EBK INTRODUCTION TO CHEMISTRY
EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 9, Problem 101QP

(a)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is 1.785 g L1 .

(a)

Expert Solution
Check Mark

Explanation of Solution

The molar mass of a gas at STP can be determined by using the value of its density. The density ρ of a gas is equal to the amount or mass m of the gas per unit volume V .

ρ=mV …… (1)

The density of the gas is 1.785 g L1 and, at STP, the volume of the gas is 22.414 L .

Substitute the given values in the above expression:

1.785 g L1=m22.414 Lm=1.785 g L1×22.414 L=40.01 g

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

n=massmolar mass

At STP, the number of moles of a gas is 1 mol and the mass of the gas is 40.01 g . So, the molar mass of the gas can be determined by substituting the values in the above expression.

1 mol=40.01 gmolar massmolar mass=40.01 g mol1

(b)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is 1.340 g L1 .

(b)

Expert Solution
Check Mark

Explanation of Solution

The density of the gas is 1.340 g L1 and, at STP, the volume of the gas is 22.414 L . Determine the mass of the gas by substituting the values in equation 1 .

1.340 g L1=m22.414 Lm=1.340 g L1×22.414 L=30.03 g

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

n=massmolar mass

At STP, the number of moles of the gas is 1 mol and the mass of the gas is 30.03 g . So, the molar mass of gas can be determined by substituting the values in the above expression.

1 mol=30.03 gmolar massmolar mass=30.03 g mol1

(c)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is 2.052 g L1 .

(c)

Expert Solution
Check Mark

Explanation of Solution

The density of the gas is 2.052 g L1 and, at STP, the volume of the gas is 22.414 L . Determine the mass of the gas by substituting the values in equation 1 .

2.052 g L1=m22.414 Lm=2.052 g L1×22.414 L=45.99 g

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

n=massmolar mass

At STP, the number of moles of the gas is 1 mol and the mass of the gas is 45.99 g . So, the molar mass of the gas can be determined by substituting the values in the above expression.

1 mol=45.99 gmolar massmolar mass=45.99 g mol1

(d)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is 0.905 g L1 .

(d)

Expert Solution
Check Mark

Explanation of Solution

The density of the gas is 0.905 g L1 and, at STP, the volume of the gas is 22.414 L . Determine the mass of the gas by substituting the values in equation 1 .

0.905 g L1=m22.414 Lm=0.905 g L1×22.414 L=20.3 g

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

n=massmolar mass

At STP, the number of moles of the gas is 1 mol and the mass of the gas is 20.03 g . So, the molar mass of the gas can be determined by substituting the values in the above expression.

1 mol=20.3 gmolar massmolar mass=20.3 g mol1

(e)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is 0.714 g L1 .

(e)

Expert Solution
Check Mark

Explanation of Solution

The density of the gas is 0.714 g L1 and, at STP, the volume of the gas is 22.414 L . Determine the mass of the gas by substituting the values in equation 1 .

0.714 g L1=m22.414 Lm=0.714 g L1×22.414 L=16 g

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

n=massmolar mass

At STP, the number of moles of the gas is 1 mol and the mass of the gas is 16 g . So, the molar mass of the gas can be determined by substituting the values in the above expression.

