   Chapter 13.6, Problem 13.9SC ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# trong>Exercise 13.9 A 2 .0 − L flask contains a mixture of nitrogen gas and oxygen gas at 25 ° C . The total pressure of the gaseous mixture is 0 91 atm , and the mixture is known to contain 0.0 5 0  mole of N 2 . Calculate the partial pressure of oxygen and the moles of oxygen present.

Interpretation Introduction

Interpretation:

Partial pressure of oxygen and the moles of oxygen present in the mixture should be determined.

Concept introduction:

The problem can be solved using ideal gas law and Daltons gas law. The total number of moles (moles of nitrogen plus moles of oxygen) can be calculated using ideal gas law as follows

PV=nRTn=PVRT

Where, P is the total pressure of the mixture

V is the volume of the mixture

R is the gas constant

T is the absolute temperature and,

n is the total number of moles of nitrogen and oxygen gas

Now,

n = nN2+ nO2

nO2=n nN2

Once nO2 is calculated, one can use Dalton’s law to find out the partial pressure of oxygen gas in the mixture as follows

PO2= xO2×P

Where PO2 is the partial pressure of oxygen gas in the mixture.

xO2 is the mole fraction of oxygen gas and,

P is the total pressure of the mixture.

Explanation

Here given,

Total pressure of the mixture, P = 0.91 atm

Volume of the mixture, V = 2 L

T = 25° C = (273+25) K = 298 K

Using the ideal gas equation,

PV=nRTn=PVRT=0.91 atm×2 L0.082 L.atm/mol.K×298 Kn=0.0745 mole

So, the total number of moles of nitrogen and oxygen gas is n = 0.0745 mole

That is,

n = nN2+ nO2

nO2=n nN2 = 0

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