ADVANCED HONORS CHEMISTRY - CHAPTER 14
NAME:
THE BEHAVIOR OF GASES
DATE:
GRAHAM'S LAW WKST - SELECT 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
)
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.
2 - AHC - Chapter 14 - Graham's Law Worksheet - Selected 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
3 - AHC - Chapter 14 - Graham's Law Worksheet - Selected 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
4 - AHC - Chapter 14 - Graham's Law Worksheet - Selected Answers - V8