Problems 1-10

**Problem #1:** If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster?

**Solution:**

Set rate_{1}= He = x

Set rate_{2}= Ar = 1The molecular weight of He = 4.00

The molecular weight of Ar = 39.95

Graham's Law is:

r_{1}over r_{2}= √MM_{2}over √MM_{1}

Substituting, we have:

x / 1 = √(39.95 / 4.00)x = 3.16 times as fast.

**Problem #2:** What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen?

**Solution:**

Set rate_{1}= other gas = 1

Set rate_{2}= H_{2}= 50The molecular weight of H

_{2}= 2.02

The molecular weight of the other gas = x.

By Graham's Law (see the answer to question #1), we have:

1 / 50 = √(2.02 / x)

x = 5050 g/mol

**Problem #3:** Two porous containers are filled with hydrogen and neon respectively. Under identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the neon to escape?

**Solution:**

Set rate_{1}= H_{2}= x

Set rate_{2}= Ne = 1The molecular weight of H

_{2}= 2.02

The molecular weight of Ne = 20.18

By Graham's Law:

x / 1 = √(20.18 / 2.02)

x = 3.16

Since the H_{2} escapes 3.16 times as fast as Ne, this calculation determines the amount of Ne leaving in 6 hours:

0.67 / 3.16 = 0.211

Calculate the time needed for half the Ne to escape, knowing that 0.211 escapes in 6 hours:

0.211 / 6 = 0.50 / xx = 14.2 hours

**Problem #4:** If the density of hydrogen is 0.090 g/L and its rate of diffusion is 5.93 times that of chlorine, what is the density of chlorine?

**Solution:**

Set rate_{1}= H_{2}= 5.93

Set rate_{2}= Cl_{2}= 1The molecular weight of H

_{2}= 2.02

The molecular weight of Cl_{2}= x.

By Graham's Law:

5.93 / 1 = √(x / 2.02)

x = 71.03 g/mol

Determine gas density using the molar volume:

71.03 g / 22.414 L = 3.169 g/L

**Problem #5:** How much faster does hydrogen escape through a porous container than sulfur dioxide?

**Solution:**

Set rate_{1}= H_{2}= x

Set rate_{2}= SO_{2}= 1The molecular weight of H

_{2}= 2.02

The molecular weight of SO_{2}= 64.06

By Graham's Law:

x / 1 = √(64.06 / 2.02)

x = 5.63 times as fast

**Problem #6:** Compare the rate of diffusion of carbon dioxide (CO_{2}) & ozone (O_{3}) at the same temperature.

**Solution:**

The molecular weight of CO_{2}= 44.0

The molecular weight of O_{3}= 48.0

Do two things:

set O_{2}rate = 1 (since it is the heavier)

assign it to be r_{2}(since r_{2}is in the denominator)

Graham's Law:

r_{1}over r_{2}= √[(molec. wt_{2}over molec. wt_{1})]

Therefore:

x over 1 = √(48 over 44)

x = 1.04

CO_{2} diffuses 1.04 times as fast as O_{3}

**Problem #7:** 2.278 x 10¯^{4} mol of an unidentified gaseous substance effuses through a tiny hole in 95.70 s. Under identical conditions, 1.738 x 10¯^{4} mol of argon gas takes 81.60 s to effuse. What is the molar mass of the unidentified substance?

**Solution:**

The first thing we need to do is compute the rate of effusion for each gas:

unknown gas: 2.278 x 10¯^{4}mol / 95.70 s = 2.380 x 10¯^{6}mol/s

argon: 1.738 x 10¯^{4}mol / 81.60 s = 2.123 x 10¯^{6}mol/s

Now, we are ready to use Graham's Law. Please note: (1) I will drop the 10¯^{6} from each rate and (2) we know the molar mass of argon from reference sources. Let argon be r_{1}:

2.123 / 2.380 = √(x / 39.948)

Square both sides and solve for x:

0.7957 = x / 39.948

x = 31.786 g/mol (31.79 to 4 significant figures)

**Problem #8:** A compound composed of carbon, hydrogen, and chlorine diffuses through a pinhole 0.411 times as fast as neon. Select the correct molecular formula for the compound:

a) CHCl_{3}

b) CH_{2}Cl_{2}

c) C_{2}H_{2}Cl_{2}

d) C_{2}H_{3}Cl

**Solution:**

Let r_{1} = 0.411; this means r_{2} (the rate of effusion for Ne) equals 1.

