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Chapter 10: Problem 88

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ; \mathrm{in}\) other words, rate is the amount that diffuses over the time it takes to diffuse.)

Short Answer

Expert verified
The molar mass of the unknown gas is approximately 348.58 g/mol.

Step by step solution

01

Understand Graham's Law

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, it can be written as: \[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \]where \( \text{Rate}_1 \) and \( \text{Rate}_2 \) are the effusion rates of two gases, and \( M_1 \) and \( M_2 \) are their molar masses, respectively.
02

Calculate Effusion Rates

The rate of effusion for a gas can be calculated using the formula:\[ \text{Rate} = \frac{\text{Volume}}{\text{Time}} \]For the unknown gas, the rate is:\[ \text{Rate}_{\text{unknown}} = \frac{1.0\,\text{L}}{105\,\text{s}} \approx 0.00952\,\text{L/s} \]For \( \mathrm{O}_2 \) gas, the rate is:\[ \text{Rate}_{\mathrm{O}_2} = \frac{1.0\,\text{L}}{31\,\text{s}} \approx 0.03226\,\text{L/s} \]
03

Apply Graham鈥檚 Law to Find Molar Mass

Using Graham's Law:\[ \frac{0.00952}{0.03226} = \sqrt{\frac{32.00}{M_1}} \]Solve this equation for \( M_1 \), where \( 32.00 \) g/mol is the molar mass of \( \mathrm{O}_2 \).
04

Solve the Molar Mass Equation

First, calculate the ratio:\[ \left(\frac{0.00952}{0.03226}\right)^2 = \frac{32.00}{M_1} \]\[ 0.0918 = \frac{32.00}{M_1} \]Rearrange to find \( M_1 \):\[ M_1 = \frac{32.00}{0.0918} \approx 348.58\,\text{g/mol} \]
05

Conclude Molar Mass

The molar mass of the unknown gas, calculated using Graham鈥檚 law, is approximately \( 348.58 \) g/mol.

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Key Concepts

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

Effusion Rates
Effusion is the process by which gas molecules escape through a tiny hole into a vacuum, and it plays a key role in understanding various properties of gases. 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. This means that lighter gases effuse faster than heavier gases.

To find the effusion rate, we use the formula:
  • Effusion Rate = Volume / Time
In this exercise, for the unknown gas, the effusion rate is calculated as approximately 0.00952 L/s. For oxygen, it is approximately 0.03226 L/s. These values indicate that the unknown gas diffuses slower than oxygen, suggesting the unknown gas is heavier. This thought process is crucial for determining the molar mass of gases based on their effusion behaviour.
Molar Mass Calculation
Calculating the molar mass of an unknown gas is an important application of Graham's Law. Using the rates derived, we apply Graham's Law to find the molar mass of the unknown gas in relation to a known gas such as oxygen. We use the equation:
  • \( \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \)
Here, \( \text{Rate}_1 \) and \( \text{Rate}_2 \) are the rates of the unknown gas and oxygen respectively.

By rearranging and solving the equation \( \frac{0.00952}{0.03226} = \sqrt{\frac{32.00}{M_1}} \), we find the molar mass of the unknown gas to be approximately 348.58 g/mol. This calculation highlights the inverse relationship between gas effusion rates and molar masses, helping us deduce the molecular weight of an unknown gas sample.
Gas Laws
Gas laws are fundamental in understanding how gases behave under different conditions. Graham's Law of Effusion is one aspect of these laws, providing insight into how effusion rates depend on molar masses. The broader set of gas laws, such as Boyle's Law, Charles's Law, and Avogadro's Law, also help predict gas behavior in various environments.

Understanding gas laws involves:
  • Predicting how gas volume changes with pressure and temperature
  • Calculating the amount of gas needed to exert a specific pressure
  • Applying gas constants to solve related chemical problems
In this particular exercise, applying Graham's Law allows us to deduce molecular properties from observed effusion rates. Mastering these laws enhances our ability to approach complex chemical reactions and predict the behavior of gas under experimental conditions.

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Most popular questions from this chapter

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000-megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, \(101.3 \mathrm{kPa}\), and \(27^{\circ} \mathrm{C}\), calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. \((\mathbf{b})\) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and \(12.16 \mathrm{MPa}\) and a density of \(1.2 \mathrm{~g} / \mathrm{cm}^{3},\) what volume does it possess? \((\mathbf{c})\) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(7.09 \mathrm{MPa},\) what volume does it occupy?

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and \(99.3 \mathrm{kPa}\), how many moles of \(\mathrm{O}_{2}\) are present? (The (b) How many mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

Determine whether each of the following changes will increase, decrease, or not affect the rate with which gas molecules collide with the walls of their container: (a) increasing the volume of the container, \((\mathbf{b})\) increasing the temperature, (c) increasing the molar mass of the gas.

The highest barometric pressure ever recorded was 823.7 torr at Agata in Siberia, Russia on December 31,1968 . Convert this pressure to \((\mathbf{a})\) atm, (b) \(\mathrm{mm} \mathrm{Hg}\) (c) pascals, (d) bars, (e) psi.
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