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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Argon is present in dry air to the extent of 0.93% by volume. What quantity of argon is present in 1.00 L of air? If you wanted to isolate 1.00 mol of argon, what volume of air would you need at 1.00 atm pressure and 25 °C?

Interpretation Introduction

Interpretation:

To calculate the quantity of argon present in 1L of air and to calculate the volume of air required to isolate 1mol of argon at 1atm pressure and 25C.

Concept introduction:

The pressure exerted by an individual gas in a mixture is known as its partial pressure.

Dalton’s law of partial pressure states that the total pressure of a mixture of gases is equal to the sum of the partial pressure of the component gases;

PTotal=P1+P2+P3+...

Dalton’s law can also be expressed using the mole fraction of gases, x;

P1=x(PTotal) (1)

The Ideal gas equation is written as,

PV=nRT (2)

Here,

The pressure of the gas is P

The volume of the gas is V

The gas constant is R

The total number of moles is n

The temperature of the system is T

Explanation

The quantity of argon present in 1L of air and the volume of air required to isolate 1mol of argon at 1atm pressure and 25C is calculated below.

Given:

The gas constant value is 0.0821atmLmol1K1.

The atomic weight of argon is 39.9gmol1.

Argon is present in dry air to the extent of 0.93% by volume. This means in 100L of air, the volume of argon is 0.93L.

From the ideal gas equation, the volume of gas is directly proportional to the total number of moles. Therefore, the number of moles of argon is 0.93mol in 100mol of dry air.

The partial pressure of argon gas at 1atm pressure is calculated from equation (1),

PAr=0.93mol100mol(1atm)=9.3×103atm

Substitute the value of partial pressure of argon in equation (2) to calculate the quantity of argon at 273K.

nV=9.3×103atm(0.0821atmLmol1K1)(273K)=0.41×103molL1

The number of moles of argon is equal to the ratio of the mass of the argon to the atomic mass of argon

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Chapter 21 Solutions

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Sect-21.11 P-2.4ACPSect-21.11 P-2.5ACPCh-21 P-1PSCh-21 P-2PSCh-21 P-3PSCh-21 P-4PSCh-21 P-5PSCh-21 P-6PSCh-21 P-7PSCh-21 P-8PSCh-21 P-9PSCh-21 P-10PSCh-21 P-11PSCh-21 P-12PSCh-21 P-13PSCh-21 P-14PSCh-21 P-15PSCh-21 P-16PSCh-21 P-17PSCh-21 P-18PSCh-21 P-19PSCh-21 P-20PSCh-21 P-21PSCh-21 P-22PSCh-21 P-23PSCh-21 P-24PSCh-21 P-25PSCh-21 P-26PSCh-21 P-27PSCh-21 P-28PSCh-21 P-29PSCh-21 P-30PSCh-21 P-31PSCh-21 P-32PSCh-21 P-33PSCh-21 P-34PSCh-21 P-35PSCh-21 P-36PSCh-21 P-37PSCh-21 P-38PSCh-21 P-39PSCh-21 P-40PSCh-21 P-41PSCh-21 P-42PSCh-21 P-43PSCh-21 P-44PSCh-21 P-45PSCh-21 P-46PSCh-21 P-47PSCh-21 P-48PSCh-21 P-49PSCh-21 P-50PSCh-21 P-51PSCh-21 P-52PSCh-21 P-53PSCh-21 P-54PSCh-21 P-55PSCh-21 P-56PSCh-21 P-57PSCh-21 P-58PSCh-21 P-59PSCh-21 P-60PSCh-21 P-61PSCh-21 P-62PSCh-21 P-63PSCh-21 P-64PSCh-21 P-65PSCh-21 P-66PSCh-21 P-67PSCh-21 P-68PSCh-21 P-69PSCh-21 P-70PSCh-21 P-71PSCh-21 P-72PSCh-21 P-73PSCh-21 P-74PSCh-21 P-75PSCh-21 P-76PSCh-21 P-77PSCh-21 P-78PSCh-21 P-79PSCh-21 P-80PSCh-21 P-81PSCh-21 P-82PSCh-21 P-83PSCh-21 P-84PSCh-21 P-85PSCh-21 P-86PSCh-21 P-87PSCh-21 P-88PSCh-21 P-89GQCh-21 P-90GQCh-21 P-91GQCh-21 P-92GQCh-21 P-93GQCh-21 P-94GQCh-21 P-95GQCh-21 P-96GQCh-21 P-97GQCh-21 P-98GQCh-21 P-99GQCh-21 P-100GQCh-21 P-101GQCh-21 P-102GQCh-21 P-103GQCh-21 P-105GQCh-21 P-106GQCh-21 P-107GQCh-21 P-108GQCh-21 P-110GQCh-21 P-111GQCh-21 P-112GQCh-21 P-113GQCh-21 P-114GQCh-21 P-115ILCh-21 P-116ILCh-21 P-117ILCh-21 P-118ILCh-21 P-119ILCh-21 P-120ILCh-21 P-121SCQCh-21 P-122SCQCh-21 P-123SCQCh-21 P-124SCQCh-21 P-125SCQCh-21 P-126SCQCh-21 P-127SCQCh-21 P-128SCQCh-21 P-129SCQCh-21 P-130SCQCh-21 P-131SCQ

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