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 10, Problem 83A

(a)

Interpretation Introduction

Interpretation: The error in the given statement needs to be identified.

Concept Introduction: According to Avogadro’s law, in 1 mol of a substance there are

  6.022×1023 units of that substance.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given statement is “one mole of any substance contains the same number of atoms”. This statement is not correct as one mole of any element contains the same number of atoms. The number of units of a substance present in 1 mol of a substance is the same. There are different types of units (representative particles) for a different substance. For elements, it is atoms; for molecules, it is molecules and for compounds, it is formula units.

Thus, the correct statement would be “one mole of any substance contains the same number of units of that substance.

(b)

Interpretation Introduction

Interpretation: The error in the given statement needs to be identified.

Concept Introduction: According to Avogadro’s law, in 1 mol of a substance there are

  6.022×1023 units of that substance.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given statement is “a mole and a molecule of a substance are identical in amount”. The statement is wrong because for a substance or molecule, in 1 mol there are 6.022×1023 number of molecules. The correct statement would be, in 1 mol of a molecule, there are 6.022×1023 number of molecules.

(c)

Interpretation Introduction

Interpretation: The error in the given statement needs to be identified.

Concept Introduction: According to Avogadro’s law, in 1 mol of a substance there are

  6.022×1023 units of that substance.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given statement is “one molar mass of CO2 contains Avogadro’s number of atoms.” According to Avogadro’s law, in 1 mol of CO2 there are 6.022×1023 number of CO2 molecules. Here, 6.022×1023 is Avogadro’s number. Here, the molar mass of carbon dioxide gives the mass in 1 mol or mass of 6.022×1023 number of CO2 molecules. Therefore, the given statement is wrong.

The correct statement would be one mole of CO2 contains Avogadro’s number of CO2 molecules.

Chapter 10 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 10.1 - Prob. 11LCCh. 10.1 - Prob. 12LCCh. 10.1 - Prob. 13LCCh. 10.1 - Prob. 14LCCh. 10.1 - Prob. 15LCCh. 10.2 - Prob. 16SPCh. 10.2 - Prob. 17SPCh. 10.2 - Prob. 18SPCh. 10.2 - Prob. 19SPCh. 10.2 - Prob. 20SPCh. 10.2 - Prob. 21SPCh. 10.2 - Prob. 22SPCh. 10.2 - Prob. 23SPCh. 10.2 - Prob. 24LCCh. 10.2 - Prob. 25LCCh. 10.2 - Prob. 26LCCh. 10.2 - Prob. 27LCCh. 10.2 - Prob. 28LCCh. 10.2 - Prob. 29LCCh. 10.2 - Prob. 30LCCh. 10.2 - Prob. 31LCCh. 10.2 - Prob. 32LCCh. 10.3 - Prob. 33SPCh. 10.3 - Prob. 34SPCh. 10.3 - Prob. 35SPCh. 10.3 - Prob. 36SPCh. 10.3 - Prob. 37SPCh. 10.3 - Prob. 38SPCh. 10.3 - Prob. 39SPCh. 10.3 - Prob. 40SPCh. 10.3 - Prob. 41SPCh. 10.3 - Prob. 42SPCh. 10.3 - Prob. 43LCCh. 10.3 - Prob. 44LCCh. 10.3 - Prob. 45LCCh. 10.3 - Prob. 46LCCh. 10.3 - Prob. 47LCCh. 10.3 - Prob. 48LCCh. 10.3 - Prob. 49LCCh. 10 - Prob. 50ACh. 10 - Prob. 51ACh. 10 - Prob. 52ACh. 10 - Prob. 53ACh. 10 - Prob. 54ACh. 10 - Prob. 55ACh. 10 - Prob. 56ACh. 10 - Prob. 57ACh. 10 - Prob. 58ACh. 10 - Prob. 59ACh. 10 - Prob. 60ACh. 10 - Prob. 61ACh. 10 - Prob. 62ACh. 10 - Prob. 63ACh. 10 - Prob. 64ACh. 10 - Prob. 65ACh. 10 - Prob. 66ACh. 10 - Prob. 67ACh. 10 - Prob. 68ACh. 10 - Prob. 69ACh. 10 - Prob. 70ACh. 10 - Prob. 71ACh. 10 - Prob. 72ACh. 10 - Prob. 73ACh. 10 - Prob. 74ACh. 10 - Prob. 75ACh. 10 - Prob. 76ACh. 10 - Prob. 77ACh. 10 - Prob. 78ACh. 10 - Prob. 79ACh. 10 - Prob. 80ACh. 10 - Prob. 81ACh. 10 - Prob. 82ACh. 10 - Prob. 83ACh. 10 - Prob. 84ACh. 10 - Prob. 85ACh. 10 - Prob. 86ACh. 10 - Prob. 87ACh. 10 - Prob. 88ACh. 10 - Prob. 89ACh. 10 - Prob. 90ACh. 10 - Prob. 91ACh. 10 - Prob. 92ACh. 10 - Prob. 93ACh. 10 - Prob. 94ACh. 10 - Prob. 95ACh. 10 - Prob. 96ACh. 10 - Prob. 97ACh. 10 - Prob. 98ACh. 10 - Prob. 99ACh. 10 - Prob. 100ACh. 10 - Prob. 101ACh. 10 - Prob. 102ACh. 10 - Prob. 103ACh. 10 - Prob. 104ACh. 10 - Prob. 105ACh. 10 - Prob. 106ACh. 10 - Prob. 108ACh. 10 - Prob. 109ACh. 10 - Prob. 110ACh. 10 - Prob. 111ACh. 10 - Prob. 112ACh. 10 - Prob. 113ACh. 10 - Prob. 114ACh. 10 - Prob. 115ACh. 10 - Prob. 116ACh. 10 - Prob. 117ACh. 10 - Prob. 118ACh. 10 - Prob. 119ACh. 10 - Prob. 120ACh. 10 - Prob. 121ACh. 10 - Prob. 122ACh. 10 - Prob. 123ACh. 10 - Prob. 124ACh. 10 - Prob. 125ACh. 10 - Prob. 126ACh. 10 - Prob. 127ACh. 10 - Prob. 128ACh. 10 - Prob. 1STPCh. 10 - Prob. 2STPCh. 10 - Prob. 3STPCh. 10 - Prob. 4STPCh. 10 - Prob. 5STPCh. 10 - Prob. 6STPCh. 10 - Prob. 7STPCh. 10 - Prob. 8STPCh. 10 - Prob. 9STPCh. 10 - Prob. 10STPCh. 10 - Prob. 11STPCh. 10 - Prob. 12STPCh. 10 - Prob. 13STPCh. 10 - Prob. 14STP
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