=
Rate of Effusion Unknown
Molar Mass He
Rate of Effusion He
2.000
=
Rate of Effusion Unknown
1
∴
2.000
=
1
Molar MassUnknown
Molar Mass He
(
Molar MassUnknown = 2.000
)(
Molar Mass He
)
⎛ 4.003 g ⎞
Molar MassUnknown = 2.000 ⎜
⎟
mol ⎠
⎝
(
)
⎛ 2.001 g ⎞
Molar MassUnknown = 2.000 ⎜
⎟
mol ⎠
⎝
(
)
Molar MassUnknown =
(
Molar MassUnknown
)
2
4.002 g
mol
⎛ 4.002 g ⎞
=⎜
⎟
mol ⎠
⎝
Molar MassUnknown =
2
16.01 g
mol
7 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8
8. An unknown gas effuses through an opening at a rate 0.3165 times that of helium gas. What is the molecular
mass of this unknown gas?
DATA TABLE
Gas 1
Gas 2
Name
Unknown
Helium
Formula
?
He
Molar Mass
⎛ g ⎞
?⎜
⎝ mol ⎟⎠
4.003 g
mol
Velocity
-
-
Molar Mass He
Rate of Effusion Unknown
=
Rate of Effusion He
Molar MassUnknown
Rate of Effusion Unknown
= 0.3165
Rate of Effusion He
∴
0.3165 =
Molar Mass He
Molar MassUnknown
Molar Mass He
Molar MassUnknown =
0.3165
Molar MassUnknown =
4.003 g
mol
0.3165
2.001 g
Molar MassUnknown =
(
Molar MassUnknown
mol
0.3165
)
2
=
6.321 g
mol
⎛ 6.321 g ⎞
=⎜
⎟
mol ⎠
⎝
Molar MassUnknown =
2
39.96 g
mol
8 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8