Problem 49 - Chemistry and Chemical Reactivity

Question

A gas whose molar mass you wish to know effuses through an opening at a rate one-third as fast as that of helium gas. What is the molar mass of the unknown gas?

Step-by-Step Solution

Answer

The molar mass of the unknown gas is 36 g/mol.

Step 1: Understanding the Problem

We know that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham's Law). Mathematically, this is expressed as \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r \) is the rate of effusion and \( M \) is the molar mass. We are given that the unknown gas effuses at a rate one-third that of helium.

Step 2: Identify Known Values

From the problem, \( r_{\text{unknown}} = \frac{1}{3} r_{\text{He}} \). We know that the molar mass of helium (\( M_{\text{He}} \)) is 4 g/mol.

Step 3: Substitute Into Graham's Law

Using Graham's Law \( \frac{r_{\text{unknown}}}{r_{\text{He}}} = \sqrt{\frac{M_{\text{He}}}{M_{\text{unknown}}}} \), and substituting the given effusion rate \( \frac{1}{3} = \sqrt{\frac{4}{M_{\text{unknown}}}} \).

Step 4: Solve the Equation

Square both sides of the equation to eliminate the square root: \( \left( \frac{1}{3} \right)^2 = \frac{4}{M_{\text{unknown}}} \). This results in \( \frac{1}{9} = \frac{4}{M_{\text{unknown}}} \).

Step 5: Cross-Multiply and Solve for M_unknown

Cross-multiply to solve for \( M_{\text{unknown}} \): \( M_{\text{unknown}} = 4 \times 9 \). Thus, \( M_{\text{unknown}} = 36 \).

Key Concepts

Effusion Rate

Effusion rate is an essential concept when studying how gases behave. It describes how quickly a gas can move through a tiny opening or small hole into a vacuum. This process is purely a physical phenomenon and is influenced by several factors.
One of the principal factors that affect effusion is the molar mass of the gas. According to Graham's Law, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means that lighter gases effuse more quickly than heavier gases.

• The formula representing Graham’s Law is as follows: Effusion Rate (r) of Gas 1 / Effusion Rate (r) of Gas 2 = \( \sqrt{\frac{M_2}{M_1}} \) where \( M \) is the molar mass.
• If a gas effuses at a rate one-third that of helium, it means that the unknown gas is heavier than helium.
• This comparison allows for the calculation of the unknown gas's molar mass.

Molar Mass Calculation

Calculating the molar mass of a gas using the rate of effusion involves a series of logical steps based on the relationship established by Graham's Law. In the exercise, you are to find the molar mass of an unknown gas that effuses one-third as quickly as helium.
Let's walk through how this calculation happens.

• Start with the given effusion rate comparison: the unknown gas has an effusion rate that is \( \frac{1}{3} \) compared to that of helium.
• Apply Graham’s Law here, substituting the known values: \( \frac{1}{3} = \sqrt{\frac{4}{M_{\text{unknown}}}} \), because helium's molar mass, \( M_{\text{He}} \), is 4 g/mol.
• To solve for \( M_{\text{unknown}} \), square both sides to remove the square root and make the equation \( \frac{1}{9} = \frac{4}{M_{\text{unknown}}} \).
• Cross-multiply to solve for \( M_{\text{unknown}} \): \( M_{\text{unknown}} = 4 \times 9 = 36 \text{ g/mol} \).

This straightforward series of steps aids in understanding how effusion rate can tell us about a gas's molar mass.

Helium

Helium is a fascinating and unique gas used often in scientific experiments due to its stable properties and low molar mass of 4 g/mol. It is classified as a noble gas, which means it is inert and doesn't react easily with other substances.
Characteristics of helium make it suitable for use in applications like

• filling balloons, due to its low density compared to other gases, which allows it to float easily
• cooling applications in scientific equipment, thanks to its low boiling point and nonreactive nature
• used as a reference in experiments involving effusion due to its consistent molar mass value

In the context of the given problem, helium is used as a standard for comparing the effusion rate due to its universally accepted molar mass.
Understanding helium's properties helps clarify why it is often used in experimental setups and calculations involving gases.