Problems 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 rate_{1}= 3.62

Set rate_{2}= x

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

MM_{2}= 2

MM_{1}= 1

These molar masses are arbitrary values, we just need MM_{2} to be twice the value for MM_{1}.

3.62 / x = √(2/1)x = 2.56 mol/min

**Problem #12:** Calculate the rate of effusion of NO_{2} compared to SO_{2} 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:rate

_{1}^{2}x MM_{1}= rate_{2}^{2}x MM_{2}Solve for the unknown

rate_{2}= √(rate_{1}^{2}x MM_{1}/ MM_{2})rate

_{2}= √(1^{2}x 46 g/mol / 64 g/mol)rate

_{2}= 0.85The rate of effusion of SO

_{2}is 0.85 times the rate of effusion of NO_{2}, which is logical because SO_{2}is more massive than NO_{2}, 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 rate_{1} 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 H_{2}, HD, and D_{2} that you want to separate into pure components. What are the various ratios of relative rates of effusion?

**Solution:**

Let us first compare H_{2} and HD to D_{2}. Since D_{2} is the heaviest molecule, it is the slowest. D_{2}'s rate (which is r_{2}) will be set to 1.

Graham's Law is: r_{1} over r_{2} = √MM_{2} over √MM_{1}

1) H_{2} : D_{2}

x/1 = √(4/2)x = 1.414

H

_{2}effuses 1.414 times faster than D_{2}

2) HD : D_{2}

x/1 = √(4/3)x = 1.155

HD effuses 1.155 times faster than D

_{2}

3) Finally, let us compare H_{2} 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 onex = 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 r

_{2}be the rate for helium. So r_{1}will be the rate for oxygen in L/hr.

2) Determine oxygen's rate of effusion:

r_{1}/r_{2}= √[MM_{2}/MM_{1}]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 10^{3} m/s. Estimate the molar mass of a gas whose molecules have an average velocity of 311 m/s.

**Solution:**

r_{1}/r_{2}= √[MM_{2}/MM_{1}]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 Cl

_{2}. The molecular weight for chlorine gas is 70.9 g/mol.

**Problem #16:** An unknown gas effuses 1.66 times more rapidly than CO_{2}. What is the molar mass of the unknown gas.

**Solution:**

r_{1}/r_{2}= √[MM_{2}/MM_{1}]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 CH

_{4}. 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:**

r_{1}/r_{2}= √[MM_{2}/MM_{1}]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:** N_{2} is contaminated with a noble gas.The contaminant effuses at 1.87x N_{2}. What is the noble gas?

**Solution:**

r_{1}/r_{2}= √[MM_{2}/MM_{1}]r

_{1}= N_{2}= 1

r_{2}= unk gas = 1.871/1.87 = √[x/28.0]

x = 8.00

Comment on this problem:

No noble gas has molar mass = 8.00However, He = 4.00, so perhaps the desired answer is He

_{2}, 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, N_{2}, effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas?

**Solution:**

(r_{1}/r_{2})^{2}= MM_{2}/MM_{1}Notice the slightly different formulation of Graham's Law.

r

_{1}= N_{2}= 1.812

r_{2}= 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:

N_{2}= 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:

r_{1}= NH_{3}= x

r_{2}= HCl = 1MM

_{1}= 17.0307 g/mol

MM_{2}= 36.4609 g/molx / 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 Br_{2}(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 Br_{2}effuses. Therefore, its rate is 1.00 mol / 10.0 min = 0.100 mol/minr

_{1}= x

r_{2}= 0.100 mol/minMM

_{1}= 39.948 g/mol

MM_{2}= 159.808 g/molx/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 I_{2} to effuse under the same conditions?

**Solution:**

r_{1}= x

r_{2}= 4.893/8.256 = 0.59266 L/minMM

_{1}= 253.8 g/mol

MM_{2}= 20.18 g/molx / 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 U^{235}F_{6} effuse than U^{238}F_{6}?

**Solution:**

1) Calculate molecular weights:

U^{235}F_{6}= 235.04393 + 6(18.99840) = 349.03433U

^{238}F_{6}= 238.05079 + 6(18.99840) = 352.04119

2) U^{238}F_{6} is heavier, so:

assign its rate to r_{2}and set the rate equal to 1

3) Solve Graham's Law:

r_{1}/ r_{2}= √[MM_{2}/ MM_{1}]x / 1 = √(352.04119 / 349.03433)

x = 1.0043

U

^{235}F_{6}effuses 1.0043 times faster than U^{238}F_{6}

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 U^{238}F_{6} and U^{235}F_{6}. Express your answer using five significant figures and as the following ratio:

rate U^{238}F_{6}/ rate U^{235}F_{6}

**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 r_{1}and MM_{1}. The U-235 values MUST be associated with r_{2}and MM_{2}:r

_{1}/ r_{2}= √[MM_{2}/ MM_{1}]x / 1 = √(349.03433 / 352.04119)

x = 0.99572

**Problem #25:** O_{3} effuses 0.8165 times as fast as O_{2}. What % of the molecules effusing first would be O_{2}?

**Solution:**

The rate of effusion of O_{2} is 1.225 times faster than O_{3}, which means that every second there will be 1225 molecules of O_{2} effusing for every 1000 molecules of O_{3}. Therefore, the percentage of O_{2} molecules is:

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

**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:

^{12}C^{16}O^{12}C^{17}O^{12}C^{18}O

**Solution:**

1) First, the molar masses are these:

^{12}C^{16}O = 28.00^{12}C^{17}O = 29.00^{12}C^{18}O = 30.00

2) Assign a relative rate of 1.000 to ^{12}C^{18}O:

The^{18}O 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 ^{12}C^{17}O to ^{12}C^{18}O:

r_{1}/ r_{2}= √(MM_{2}/ MM_{1})x / 1 = √(30/29)

x = 1.017

4) Compare ^{12}C^{16}O to ^{12}C^{18}O:

r_{1}/ r_{2}= √(MM_{2}/ MM_{1})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.