Inserting values into Graham's Law yields:

0/411 / 1 = √(20.18 /x )the 20.18 is the atomic weight of Ne.

Squaring both sides gives:

0.16892 = 20.18 / x

Solving for x yields:

x = 119.46 g/mol

Examining the formulas for the possible answers, we see that answer a (CHCl_{3}) gives a molecular weight of about 119.5.

**Problem #9:** Which pair of gases contains one which effuses at twice the rate of the other in the pair?

A. He and Ne

B. Ne and CO_{2}

C. He and CH_{4}

D. CO_{2}and HCl

E. CH_{4}and HCl

**Solution:**

1) We can solve this problem by solving a fake problem:

Set rate_{1}= 2

Set rate_{2}= 1

We now have a gas (rate_{1}) effusing twice as fast as another gas (rate_{2}). We now want to know how much heavier the slower gas is.

Set MM_{1}= 1

Set MM_{2}= x

Our faster gas (rate_{1}) is also our lighter gas (MM_{1}). We know want to know the molar mass (MM_{2}) of our heavier, slower (rate_{2}) gas.

Notice how I set the lighter gas' mass equal to 1. I could have used any number, all I need to know is how many times larger the mass of the slower gas is.

2) Use Graham's Law:

2/1 = √(x/1)x = 4

Our heavier gas is four times heaver than the lighter gas (remember that the lighter is effusing twice as fast as the heavier gas).

3) Answer the question:

We look for a pair of gases in which the heavier gas is four times as heavy as the lighter gas. We find the only choice which satisfies that criterion is answer c.

To continue the 'twice as' theme, you could solve this problem, if you wish:

Oxygen weighs approximately twice as much as methane. Under the same conditions of temperature and pressure, how much faster does a sample of methane effuse than a sample of oxygen?

**Problem #10:** If a molecule of CH_{4} diffuses a distance of 0.530 m from a point source, calculate the distance (in meters) that a molecule of N_{2} would diffuse under the same conditions for the same period of time.

**Solution:**

Assume the gases each diffuse in one second, in order to create a rate.

Set rate_{1}= N_{2}= x

Set rate_{2}= CH_{4}= 0.530 m/sThe molecular weight of N

_{2}= 28.0

The molecular weight of CH_{4}= 16.0

Graham's Law is:

r_{1}over r_{2}= √MM_{2}over √MM_{1}

Substituting, we have:

x / 0.530 = √(16.0 / 28.0)x = 0.400 m/s

**Bonus Problem:** Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of nitrogen at the same temperature and pressure, effuses through this apparatus in 8.50 minutes.

**Solution:**

1) Let us assume 1.00 L of each gas effused. This allows us to calculate rates:

unknown ---> 1.00 L / 6.60 min = 0.1515152 L/min

N_{2}---> 1.00 L / 8.50 min = 0.117647 L/minIt is a common mistake in this type of problem to use the times directly in the solution to the problem. This is a mistake that the ChemTem has made. Oops!

2) Next, we use Graham's Law:

r_{1}/r_{2}= SQRT(MM_{2}/ MM_{1})I will call the unknown r

_{2}because that will put its molar mass in the numerator (making for an easier solution sequence).0.117647 / 0.1515152 = SQRT( x / 28.014)

0.881175 = x / 28.014

x = 24.685 g/mol

3) Since we know everything was at STP, the density is as follows:

a) using molar volume:24.685 g/mol / 22.414 L/mol = 1.10 g/Lb) or, using PV = nRT:

(1.00 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (273 K)n = 0.044638 mol (this is the moles of gas in 1.00 L)

24.685 g/mol times 0.044638 mol = 1.10 g

Since the grams are in 1.00 L, the density is 1.10 g/L