### Graham's LawProblems 11-25

Problem #11: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.62 mol/min?

Solution:

Set rate1 = 3.62
Set rate2 = x

The gas that has twice the molar mass is the one whose rate we are trying to determine.

MM2 = 2
MM1 = 1

These molar masses are arbitrary values, we just need MM2 to be twice the value for MM1.

3.62 / x = √(2/1)

x = 2.56 mol/min

Problem #12: Calculate the rate of effusion of NO2 compared to SO2 at the same temperature and pressure.

Solution:

The rates of effusion of two gases are inversely proportional to the square roots of their molar masses -- Graham's law.

We can state Graham's law like this:

rate12 x MM1 = rate22 x MM2

Solve for the unknown

rate2 = √(rate12 x MM1 / MM2)

rate2 = √(12 x 46 g/mol / 64 g/mol)

rate2 = 0.85

The rate of effusion of SO2 is 0.85 times the rate of effusion of NO2, which is logical because SO2 is more massive than NO2, and moves more slowly, on average.

This is not my (the ChemTeam's) solution, but it is rather nice, so I decided to copy it as is. Notice how the solution assigns rate1 to be equal to 1. You might wish to rearrange the writer's Graham's law equation into the one the ChemTeam tends to use.

Problem #13: Assume you have a sample of hydrogen gas containing H2, HD, and D2 that you want to separate into pure components. What are the various ratios of relative rates of effusion?

Solution:

Let us first compare H2 and HD to D2. Since D2 is the heaviest molecule, it is the slowest. D2's rate (which is r2) will be set to 1.

Graham's Law is: r1 over r2 = √MM2 over √MM1

1) H2 : D2

x/1 = √(4/2)

x = 1.414

H2 effuses 1.414 times faster than D2

2) HD : D2

x/1 = √(4/3)

x = 1.155

HD effuses 1.155 times faster than D2

3) Finally, let us compare H2 to HD. This may be solved two different ways:

x/1 = √(3/2)

x = 1.225

or, use a ratio and proportion:

1.414 is to 1.155 as x is to one

x = 1.224

Problem #14: A 3.00 L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane in 25 hours. A 3.00 L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse though the membrane?

Solution:

1) Determine helium's rate of effusion:

1.50 L per 25 hr = 0.0600 L/hr.

Let r2 be the rate for helium. So r1 will be the rate for oxygen in L/hr.

2) Determine oxygen's rate of effusion:

r1/r2 = √[MM2/MM1]

x / 0.0600 = √[4/32]

x = 0.0212132 L/hr

3) Determine time for half of oxygen's 3.00 liters to effuse:

1.50 L divided by 0.0212132 L/hr = 70.7 hrs

Problem #15: At a certain temperature, hydrogen molecules move at an average velocity of 1.84 x 103 m/s. Estimate the molar mass of a gas whose molecules have an average velocity of 311 m/s.

Solution:

r1/r2 = √[MM2/MM1]

1840 / 311 = √[x / 2.02]

Divide, square both sides, multiply by 2.02

x = 70.7 g/mol

Although the question does not ask for the identity of the gas, we could identify it tentatively (based on just the data we have) as Cl2. The molecular weight for chlorine gas is 70.9 g/mol.

Problem #16: An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas.

Solution:

r1/r2 = √[MM2/MM1]

1 / 1.66 = √[x / 44.01]

Divide, square both sides, multiply by 44.01

x = 16.0 g/mol

Although the question does not ask for the identity of the gas, we could identify it tentatively (based on just the data we have) as CH4. The molecular weight for methane gas is 16.043 g/mol.

Problem #17: A sample of hydrogen gas effuse through a porous container 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Solution:

r1/r2 = √[MM2/MM1]

9 / 1 = √[x / 2.02]

x = 163.62 g/mol

I always try and set up these problems so that the x is in the numerator of the right-hand side of the equation. Makes for a slightly easier solution path.

Sorry. I don't know what compound this gas is.

Problem #18: N2 is contaminated with a noble gas.The contaminant effuses at 1.87x N2. What is the noble gas?

Solution:

r1/r2 = √[MM2/MM1]

r1 = N2 = 1
r2 = unk gas = 1.87

1/1.87 = √[x/28.0]

x = 8.00

Comment on this problem:

No noble gas has molar mass = 8.00

However, He = 4.00, so perhaps the desired answer is He2, which does not exist.

The ChemTeam's personal opinion is that the writer (NOT the ChemTeam!) used 14.0 rather than 28.0 in designing the problem. Using 14.0 gives an answer of 4.00, the atomic weight of He.

Problem #19: In an effusion experiment, it was determined that nitrogen gas, N2, effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas?

Solution:

(r1/r2)2 = MM2/MM1

Notice the slightly different formulation of Graham's Law.

r1 = N2 = 1.812
r2 = unk gas = 1

(1.812/1)2 = (x/28.014]

x = 91.98 g/mol

Problem #20: Why are the rates of diffusion of nitrogen gas and carbon monoxide almost identical at the same temperature?

Solution:

1) The speed of a gas is given by:

v = √(3RT/M)

where M is the molecular weight of the gas in kg/mol.

