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 152QP

(a)

Interpretation Introduction

Interpretation:

The partial pressure of the water vapour is to be determined.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given Information: Table 9.2

The partial pressure of a gas in a mixture is defined as the force exerted by the molecules of that gas per unit area if it alone occupies the entire volume of the container in which the mixture is kept. It depends upon the mole fraction of the gas. The partial pressure of water vapor at different temperatures is standardized in table 9.2.

At 50.0°C , the partial pressure of water vapor is 92.5 torr .

(b)

Interpretation Introduction

Interpretation:

The partial pressure of oxygen using the ratio of H2O and O2 gases in the molecular level image and Dalton’s law of partial pressure is to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The molecular level image shows that there are O2 gas molecules as 5 times of H2O molecules. So, the partial pressure of the O2 gas is also 5 times of H2O molecules. The partial pressure of water vapor is 92.5 torr . So, the partial pressure of O2 gas is,

PO2=5×92.5 torr=462.5 torr

(c)

Interpretation Introduction

Interpretation:

The total pressure of the gas collected is to be determined.

(c)

Expert Solution
Check Mark

Explanation of Solution

Dalton’s law of partial pressure states that in a mixture, the total pressure exerted by the mixture of gases is equal to the sum of partial pressures exerted by each individual gas.

Ptotal=P1+P2+........+Pn

The partial pressure of water vapor is 92.5 torr and the partial pressure of O2 gas is 462.5 torr . So, the total pressure of the gas collected is determined by using Dalton’s law of partial pressure.

Ptotal=PO2+PH2O=462.5 torr+92.5 torr=555 torr

(d)

Interpretation Introduction

Interpretation:

The number of moles of oxygen gas collected is to be determined.

(d)

Expert Solution
Check Mark

Explanation of Solution

The ideal gas equation states that the pressure P and volume V of the gas is directly proportional to the number of moles n and temperature T of the gas.

PVnTPV=nRT … (1)

Here, R is a universal gas constant.

The volume of the collected gas is 120.0 mL . The partial pressure of O2 gas is 462.5 torr and temperature is 50.0°C .

Convert mL into L units as follows:

1 L=1000 mL1 mL=1 L1000 mL

Convert 120.0 mL into L units as follows:

120 mL=1 L1000 mL×120 mL=0.12 L

Convert temperature units from degree Celsius to Kelvin as follows:

T°C=TK+273.15

Convert temperature 50.0°C into Kelvin as follows:

T=50°C+273.15=323.15 K

Convert pressure units from torr to atm as follows:

1 atm=760 torr1 torr=1 atm760 torr

Convert pressure 462.5 torr into atm as follows:

462.5 torr=1 atm760 torr×462.5 torr=0.60855 atm

Substitute P as 0.60855 atm , V as 0.12 L , R as 0.08206 L atm mol1 K1 and T as 323.15 K in equation (1) as follows:

0.60855 atm×0.12 L=n×0.08206 L atm mol1 K1×323.15 Kn=0.60855 atm×0.12 L0.08206 L atm mol1 K1×323.15 K=0.002754 mol

So, the number of moles of oxygen gas collected is 0.002754 mol .

(e)

Interpretation Introduction

Interpretation:

The number of moles of KClO3 reacted is to be determined.

(e)

Expert Solution
Check Mark

Explanation of Solution

The reaction is as follows:

2KClO3sheat2KCls+3O2g

According to the reaction,

2 mol KClO3 produces=3 mol O21 mol O2 is produced by=2 mol KClO33 mol O2

So, 0.002754 mol O2 is produced by,

0.002754 mol O2=2 mol KClO33 mol O2×0.002754 mol O2=0.00184 mol KClO3

The number of moles of KClO3 reacted is 0.00184 mol .

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Chapter 9 Solutions

EBK INTRODUCTION TO CHEMISTRY

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