1 The Mole Concept 2.1The Mole 2.2Molar Volume and Avogadro’s Law 2.3Ideal Gas Equation 2.4 Determination of Molar Mass 2.5Dalton’s Law of Partial Pressures

1

The Mole ConceptThe Mole Concept

2.12.1 The MoleThe Mole

2.22.2 Molar Volume and Avogadro’s LawMolar Volume and Avogadro’s Law

2.32.3 Ideal Gas EquationIdeal Gas Equation

2.4 2.4 Determination of Molar MassDetermination of Molar Mass

2.52.5 Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

22

2

2.2.11 The MoleThe Mole

3P. 3 / 66

A burrowing mammal with fossorial forefeet

A small congenital pigmented spot on the skin

An undercover agent, a counterspy, a double agent

A breakwater

Mole

4

A mole is the number of atoms in exactly 12.00 g ofpure isotope.C12

6

This number, known as the Avogadro’s constant,

can be determined by mass spectrometry.

VB

kem 2

B is the magnetic field strength.

V is the accelerating potential.

k is a constant of the instrument.

5

VB

kem 2

At fixed e, k, B and V

m can be determined.

6

Q.5

Mass of one mole of atoms = 12.00 g mol1

C126

= Mass of an Avogadro’s number of atoms

C126

= Avogadro’s number 1.992648 10-23 g

Avogadro’s number g101.992648

mol g 12.0023

1

= 6.022 1023

mol1

7

2.1 The mole (SB p.18)

What is “mole”?What is “mole”?

ItemUnit used to

count

No. of items per

unit

Shoes pairs 2

Eggs dozens 12

Paper reams 500

Particles in Chemistry

moles6.022

1023

for counting particles like atoms, ions, molecules

for counting common objects

8P. 8 / 66

1 mole ~

~602,200,000,000,000,000,000,000 mol1million

billionquadrillion

trillionquintillion

sextillion

千進制602.2 sextillions

9

P. 9 / 66

The fastest supercomputer can count 1.7591015 atoms per second.

Calculate the time taken for the superconductor to count 1 mole of carbon-12 atoms.

s 103.424101.759106.022 8

15

23

10.85 years

10P. 10 / 66

We can count the number of coins by weighing if the mass of one coin is known.

Similarly, we can count the number of 12C by weighing if the mass of one 12C is known.

23106.02particles of no.

moles of no.

mass molarmass

11

Molar mass is the mass, in grams, of 1 mole of a substance

12

Relative atomic mass

12.01101.121.12

13.003101.12100.00

12.000

Q.6

Relative isotopic mass

Relative intensity

12.000 100.00

13.003 1.12

C12

C13

13

Q.6


Relative intensity

12.000 100.00

13.003 1.12

C12

C13

Molar mass of carbon

= 12.01 g mol1

14

Q.6


Relative intensity

12.000 100.00

13.003 1.12

C12

C13

Relative isotopic mass is not exactly

equal to mass number of the isotope

15

Number of moles of a substance

number of particles

6.022 1023 mol1

mass

molar mass= =

Q.7

Number of moles of oxygen atoms

=number of oxygen atoms

6.022 1023 mol1

2 g

16 g mol1=

16

Q.7

Number of moles of oxygen atoms

=number of oxygen atoms

6.022 1023 mol1

2 g

16. g mol1=

Number of oxygen atoms = 2310022.616

2

Number of atoms

O17 %04.010022.616

2 23

= 3.011 1019

17

Molar mass is the same as the relative atomic mass in grams.

Molar mass is the same as the relative molecular mass in grams.

Molar mass is the same as the formula mass in grams.


Example 2-1AExample 2-1A Example 2-1BExample 2-1B Example 2-1CExample 2-1C

Example 2-1DExample 2-1D Example 2-1EExample 2-1E Check Point 2-1Check Point 2-1

18

2.2.22Molar Volume Molar Volume

and and Avogadro’s Avogadro’s

LawLaw

19

What is molar volume of What is molar volume of gases?gases?

Volume occupied by one mole of molecules of a gas.

2.2 Molar volume and Avogadro’s law (SB p.24)

20


Depends on T & P

Two sets of conditions


21


at 298 K & 1 atm

(Room temp & pressure / R.T.P.)


22


at 273 K & 1 atm

(Standard temp & pressure / S.T.P.)


22.4 dm3 22.4 dm322.4 dm3

23

GasMolar mass/g

Molar volume at R.T.P./dm3

Molar volume at S.T.P./dm3

O2 32 24.0 22.397

N2 28 24.0 22.402

H2 2 24.1 22.433

He 4 24.1 22.434

CO2 44 24.3 22.260

17 24.1 22.079

NH3

Not constant~ 24 ~ 22.4

24

Avogadro’s LawAvogadro’s Law2.2 Molar volume and Avogadro’s law (SB p.24)

Equal volumes of ALL gases at the same temperature and pressure

contain the same number of moles of molecules.

