Chapter 10, Problem 83GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Carbon dioxide, CO2, was shown lo effuse through a porous plate at the rate of 0.033 mol/min. The same quantity of an unknown gas, 0.033 moles, is found to effuse through the same porous barrier in 104 seconds. Calculate the molar mass of the unknown gas.

Interpretation Introduction

Interpretation:

The molar mass of the unknown gas has to be calculated.

Concept introduction:

Effusion:

The movement of gas particles through a small hole is defined as effusion.  The rate of effusion for a gas is inversely proportional to the square root of the mass of gas particles.

The root mean square velocity μ is defined as the measure of velocity of particle in gas.  It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

μrms=(3RTM)1/2 (1)

(gas constant)R=8.314JKmolM=Molarmass

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Explanation

Given,

â€‚Â rateÂ ofÂ CO2â€Šâ€Š=â€Šâ€Š0.033mol/minmassÂ ofÂ unknownÂ gasâ€Šâ€Š=Â â€Šâ€Š0.033molesrateÂ ofÂ unknownÂ gasÂ =104s

The molar mass of helium gas is 4Â g/mol

Using Grahamâ€™s Law of effusion the molar mass of the unknown gas is calculated as follows,

â€‚Â rateÂ ofÂ CO2Â rateÂ ofÂ unkonwnÂ gasÂ =Â molarÂ massÂ ofÂ unknownmolarÂ massÂ ofÂ CO20.033mol/min104Â sÂ =Â molarÂ massÂ ofÂ unknown44molarÂ massÂ ofÂ unknownâ€Šâ€Š=â€Šâ€Š3Ã—44â€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š

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