Problem 82 - General Chemistry: Principles and Modern Applications

Question

A sample of \(\mathrm{N}_{2}(\mathrm{g})\) effuses through a tiny hole in \(38 \mathrm{s}\) What must be the molar mass of a gas that requires \(64 \mathrm{s}\) to effuse under identical conditions?

Step-by-Step Solution

Answer

The molar mass of the unknown gas, that effuses in 64 seconds under the same conditions as nitrogen, is about 47.3 g/mol.

Step 1: Understanding Graham's law of effusion

Graham’s law of effusion states that, under the same conditions of temperature and pressure, the rate of effusion of different gases is inversely proportional to the square root of their molar masses. Mathematically this can be written as: \( r_1 / r_2 = \sqrt {M_2 / M_1} \) where r is the rate of effusion and M is the molar mass.

Step 2: Apply the rates and molar mass given

Here, the rate of effusion is proportional to 1/time required. So we can express \( r_1 = 1/t_1 \) and \( r_2 = 1/t_2 \). Given in the problem, \( r_1 = 1/38s \) and \( r_2 = 1/64s \). The first gas is nitrogen with a molar mass \( M_1 = 28 g/mol \). Substituting these into Graham’s law equation we get: \( (1/38s) / (1/64s) = \sqrt {M_2 / 28 g/mol} \)

Step 3: Solve the equation

Rearranging to solve for \( M_2 \), we get: \( M_2 = [(64/38)^2] * 28 g/mol \). Solving this, we get \( M_2 = 47.3 g/mol \). So the molar mass of the unknown gas is about 47.3 g/mol.

Key Concepts

Rate of Effusion

The rate of effusion refers to how quickly a gas escapes through a small hole into a vacuum. Understanding this concept is key to predicting how different gases behave under similar conditions.

According to Graham's law of effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means that lighter gases will effuse more quickly than heavier gases. When calculating the rate of effusion, we often express it as \( r = \frac{1}{t} \), where \( t \) is the time it takes for the gas to effuse.

For example, if one gas takes 38 seconds to effuse and another takes 64 seconds, the rate of effusion can help us determine relative parameters such as molar mass.

Molar Mass

Molar mass is a critical concept in understanding effusion and is defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It is expressed in grams per mole (g/mol).

In problems involving effusion, like the sample problem involving nitrogen gas, knowing the molar mass allows us to compare rates of effusion between different gases. Graham's law uses the molar masses of gases to calculate the relative rate of effusion. The key relationship here is that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. So, if a lighter gas (lower molar mass) will effuse faster, a heavier one will take a longer time.

Understanding how to accurately calculate and compare molar masses is essential in solving problems involving gas effusion.

Nitrogen Gas Effusion

Nitrogen gas, with a chemical formula of \(N_2\), is commonly used in examples studying effusion due to its well-known properties and straightforward molar mass calculation. Given that nitrogen has a molar mass of approximately 28 g/mol, it can often serve as a standard for comparing the effusion rates of other gases.

In the provided problem, the effusion of nitrogen is compared with an unknown gas under the same conditions of temperature and pressure. By using Graham’s law, students can find the unknown gas’s molar mass by observing the time each gas takes to effuse. If nitrogen effuses in 38 seconds, and the unknown gas takes 64 seconds, you can use these time measurements inline with Graham’s law to solve for the unknown gas's molar mass.

This process reinforces understanding about how gas properties like molar mass influence effusion and highlights practical applications of theoretical gas laws.