Equal volumes of ALL gases at the same temperature and pressure

contain the same number of moles of molecules.

At fixed T & P, V nIf n = 1, V = molar volume

25

Avogadro’s LawAvogadro’s Law

)mol (dm volume molar)(dm gas of volume

molecules gas of moles of no.

1-3

3

R.T.P. at mol dm 24

)(dm gas of volume 1-3

3


S.T.P. at mol dm 22.4

)(dm gas of volume 1-3

3

V n

V = Vm n

26


Interconversions involving Interconversions involving number of molesnumber of moles

Example 2-2AExample 2-2A Example 2-2BExample 2-2B Example 2-2CExample 2-2C

Example 2-2DExample 2-2D Check Point 2-2Check Point 2-2

27

2.2.33 Ideal Gas Ideal Gas

EquationEquation

28

Boyle’s lawBoyle’s law

2.3 Ideal gas equation (SB p.27)

At fixed n and T,

PV = constant or

P1

V

n = number of moles of gas molecules

29


Schematic diagrams explaining Boyle’s law

30

1/P

31


A graph of volume against the reciprocal of pressure for a gas at constant temperature

32

At fixed n and P,


Charles’ lawCharles’ law

TV T is the absolute temperature in Kelvin, K

33


Schematic diagrams explaining Charles’ law

34


A graph of volume against temperature for a gas at constant pressure

0oCTemperature / oC

Vo

lum

e

-273.15 oC

35


A graph of volume against absolute temperature for a gas at constant pressure

/ K

36

PV = nRTPV = nRT


Ideal gas equationIdeal gas equation

P

RnTV

R is the same for all gases

R is known as the universal gas constant

nV Avogadro’s law

P

1V Boyle’s law

Charles’ lawTV

Ideal gas equation

37


Relationship between the ideal gas equation and the individual gas laws

38

At fixed n,

nRT

PVa constant

......3

33

2

22

1

11 T

VP

T

VP

T

VP= a constant

Ideal gas behaviour is assumed in all gas laws

39

PV = nRTPV = nRT

2.3 Ideal gas equation (SB p.27)Gas Gas

lawslawsnV Avogadro’s law

P

1V Boyle’s law

Charles’ lawTV

Ideal gas equation

40

What is the difference between a theory and a law?


A law describes what happens under a given set of circumstances.

A theory attempts to explain why that behaviour occurs.

Gas laws vs kinetic theory of gases

41

Ideal gas behaviour

1. Gas particles are in a state of constant and random motion in all directions, undergoing frequent collisions with one another and with walls of the container.

2. Gas particles are treated as point masses, i.e. they do not occupy volume.

Four assumptions as stated in kinetic theory of gases

Volume of a gas = capacity of the vessel

42

Ideal gas behaviour

4. Collisions between gas particles are perfectly elastic, i.e. kinetic energy is conserved.

3. There is no interaction among gas particles.

43

The ideal gas equation is obeyed byreal gases only at

(i) low pressure

(ii) high temperature

(less deviation from 24 dm3 at R.T.P.)

44

(i) At low pressure, gas particles are so far apart that

(1) any interaction among them becomes negligible (assumption 3)

(2) the volume occupied by the gaseous molecules becomes negligible when compared with that of the container (assumption 2)

45

At high temperature,

gaseous molecules possess sufficient energy to overcome intermolecular interactions readily. (assumption 3)

46


(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to?

2

2

1

1

TVP

TVP

2Tatm 10

20)K(273atm 5

T2 = 586 K

47


(c) A balloon is filled with helium at 25 oC. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How many moles of helium have been introduced into the balloon?

nRTPV

K 298 mol K dm atm 0.082 n dm 0.450 atm 1.5 -1-133

n = 0.0276 molOrK 298 mol K J 8.314 n m 10 450Nm 101325 1.5 -1-13-6-2

n = 0.0276 mol

48


(d)25.8 cm3 sample of a gas has a pressure of 690 mmHg and a temperature of 17 oC. What is the volume of the gas if the pressure is changed to 1.85 atm and the temperature to 345 K?

(1 atm = 760 mmHg)

2

22

1

11

TVP

TVP

K 345V atm 1.85

K 17)(273

cm 25.82

3mmHg 760mmHg 690

V2 = 15.1 cm3

49

For one mole of an ideal gas at S.T.P.,

P = 1 atm or 101,325 Nm-2 (Pa)

V = 22.4 dm3 or 0.0224 m3

n = 1 mol

T = 273K


Q.8Calculate the universal gas constant at S.T.P.

