   # The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane (CH 4 ) gas is 47.8 mL/min. What is the molar mass of the unknown gas? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 112E
Textbook Problem
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## The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane (CH4) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

Interpretation Introduction

Interpretation: From the given rate of effusion of two gases, molar mass of the unknown gas should be determined.

Concept introduction:

Effusion is used to describe the passage of a gas through a tiny particle into an evacuated chamber.

The rate of effusion is the measure speed at which the gas is transferred to the chamber.

According to Thomas Graham the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

The relative rate of effusion of two gases at the same temperature and pressure are the inverse ratio of the square root of the masses of the gases particles. That is,

Rateofeffusionforgas1Rateofeffusionforgas2=M2M1orRate1Rate2=(M2M1)1/2

M1andM2arethemolarmassesoftwogases

This equation is known as Graham’s law of effusion.

### Explanation of Solution

Explanation

To determine: the molar mass of the unknown gas

The molar mass of the unknown gas M1=63.7g/mol

According to Graham’s law of effusion,

Rate1Rate2=(M2M1)1/2

Here,Gas1=unknowngasGas2=methanegasM2=molarmassof CH4=16.04g/molM1=molarmassofunknowngasRate1Rate2=(M2M1)1/2Rate1=24.0mL/minRate2=47.8mL/minSince,therateofeffusionof particular gastomethaneis24.0:47

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