   # The halothane-oxygen mixture described in this Example is placed in a 5.00-L tank at 25.0 °C. What is the total pressure (in mm Hg) of the gas mixture in the tank? What are the partial pressures (in mm Hg) of the gases? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 10.5, Problem 1CYU
Textbook Problem
391 views

## The halothane-oxygen mixture described in this Example is placed in a 5.00-L tank at 25.0 °C. What is the total pressure (in mm Hg) of the gas mixture in the tank? What are the partial pressures (in mm Hg) of the gases?

Interpretation Introduction

Interpretation:

The partial pressures of halothane vapour and oxygen gas and total pressure of the gas mixture has to be given if the halothane-oxygen mixture in example-10 is placed in a 5 L tank.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture.
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

xa=nantotal

• pa=xa×Ptotal

Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

### Explanation of Solution

The partial pressures of given amount of halothane vapour and oxygen gas in the mixture and also the total pressure of the gas mixture.

Using the data from example-10 we could find out the number of moles of halothane and oxygen vapour.  [In example-10 it is given that 15gofhalothaneand23.5goxygenispresent ]

Given that:

V=5LT=25°C=298K

Numberofmoles=massmolarmass

No.ofmolesofC2HBrClF3=15g197.4g=0.07599molNo.ofmolesofO2=23.5g32g=0.7344mol

Now we have number of moles, volume and temperature and using these we can find out the pressure.

PV=nRT

Phalothane×V=nRTPhalothane=nRTV=0.0759×0.082L.atm/mol.K×298K5L=0.3719atm1atm=760mmHgPhalothane=0

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