50


0.082

K 273mol 1dm 22.4atm 1

nTPV

R3

atm dm3 K1 mol1

K 273mol 1

m 0.0224Nm 101325R

32

= 8.314 Nm K1 mol1

= 8.314 J K1 mol1

Or,

51

Q.9

PV = nRT

RTMm

PV

MRT

Vm

P

MρRT

P

ρRTM

m = mass of the gas

M = molar mass of the gas

52

2.2.44DeterminatioDeterminatio

n of Molar n of Molar MassMass

53

1. Mass Spectrometry

2. Density Measurement

PρRT

M

Determination of Molar Mass

54

(Mass of syringe + liquid) before injection (m1)

= 38.545 g

2.4 Determination of molar mass (SB p.32)

Vm

ρ

55

(Mass of syringe + liquid) after injection (m2)

= 38.260 g


Vm

ρ

56

Mass of liquid injected (m1 – m2)

= 38.545 g – 38.260 g = 0.285 g


Vm

ρ

57

Volume of air in syringe before injection (V1)

= 10.5 cm3


Vm

ρ

58

Volume of air + vapour in syringe after injection (V2)

= 146.6 cm3


Vm

ρ

59

Volume of vapour in syringe (V2 – V1)

= 146.6 cm3 - 10.5 cm3 = 136.1 cm3


Vm

ρ

60


Once m and V of the vapour are known,

Vm

ρdensity( )can be determined

61

Temperature = 273 + 65 = 338 K


62

Pressure = 1 atm


1 atm

63


Molar mass

PRT

VVmm

PRT

Vm

PρRT

12

12

atm 1

K) (338mol K dm atm 0.082dm10136.1

g 0.285 113

33

= 58.0 g mol1

Relative molecular mass = 58.0

Q.10

64

R = 8.314 J K1mol1 = 0.082 atm dm3 K1mol1

1m3 = 103 dm3 = 106 cm3

1 atm = 760 mmHg = 101325 Nm2 = 101325 Pa

2.5 Dalton’s law of partial pressures (SB p.35)

Unit conversions : -

65

(a)0.204 g of phosphorus vapour occupies a volume of 81.0 cm3 at 327 oC and 1.00 atm. Determine the molar mass of phosphorus.


PρRT

M

atm 1.00

K 327)(273 mol K dm atm 0.082 -1-13

dm 0.0810

g 0.2043

= 124 g mol-1

66

(b) A sample of gas has a mass of 12.0 g and occupies a volume of 4.16 dm3 measured at 97 oC and 1.62 atm. Calculate the molar mass of the gas.

(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)


PρRT

M

2-

-1-1

m 10 4.16

g 12.0

Nm 101325 1.62

K 97)(273 mol K J 8.31433-

= 54.1 g mol-1

Nm

67

(c) A sample of 0.037 g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium.


RTPV

n

K 25)(273 mol K dm atm 0.082

0.0382dm1-1-3

3mmHg 760mmHg 740

= 1.52103 mol

68



Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 1.52103

mol1.52103 mol

Mg of mass molarg0.037

Mg of mass molar

mass mol 101.52 3

1-3- mol g 24.3

mol 10 1.52g0.037

Mg of mass molar

69

2.2.55 Dalton’s Law Dalton’s Law

of Partial of Partial PressuresPressures

70

At fixed T & n,

PV = constant(15 atm)(5 dm3) = (PA)(15 dm3)

PA = 5 atm

empty

Experiment 1

Tap opened

Gas A

71

At fixed T & n

PV = constant(12 atm)(10 dm3) = (PB)(15 dm3)

PB = 8 atm

12 atm

Gas Bempty

Tap opened

Gas B

Experiment 2

72

The total pressure PT = 13 atm

= 5 atm + 8 atm

= PA + PB

Gas B

12 atm

Partial pressures of gases A & B

Experiment 3

Tap opened

Gas A + Gas B

73

Gas B

12 atm

Tap opened

Partial pressure of a constituent gas in a mixture is the pressure that the gas would exert if it were present alone under the same conditions

PA = 5 atm

PB = 8 atm

74

In a mixture of gases which do not react chemically, the total pressure of the mixture is the sum of the partial pressures of the component gases (the sum of the pressure that each gas would exert if it were present alone under the same conditions).

In a mixture of gases which do not react chemically, the total pressure of the mixture is the sum of the partial pressures of the component gases (the sum of the pressure that each gas would exert if it were present alone under the same conditions).

