2. A sample of He gas effused 5.3 times faster than an unknown gas, under the same conditions. Calculate the molecular mass of the unknown gas.
2. A sample of He gas effused 5.3 times faster than an unknown gas, under the same conditions. Calculate the molecular mass of the unknown gas.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Effusion and Molecular Mass Calculation**
**Problem Statement:**
A sample of He gas effused 5.3 times faster than an unknown gas, under the same conditions. Calculate the molecular mass of the unknown gas.
**Explanation:**
Effusion is a process by which gas particles pass through a tiny opening. Graham’s law of effusion states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass (molecular mass).
Mathematically, this relationship can be expressed as:
\[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \]
Where:
- \( Rate_1 \) and \( Rate_2 \) are the effusion rates of gas 1 and gas 2, respectively.
- \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2, respectively.
Given data:
- The effusion rate of Helium (He) gas is 5.3 times faster than that of the unknown gas.
So we have:
\[ \frac{Rate_{He}}{Rate_{unknown}} = 5.3 \]
We need to find the molar mass (molecular mass) of the unknown gas (\( M_{unknown} \)).
Using Graham's law, we substitute the information into the equation:
\[ 5.3 = \sqrt{\frac{M_{unknown}}{M_{He}}} \]
The molar mass of helium (He) is 4 g/mol.
Squaring both sides to solve for \( M_{unknown} \):
\[ (5.3)^2 = \frac{M_{unknown}}{4} \]
\[ 28.09 = \frac{M_{unknown}}{4} \]
\[ M_{unknown} = 28.09 \times 4 \]
\[ M_{unknown} = 112.36 \]
Therefore, the molecular mass of the unknown gas is 112.36 g/mol.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09cc29b2-b8f1-46a4-8709-ee61d656c9a7%2F0c555786-db05-48d2-a6d4-56050d5e6e75%2F8ytmo8n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Effusion and Molecular Mass Calculation**
**Problem Statement:**
A sample of He gas effused 5.3 times faster than an unknown gas, under the same conditions. Calculate the molecular mass of the unknown gas.
**Explanation:**
Effusion is a process by which gas particles pass through a tiny opening. Graham’s law of effusion states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass (molecular mass).
Mathematically, this relationship can be expressed as:
\[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \]
Where:
- \( Rate_1 \) and \( Rate_2 \) are the effusion rates of gas 1 and gas 2, respectively.
- \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2, respectively.
Given data:
- The effusion rate of Helium (He) gas is 5.3 times faster than that of the unknown gas.
So we have:
\[ \frac{Rate_{He}}{Rate_{unknown}} = 5.3 \]
We need to find the molar mass (molecular mass) of the unknown gas (\( M_{unknown} \)).
Using Graham's law, we substitute the information into the equation:
\[ 5.3 = \sqrt{\frac{M_{unknown}}{M_{He}}} \]
The molar mass of helium (He) is 4 g/mol.
Squaring both sides to solve for \( M_{unknown} \):
\[ (5.3)^2 = \frac{M_{unknown}}{4} \]
\[ 28.09 = \frac{M_{unknown}}{4} \]
\[ M_{unknown} = 28.09 \times 4 \]
\[ M_{unknown} = 112.36 \]
Therefore, the molecular mass of the unknown gas is 112.36 g/mol.
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