1

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

Molar Mass

64 gmol

84 gmol

Velocity ?

msec

⎛⎝⎜

⎞⎠⎟

750 msec

VelocitySulfur Dioxide

VelocityKrypton

= Molar MassKrypton

Molar MassSulfur Dioxide

Velocity Krypton( ) VelocitySulfur Dioxide

Velocity Krypton

⎛

⎝⎜

⎞

⎠⎟ = VelocityKrypton( ) Molar MassKrypton

Molar MassSulfur Dioxide

⎛

⎝⎜⎜

⎞

⎠⎟⎟

VelocitySulfur Dioxide =

VelocityKrypton( ) Molar MassKrypton( )Molar MassSulfur Dioxide

VelocitySulfur Dioxide =

750 msec

⎛⎝⎜

⎞⎠⎟

84 gmol

⎛

⎝⎜

⎞

⎠⎟

64 gmol

VelocitySulfur Dioxide = 860 meters

sec

2 - AHC - Chapter 14 - Graham's Law Worksheet - Answers - V8

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 gmol

32.0 gmol

Velocity

300. msec

?

msec

⎛⎝⎜

⎞⎠⎟

VelocityFluorine

VelocityMethanol

= Molar MassMethanol

Molar MassFluorine

Cross Multiply To Get The Variable in the Numerator.

VelocityFluorine( ) Molar MassFluorine( ) = VelocityMethanol( ) Molar MassMethanol( )

VelocityFluorine( ) Molar MassFluorine( )Molar MassMethanol

=VelocityMethanol( ) Molar Mass Methanol( )

Molar Mass Methanol

VelocityMethanol =

VelocityFluorine( ) Molar MassFluorine( )Molar MassMethanol

VelocityMethanol =

300. msec

⎛⎝⎜

⎞⎠⎟

38.0 gmol

⎛

⎝⎜

⎞

⎠⎟

32.0 gmol

VelocityMethanol = 327 meters

sec

3 - 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 gmol

131 gmol

Velocity - -

RateCarbon Dioxide

RateXenon

= Molar MassXenon

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

RateXenon

= x

RateCarbon Dioxide

RateXenon

= Molar MassXenon

Molar MassCarbon Dioxide

x =

Molar MassXenon

Molar MassCarbon Dioxide

x =

131 gmol

44.0 gmol

= 1.73

Carbon dioxide will diffuse 1.73 times faster than xenon.

4 - 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 gmol

28.0 gmol

Velocity - -

RateAmmonia

RateNitrogen

= Molar MassNitrogen

Molar MassAmmonia

Since we are solving for the ratio of the rates of diffusion, the left side of the equation can be expressed as one variable:

RateAmmonia

RateNitrogen

= x

x =

Molar MassNitrogen

Molar MassAmmonia

x =

28.0 gmol

17.0 gmol

= 1.28

Ammonia will diffuse 1.28 times faster than nitrogen.

5 - 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 gmol

222 gmol

Velocity

1,000. msec

?

msec

⎛⎝⎜

⎞⎠⎟

VelocityHe

VelocityRn

= Molar MassRn

Molar MassHe

Cross Multiply To Get The Variable in the Numerator.

VelocityHe( ) Molar MassHe( ) = VelocityRn( ) Molar MassRn( )

VelocityHe( ) Molar MassHe( )Molar MassRn

=VelocityRn( ) Molar Mass Rn( )

Molar Mass Rn

VelocityRn =

VelocityHe( ) Molar MassHe( )Molar MassRn

VelocityRn =

1,000. msec

⎛⎝⎜

⎞⎠⎟

4.00 gramsmole

⎛

⎝⎜

⎞

⎠⎟

222 gramsmole

VelocityRn = 134 m

sec

6 - 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 gmol

46.1 gmol

Velocity

500. msec

?

msec

⎛⎝⎜

⎞⎠⎟

VelocityN2

VelocityC2 H5OH

= Molar MassC2H5OH

Molar MassN2

Cross Multiply To Get The Variable in the Numerator.

VelocityN2( ) Molar MassN2( ) = VelocityC2 H5OH( ) Molar MassC2H5OH( )

VelocityN2( ) Molar MassN2( )

Molar MassC2H5OH

= VelocityC2 H5OH( ) Molar Mass C2H5OH( )

Molar Mass C2H5OH

VelocityC2 H5OH = VelocityN2( ) Molar MassN2( )

Molar MassC2H5OH

VelocityC2 H5OH =

500. msec

⎛⎝⎜

⎞⎠⎟

28.0 gramsmole

⎛

⎝⎜

⎞

⎠⎟

46.1 gramsmole

VelocityC2 H5OH = 390. m

sec

7 - 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 gmol

?

gmol

⎛⎝⎜

⎞⎠⎟

Velocity - -

Rate of EffusionHe

Rate of EffusionUnknown

= Molar MassUnknown

Molar MassHe

Rate of EffusionHe

Rate of EffusionUnknown

= 2.0001

∴ 2.000

1 =

Molar MassUnknown

Molar MassHe

Molar MassUnknown = 2.000( ) Molar MassHe( )

Molar MassUnknown = 2.000( ) 4.003 g

mol

⎛

⎝⎜

⎞

⎠⎟

Molar MassUnknown = 2.000( ) 2.001 g

mol

⎛

⎝⎜

⎞

⎠⎟

Molar MassUnknown =

4.002 gmol

Molar MassUnknown( )2

= 4.002 g

mol

⎛

⎝⎜

⎞

⎠⎟

2

Molar MassUnknown = 16.01 g

mol

8 - 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 ?

gmol

⎛⎝⎜

⎞⎠⎟

4.003 gmol

Velocity - -

Rate of EffusionUnknown

Rate of EffusionHe

= Molar MassHe

Molar MassUnknown

Rate of EffusionUnknown

Rate of EffusionHe

= 0.3165

∴ 0.3165 =

Molar MassHe

Molar MassUnknown

Molar MassUnknown =

Molar MassHe

0.3165

Molar MassUnknown =

4.003 gmol

0.3165

Molar MassUnknown =

2.001 gmol

0.3165 =

6.321 gmol

Molar MassUnknown( )2

= 6.321 g

mol

⎛

⎝⎜

⎞

⎠⎟

2

Molar MassUnknown = 39.96 g

mol