Chapter 21, Problem 87PS

### Chemistry & Chemical Reactivity

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

Chapter
Section

### 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 1â€‰L of air and the volume of air required to isolate 1â€‰mol of argon at 1â€‰atm pressure and 25â€‰âˆ˜C is calculated below.

Given:

The gas constant value is 0.0821â€‰atmâ‹…Lâ‹…molâˆ’1â‹…Kâˆ’1.

The atomic weight of argon is 39.9â€‰gâ‹…molâˆ’1.

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

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 0.93â€‰mol in 100â€‰mol of dry air.

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

PAr=0.93â€‰mol100â€‰mol(1â€‰atm)=9.3Ã—10âˆ’3â€‰atm

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

nV=9.3Ã—10âˆ’3â€‰atm(0.0821â€‰atmâ‹…Lâ‹…molâˆ’1â‹…Kâˆ’1)(273â€‰K)=0.41Ã—10âˆ’3â€‰molâ‹…Lâˆ’1

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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