2) The molecular weights are:

N2 = 0.028014 kg/mol
CO = 0.028010 kg/mol

Without solving the formula for the speeds, you should be able to see that the speeds will be nearly identical. Two values (R and T) are going to be same for each gas and the values for M are very nearly the same.

The diffusion rates for nitrogen gas and carbon monoxide gas are very nearly the same at the same temperature because the two substances have very nearly the same molecular weights.

Problem #21: In running a diffusion experiment, ammonia is found to diffuse 30.0 cm during the time hydrogen chloride moves 20.0 cm. Calculate the percentage deviation from Graham's Law.

Solution:

The experimentally determined ratio 30.0/20.0 is 1.50.

1) What ratio is predicted by Graham's Law:

r1 = NH3 = x
r2 = HCl = 1

MM1 = 17.0307 g/mol
MM2 = 36.4609 g/mol

x / 1 = √[36.4609 / 17.0307]

x = 1.463

Ammonia diffuses 1.463 times faster than HCl.

2) Percent deviation is:

(1.50 - 1.463) / 1.50 = 2.47%

Problem #22: A sample of Br2(g) take 10.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane?

Solution:

Let us assume that 1.00 mole of Br2 effuses. Therefore, its rate is 1.00 mol / 10.0 min = 0.100 mol/min

r1 = x
r2 = 0.100 mol/min

MM1 = 39.948 g/mol
MM2 = 159.808 g/mol

x/0.100 = √[159.808/39.948]

x/0.100 = 2.00

x = 0.200 mol/min

1.00 mole of Ar effuses in 5.00 minutes

Problem #23: At a particular pressure and temperature, it takes just 8.256 min for a 4.893 L sample of Ne to effuse through a porous membrane. How long would it take for the same volume of I2 to effuse under the same conditions?

Solution:

r1 = x
r2 = 4.893/8.256 = 0.59266 L/min

MM1 = 253.8 g/mol
MM2 = 20.18 g/mol

x / 0.59266 = √[20.18 / 253.6]

x / 0.59266 = 0.2821

x = 0.16719 L/min

4.893 L / 0.16719 L/min = 29.27 min

Problem #24a: How much faster does U235F6 effuse than U238F6?

Solution:

1) Calculate molecular weights:

U235F6 = 235.04393 + 6(18.99840) = 349.03433

U238F6 = 238.05079 + 6(18.99840) = 352.04119

2) U238F6 is heavier, so:

assign its rate to r2 and set the rate equal to 1

3) Solve Graham's Law:

r1 / r2 = √[MM2 / MM1]

x / 1 = √(352.04119 / 349.03433)

x = 1.0043

U235F6 effuses 1.0043 times faster than U238F6

The following problem is worded so as to use the exact reverse ratio in problem 24a.

Problem #24b: Calculate the ratio of effusion rates for U238F6 and U235F6. Express your answer using five significant figures and as the following ratio:

rate U238F6 / rate U235F6

Solution:

Specifying the form of the ratio forces the numbers to be placed in certain places in Graham's law The U-238 values MUST be associated with r1 and MM1. The U-235 values MUST be associated with r2 and MM2:

r1 / r2 = √[MM2 / MM1]

x / 1 = √(349.03433 / 352.04119)

x = 0.99572

Problem #25: O3 effuses 0.8165 times as fast as O2. What % of the molecules effusing first would be O2?

Solution:

The rate of effusion of O2 is 1.225 times faster than O3, which means that every second there will be 1225 molecules of O2 effusing for every 1000 molecules of O3. Therefore, the percentage of O2 molecules is:

[1225 / (1225 + 1000)] x 100 = 55%

Problem #26: HCl and NH3 diffuse through a tube and a white disc of NH4Cl is formed. Where in the tube?

Solution:

1) Write Graham's Law:

r1 / r2 = √[MM2 / MM1]

(a) assign r2 to be a value of 1.000
(b) Assign r2 to be HCl (the heavier of the two)

3) Solve Graham's Law for r1:

r1 / 1.00 = √[36.4609 / 17.0307]

r1 = 1.463

(a) Those numbers mean that, for every amount the HCl (slower because it weighs more) moves, the NH3 (faster because it weighs less) will move 1.463 times that amount.
(b) If we were to specify a length of tube, we could determine where the disc would form.

5) An example:

Set tube length to be 2.500 meters.

x + 1.463x = 2.500

x = 1.015 m <--- the distance the HCl moves

(1.463) (1.015) = 1.485 m <--- the distance the NH3 moves

Bonus Problem: One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of:

12C16O
12C17O
12C18O

Solution:

1) First, the molar masses are these:

12C16O = 28.00
12C17O = 29.00
12C18O = 30.00

2) Assign a relative rate of 1.000 to 12C18O:

The 18O form of CO is picked because it is the heaviest, therefore the slowest. The other two will have a relative rate slightly greater than 1.

This is done purely for convenience. Any of the three forms of CO could be assigned a rate of 1.000.

3) Compare 12C17O to 12C18O:

r1 / r2 = √(MM2 / MM1)

x / 1 = √(30/29)

x = 1.017

4) Compare 12C16O to 12C18O:

r1 / r2 = √(MM2 / MM1)

x / 1 = √(30/28)

x = 1.035

You may wish to ponder this: name some advantages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.