Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 14, Problem 116A
Interpretation Introduction

Interpretation: To determine the significance of the volume 22.4 Liters.

Concept Introduction: This standard volume is given by Avogadro’s hypothesis which states that at same conditions of temperature and pressure, equal volume of different gases will have same number of molecules. This volume is derived from the ideal gas equation when pressure is 1 atm and temperature is 273.15 K.

  V=nRTPn=1molR=0.082057Latmmol-1K-1T=273.15KP=1atm

  V=1mol×0.082057Latmmol-1K-1×273.15K1atm=22.4L

Expert Solution & Answer
Check Mark

Answer to Problem 116A

This volume implies that there is a direct relationship between the amount of the gas and the volume.

Explanation of Solution

This volume is very important because it states that there are same number of molecules at equal volumes of different gases. This volume will give the moles of compound or element present in a gas molecule as 1 mole of any gas will occupy 22.4 L. This volume is also useful in determining the relative molar masses and molecular formula of gaseous substance.

Conclusion

This was given by Avogadro through his hypothesis. He stated that the volume of gas is independent of the mass or size of the gas molecules and also stated that the volume of a gas will be equal at equal temperature and pressure.

Chapter 14 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 14.2 - Prob. 11SPCh. 14.2 - Prob. 12SPCh. 14.2 - Prob. 13SPCh. 14.2 - Prob. 14SPCh. 14.2 - Prob. 15SPCh. 14.2 - Prob. 16SPCh. 14.2 - Prob. 17LCCh. 14.2 - Prob. 18LCCh. 14.2 - Prob. 19LCCh. 14.2 - Prob. 20LCCh. 14.2 - Prob. 21LCCh. 14.2 - Prob. 22LCCh. 14.2 - Prob. 23LCCh. 14.2 - Prob. 24LCCh. 14.2 - Prob. 25LCCh. 14.3 - Prob. 26SPCh. 14.3 - Prob. 27SPCh. 14.3 - Prob. 28SPCh. 14.3 - Prob. 29SPCh. 14.3 - Prob. 30LCCh. 14.3 - Prob. 31LCCh. 14.3 - Prob. 32LCCh. 14.3 - Prob. 33LCCh. 14.3 - Prob. 34LCCh. 14.3 - Prob. 35LCCh. 14.3 - Prob. 36LCCh. 14.4 - Prob. 37SPCh. 14.4 - Prob. 38SPCh. 14.4 - Prob. 39SPCh. 14.4 - Prob. 40LCCh. 14.4 - Prob. 41LCCh. 14.4 - Prob. 42LCCh. 14.4 - Prob. 43LCCh. 14.4 - Prob. 44LCCh. 14.4 - Prob. 45LCCh. 14.4 - Prob. 46LCCh. 14 - Prob. 47ACh. 14 - Prob. 48ACh. 14 - Prob. 49ACh. 14 - Prob. 50ACh. 14 - Prob. 51ACh. 14 - Prob. 52ACh. 14 - Prob. 53ACh. 14 - Prob. 54ACh. 14 - Prob. 55ACh. 14 - Prob. 56ACh. 14 - Prob. 57ACh. 14 - Prob. 58ACh. 14 - Prob. 59ACh. 14 - Prob. 60ACh. 14 - Prob. 61ACh. 14 - Prob. 62ACh. 14 - Prob. 63ACh. 14 - Prob. 64ACh. 14 - Prob. 65ACh. 14 - Prob. 66ACh. 14 - Prob. 67ACh. 14 - Prob. 68ACh. 14 - Prob. 69ACh. 14 - Prob. 70ACh. 14 - Prob. 71ACh. 14 - Prob. 72ACh. 14 - Prob. 73ACh. 14 - Prob. 74ACh. 14 - Prob. 75ACh. 14 - Prob. 76ACh. 14 - Prob. 77ACh. 14 - Prob. 78ACh. 14 - Prob. 79ACh. 14 - Prob. 80ACh. 14 - Prob. 81ACh. 14 - Prob. 82ACh. 14 - Prob. 83ACh. 14 - Prob. 84ACh. 14 - Prob. 85ACh. 14 - Prob. 86ACh. 14 - Prob. 87ACh. 14 - Prob. 88ACh. 14 - Prob. 89ACh. 14 - Prob. 90ACh. 14 - Prob. 91ACh. 14 - Prob. 92ACh. 14 - Prob. 93ACh. 14 - Prob. 94ACh. 14 - Prob. 95ACh. 14 - Prob. 96ACh. 14 - Prob. 97ACh. 14 - Prob. 98ACh. 14 - Prob. 99ACh. 14 - Prob. 100ACh. 14 - Prob. 101ACh. 14 - Prob. 102ACh. 14 - Prob. 106ACh. 14 - Prob. 107ACh. 14 - Prob. 108ACh. 14 - Prob. 109ACh. 14 - Prob. 110ACh. 14 - Prob. 111ACh. 14 - Prob. 112ACh. 14 - Prob. 113ACh. 14 - Prob. 114ACh. 14 - Prob. 115ACh. 14 - Prob. 116ACh. 14 - Prob. 117ACh. 14 - Prob. 118ACh. 14 - Prob. 119ACh. 14 - Prob. 120ACh. 14 - Prob. 121ACh. 14 - Prob. 122ACh. 14 - Prob. 123ACh. 14 - Prob. 1STPCh. 14 - Prob. 2STPCh. 14 - Prob. 3STPCh. 14 - Prob. 4STPCh. 14 - Prob. 5STPCh. 14 - Prob. 6STPCh. 14 - Prob. 7STPCh. 14 - Prob. 8STPCh. 14 - Prob. 9STPCh. 14 - Prob. 10STPCh. 14 - Prob. 11STPCh. 14 - Prob. 12STP
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