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 70A

(a)

Interpretation Introduction

Interpretation: The volume of 835 g of SO3 at STP needs to be determined.

Concept Introduction: The pressure, volume, number of moles, and temperature are related to each other as follows:

  PV=nRT

Here, P is pressure, V is volume, n is the number of moles, R is the Universal gas constant and T is temperature.

(a)

Expert Solution
Check Mark

Explanation of Solution

The mass of SO3 is 835 g. The number of moles can be calculated as follows:

  n=mM

Here, m is mass and M is molar mass.

Molar mass of SO3 is 80.06 g/mol.

Substitute the values,

  n=835 g80.06 g/mol=10.43 mol

Now, at STP the value of temperature and pressure is 273.15 K and 1 atm respectively.

The volume is calculated as follows:

  V=nRTP

Substitute the values,

  V=10.43 mol0.082 L atm/mol K273.15 K1 atm=233.2 L

Thus, a volume of 835 g of SO3 at STP 233.2 L .

(b)

Interpretation Introduction

Interpretation: The mass in grams of a molecule of aspirin needs to be calculated.

Concept Introduction: Mass of a substance can be calculated from the number of moles and molar mass as follows:

  m=n×M

Here, n is the number of moles and M is the molar mass.

(b)

Expert Solution
Check Mark

Explanation of Solution

According to Avogadro’s law, 1 mol of aspirin there are 6.022×1023 molecules thus, the number of moles of 1 molecule of aspirin will be:

  n=1 molecule1 mol6.022×1023 molecules

Or,

  n=1.66×1024 mol

Now, the molar mass of aspirin C9H8O4 is 180.16 g/mol thus, mass can be calculated as follows:

  m=n×M

Substitute the values,

  m=1.66×1024 mol180.16 g/mol=3.0×1022 g

Thus, the mass of 1 molecule of aspirin is 3.0×1022 g.

(c)

Interpretation Introduction

Interpretation: The number of atoms in 5.78 mol of NH4NO3 .

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 number of moles of NH4NO3 is 5.78 mol. Here, NH4NO3 is an ionic compound thus, the representative particles are formula units. The number of atoms can be calculated by multiplying the number of representative particles with the total number of moles of each element present in the compound.

First, calculate the number of formula units as follows:

  N=5.78 mol6.022×1023 formula units1 mol=3.48×1024 formula units

Now, the number of atoms can be calculated as follows:

  N=93.48×1024=3.132×1025

Thus, the number of atoms in 5.78 mol of NH4NO3 is 3.132×1025 atoms.

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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