1 mol=16 gmolar massmolar mass=16 g mol1

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Chapter 9 Solutions

EBK INTRODUCTION TO CHEMISTRY

Ch. 9 - Prob. 6PPCh. 9 - Prob. 7PPCh. 9 - Prob. 8PPCh. 9 - Prob. 9PPCh. 9 - Prob. 10PPCh. 9 - Prob. 11PPCh. 9 - Prob. 12PPCh. 9 - Prob. 13PPCh. 9 - Prob. 14PPCh. 9 - Prob. 15PPCh. 9 - Prob. 16PPCh. 9 - Prob. 17PPCh. 9 - Prob. 18PPCh. 9 - Prob. 1QPCh. 9 - Prob. 2QPCh. 9 - Prob. 3QPCh. 9 - Prob. 4QPCh. 9 - A series of organic compounds called the alkanes...Ch. 9 - Prob. 6QPCh. 9 - Prob. 7QPCh. 9 - Prob. 8QPCh. 9 - Prob. 9QPCh. 9 - Prob. 10QPCh. 9 - Prob. 11QPCh. 9 - Prob. 12QPCh. 9 - Prob. 13QPCh. 9 - Prob. 14QPCh. 9 - Prob. 15QPCh. 9 - Prob. 16QPCh. 9 - Prob. 17QPCh. 9 - Prob. 18QPCh. 9 - Prob. 19QPCh. 9 - Prob. 20QPCh. 9 - Prob. 21QPCh. 9 - Prob. 22QPCh. 9 - Prob. 23QPCh. 9 - Prob. 24QPCh. 9 - Prob. 25QPCh. 9 - Prob. 26QPCh. 9 - Prob. 27QPCh. 9 - Prob. 28QPCh. 9 - Prob. 29QPCh. 9 - Prob. 30QPCh. 9 - Prob. 31QPCh. 9 - Prob. 32QPCh. 9 - Prob. 33QPCh. 9 - Prob. 34QPCh. 9 - Prob. 35QPCh. 9 - Prob. 36QPCh. 9 - Prob. 37QPCh. 9 - Prob. 38QPCh. 9 - Prob. 39QPCh. 9 - Prob. 40QPCh. 9 - Prob. 41QPCh. 9 - Prob. 42QPCh. 9 - Prob. 43QPCh. 9 - Prob. 44QPCh. 9 - Prob. 45QPCh. 9 - Prob. 46QPCh. 9 - Prob. 47QPCh. 9 - Prob. 48QPCh. 9 - Prob. 49QPCh. 9 - Prob. 50QPCh. 9 - Prob. 51QPCh. 9 - Prob. 52QPCh. 9 - Prob. 53QPCh. 9 - Prob. 54QPCh. 9 - Prob. 55QPCh. 9 - Prob. 56QPCh. 9 - Prob. 57QPCh. 9 - Prob. 58QPCh. 9 - Prob. 59QPCh. 9 - Prob. 60QPCh. 9 - Prob. 61QPCh. 9 - Prob. 62QPCh. 9 - Prob. 63QPCh. 9 - Prob. 64QPCh. 9 - Prob. 65QPCh. 9 - Prob. 66QPCh. 9 - Prob. 67QPCh. 9 - Prob. 68QPCh. 9 - Prob. 69QPCh. 9 - Prob. 70QPCh. 9 - Prob. 71QPCh. 9 - Prob. 72QPCh. 9 - Prob. 73QPCh. 9 - Prob. 74QPCh. 9 - Prob. 75QPCh. 9 - Prob. 76QPCh. 9 - Prob. 77QPCh. 9 - Prob. 78QPCh. 9 - Prob. 79QPCh. 9 - Prob. 80QPCh. 9 - Prob. 81QPCh. 9 - Prob. 82QPCh. 9 - Prob. 83QPCh. 9 - Prob. 84QPCh. 9 - Prob. 85QPCh. 9 - Prob. 86QPCh. 9 - Prob. 87QPCh. 9 - Prob. 88QPCh. 9 - Prob. 89QPCh. 9 - Prob. 90QPCh. 9 - Prob. 91QPCh. 9 - Prob. 92QPCh. 9 - Prob. 93QPCh. 9 - Prob. 94QPCh. 9 - Prob. 95QPCh. 9 - Prob. 96QPCh. 9 - Prob. 97QPCh. 9 - Prob. 98QPCh. 9 - Prob. 99QPCh. 9 - Prob. 100QPCh. 9 - Prob. 101QPCh. 9 - Prob. 102QPCh. 9 - Prob. 103QPCh. 9 - Prob. 104QPCh. 9 - Prob. 105QPCh. 9 - Prob. 106QPCh. 9 - Prob. 107QPCh. 9 - Prob. 108QPCh. 9 - Prob. 109QPCh. 9 - Prob. 110QPCh. 9 - Prob. 111QPCh. 9 - Prob. 112QPCh. 9 - Prob. 113QPCh. 9 - Prob. 114QPCh. 9 - Prob. 115QPCh. 9 - Prob. 116QPCh. 9 - Prob. 117QPCh. 9 - Prob. 118QPCh. 9 - Prob. 119QPCh. 9 - Prob. 120QPCh. 9 - Prob. 121QPCh. 9 - Prob. 122QPCh. 9 - Prob. 123QPCh. 9 - Prob. 124QPCh. 9 - Prob. 125QPCh. 9 - Prob. 126QPCh. 9 - Prob. 127QPCh. 9 - Prob. 128QPCh. 9 - Prob. 129QPCh. 9 - Prob. 130QPCh. 9 - Prob. 131QPCh. 9 - Prob. 132QPCh. 9 - Prob. 133QPCh. 9 - Prob. 134QPCh. 9 - Prob. 135QPCh. 9 - Prob. 136QPCh. 9 - Prob. 137QPCh. 9 - Prob. 138QPCh. 9 - Prob. 139QPCh. 9 - Prob. 140QPCh. 9 - Prob. 141QPCh. 9 - Prob. 142QPCh. 9 - Prob. 143QPCh. 9 - Prob. 144QPCh. 9 - Prob. 145QPCh. 9 - Prob. 146QPCh. 9 - Prob. 147QPCh. 9 - Prob. 148QPCh. 9 - Prob. 149QPCh. 9 - Prob. 150QPCh. 9 - Prob. 151QPCh. 9 - Prob. 152QPCh. 9 - Prob. 153QPCh. 9 - Prob. 154QPCh. 9 - Prob. 155QPCh. 9 - Prob. 156QPCh. 9 - Prob. 157QPCh. 9 - Prob. 158QPCh. 9 - Prob. 159QPCh. 9 - Prob. 160QPCh. 9 - Prob. 161QPCh. 9 - Prob. 162QPCh. 9 - Prob. 163QPCh. 9 - Prob. 164QPCh. 9 - Prob. 165QPCh. 9 - Butane burns with oxygen according to the...Ch. 9 - Prob. 167QP
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