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Chapter 13: Problem 55

It takes 145 seconds for \(1.00\) milliliter of \(\mathrm{N}_{2}(g)\) to effuse from a certain porous container. Given that it takes 230 seconds for \(1.00\) milliliter of an unknown gas to effuse under the same temperature and pressure, calculate the molecular mass of the unknown gas.

Short Answer

Expert verified
The molecular mass of the unknown gas is approximately 70.85 g/mol.

Step by step solution

01

Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, we can write the law as \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r \) represents the rate of effusion and \( M \) represents molar mass.
02

Identify Given Values and Variables

We know the effusion times for the known gas, \( N_2 \), and the unknown gas: 145 seconds for \( N_2 \) and 230 seconds for the unknown gas. To find the rates of effusion, use the formula for rate \( r = \frac{1}{t} \). Let \( M_1 \) be the molar mass of \( N_2 \) and \( M_2 \) the molar mass of the unknown gas.
03

Calculate Rates of Effusion

The rate of effusion for \( N_2 \) (\( r_1 \)) is \( \frac{1}{145} \), and for the unknown gas (\( r_2 \)) it is \( \frac{1}{230} \).
04

Plug Values into Graham's Law

Using Graham's Law: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \). With \( r_1 = \frac{1}{145} \) and \( r_2 = \frac{1}{230} \), the equation becomes \( \frac{\frac{1}{145}}{\frac{1}{230}} = \sqrt{\frac{M_2}{M_1}} \).
05

Simplify and Solve for \( M_2 \)

Calculate the left-hand side: \( \frac{230}{145} \), which simplifies to approximately 1.5862. We have \( \sqrt{\frac{M_2}{28.02}} = 1.5862 \). Square both sides to eliminate the square root: \( \frac{M_2}{28.02} = 1.5862^2 \). Then calculate \( M_2 = 28.02 \times 1.5862^2 \), solving for \( M_2 \).
06

Final Calculation for Molecular Mass

Perform the final calculation: \( 28.02 \times (1.5862)^2 = 70.85 \). Therefore, the molecular mass of the unknown gas is approximately \( 70.85 \text{ g/mol} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Effusion
The rate of effusion is a measure of how quickly a gas escapes through a small opening into a vacuum. This concept is important when dealing with gaseous substances in scientific experiments. The faster the rate, the more quickly the gas effuses.

In this context, Graham's Law of Effusion helps us understand that the rate of effusion is inversely related to the square root of the molecular mass of the gas. Simply put, lighter gases effuse faster than heavier ones.

To calculate the rate of effusion, use the formula:
  • Rate of effusion, \( r = \frac{1}{t} \)
where \( t \) is the time taken for a specific volume of gas to effuse. This means that if it takes more time for a gas to effuse, its rate of effusion is slower.
Molecular Mass
Molecular mass, often expressed in grams per mole (g/mol), represents the mass of a molecule of a substance. It plays a critical role in the behavior of gases, including their effusion rates.

Understanding the molecular mass of a gas allows scientists to predict how it will behave under different conditions, such as temperature and pressure.
In our exercise, we need the molecular mass of nitrogen, \( \mathrm{N}_{2} \), as a reference. Nitrogen has a molecular mass of approximately 28.02 g/mol.
  • This serves as \( M_1 \) in Graham鈥檚 Law of Effusion.
  • The molecular mass of the unknown gas, \( M_2 \), is what we calculate using the law.
Since Graham's Law shows us that lighter molecules move faster, by knowing nitrogen's molecular mass, we can determine the unknown gas's mass through effusion rates.
Effusion Calculation Steps
To determine the molecular mass of the unknown gas using Graham's Law, we follow a series of straightforward steps.
First, understand the variables that are given:
Effusion times for both \( \mathrm{N}_{2} \) and the unknown gas need to be established as initial inputs.
  • For \( \mathrm{N}_{2} \), the time was 145 seconds.
  • For the unknown gas, the time was 230 seconds.
With these times, calculate the rates of effusion:
  • For \( \mathrm{N}_{2} (r_1) = \frac{1}{145} \)
  • For the unknown gas \( (r_2) = \frac{1}{230} \)
Plug these values into Graham's Law, \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \).
  • Replace \( r_1 \) and \( r_2 \) with their computed values.
  • Substitute \( M_1 \) with nitrogen's molecular mass, 28.02 g/mol.
The result, when solved step by step, will give us the value of \( M_2 \), representing the molecular mass of the unknown gas.

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Most popular questions from this chapter

A website claims that it requires 10000 gallons of air to burn one gallon of gasoline. Using octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), as the chemical formula of gasoline and the fact that air is \(21 \%\) oxygen by volume, calculate the volume of air at \(0^{\circ} \mathrm{C}\) and \(1.00\) bar that is required to burn a gallon of gasoline. Is the information on the website correct? Take the density of octane to be \(0.70 \mathrm{~g} \cdot \mathrm{mL}^{-1}\).

Convert the following temperatures to the Kelvin scale: (a) \(-183^{\circ} \mathrm{C}\) (the melting point of oxygen) (b) \(6000^{\circ} \mathrm{C}\) (temperature at the surface of the sun) (c) \(-269^{\circ} \mathrm{C}\) (the boiling point of helium) (d) \(800^{\circ} \mathrm{C}\) (the melting point of sodium chloride)

While working after school in his high school chemistry laboratory, Joel Hildebrand (Chapter 12 Frontispiece) showed that the formula for a certain oxide of nitrogen published in a college chemistry textbook as \(\mathrm{N}_{2} \mathrm{O}_{2}\) was wrong. He did so by demonstrating that two volumes of the oxide in question combined with one volume of oxygen to form one volume of the brown gas, dinitrogen tetroxide. Use Gay-Lussac's law of combining volumes to identify the correct formula for this oxide of nitrogen.

On a hot, humid day the partial pressure of water vapor in the atmosphere is typically 30 to 40 Torr. Suppose that the partial pressure of water vapor is 35 Torr and that the temperature is \(35^{\circ} \mathrm{C}\). If all the water vapor in a room that measures \(3.0\) meters by \(5.0\) meters by \(6.0\) meters were condensed, how many milliliters of \(\mathrm{H}_{2} \mathrm{O}(l)\) would be obtained?

A mixture of zinc and aluminum with a total mass of \(5.62\) grams reacts completely with hydrochloric acid, liberating \(2.67\) liters of hydrogen gas at \(23^{\circ} \mathrm{C}\) and 773 Torr. Calculate the mass percentage of zinc in the mixture.
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