PT = PA + PB + PC

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures


75

Consider a mixture of gases A, B and C at fixed T & V.nA, nB and nC are the numbers of moles of each gas.The total number of moles of gases in the mixturenT = nA + nB + nC


Multiply by the constant RT/V

If gases A, B and C obey ideal gas behaviourPtotal = PA + PB + PC

VRTn

VRT

)nn(n TCBA

nT(RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V)

Derivation from ideal gas equation

76

Partial Pressures and Mole FractionsConsider a mixture of two gases A and B in a container of capacity V at temperature T

RTnVP AA RTnVP BB RTnVP TT

RTnRTn

VPVP

T

A

T

A

RTnRTn

VPVP

T

B

T

B

ABA

A

T

A Xnn

nPP

BBA

B

T

B Xnn

nPP

PA = PTXA PB = PTXB 1XX BA

Mole fractions of A & B

77

Consider a mixture of gases A, B, C, D,…

1...XXXX DCBA

PA = PTXA

PB = PTXB

PC = PTXC

PD = PTXD

…

78

Q.11At fixed T & n, PV = constant

For N2, P1V1 = P2V2

(0.20 Pa)(1.0 dm3) = P2(4.0 dm3) P2 = 0.05 Pa

For O2, P1’V1’ = P2’V2’

(0.40 Pa)(2.0 dm3) = P2’(4.0 dm3) P2’ = 0.2 Pa

By Dalton’s law of partial pressures22 ONT PPP = 0.05 Pa + 0.2 Pa = 0.25 Pa

79

Q.12

At –40oC, only N2 exists as a gas in the mixture

For a given amount of N2 at fixed V, P T

2

1

2

1

TT

PP

K 40)(273K 200)(273

atm 1.50P1

2N1 Patm 3.05P At 200oC

propaneNT PPatm 4.50P2

atm 1.45 atm 3.05) - (4.50 P - P P2NTpropane

80

At fixed T & V,

0.322atm 4.50atm 1.45

P

PX

T

propanepropane

propaneTpropane XPP

81

Q.13(a)

24NHNHT Nm 109.8PPPP

223

At fixed P & T, V n

T

NH

T

NH

V

V

n

n33

)(20%)Nm 10(9.8XP P 24NHTNH 33

= 1.96 104 Nm2

3NHX =

20%

82

Q.13(a)

55%V

V

n

nX

T

H

T

HH

22

2

)(55%)Nm 10(9.8XPP 24HTH 22

= 5.39 104 Nm2

83

Q.13(a)

25%V

V

n

nX

T

N

T

NN

22

2

)(25%)Nm 10(9.8XPP 24NTN 22

= 2.45 104 Nm2

84

Q.13(b)

223 NHNHT PPPP

22 NH PP

= 5.39 104 Nm2 + 2.45 104 Nm2

= 7.84 104 Nm2

NH3 is removed

but PT changes

Note : remain unchanged, 22 NH P & P

85


(c)The valve between a 6 dm3 vessel containing gas A at a pressure of 7 atm and an 8 dm3 vessel containing gas B at a pressure of 9 atm is opened. Assuming that the temperature of the system remains constant and there is no reaction between the gases, what is the final pressure of the system?

86


2211 VPVP

atm 3 dm 8)(6

dm 6 atm7

VVP

P A gas of pressure Partial 3

3

2

112

atm 5.1 dm 8)(6

dm 8 atm 9

VVP

P B gas of pressure Partial 3

3

2

112

Total pressure = PA + PB = (3 + 5.1) atm = 8.1 atm

87


(d)2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are introduced into a 15 dm3 vessel at 100 oC.

(i) What are the mole fractions of helium, nitrogen and argon in the system?

mol 0.50 mol g 4.0g 2.0

mass molar

mass He of moles of no. 1-

mol 0.11 mol g 28.0g 3.0

mass molar

mass N of moles of no. 1-2

mol 0.10 mol g 39.9g 4.0

mass molar

mass Ar of moles of no. 1-

Total no. of moles = (0.50 + 0.11 + 0.10) mol = 0.71 mol

0.700.710.50

XHe 0.150.710.11

X2N 0.14

0.710.10

XAr

88


(d)2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are introduced into a 15 dm3 vessel at 100 oC.

(ii) Calculate the total pressure of the system, and hence the partial pressures of helium, nitrogen and argon.

atm 1.45 dm 15

K 373mol K dm atm 0.082 mol 0.71VRTn

P 3

-1-13T

T

atm 1.0 0.710.50

atm 1.45 XP P HeTHe

atm 0.22 0.710.11

atm 1.45 XP P22 NTN

atm 0.20 0.710.10

atm 1.45 XP P ArTAr

89

The END

90

What is the mass of 0.2 mol of calcium carbonate?


Back

The chemical formula of calcium carbonate is CaCO3.

Molar mass of calcium carbonate = (40.1 + 12.0 + 16.0 3) g mol-1

= 100.1 g mol-1

Mass of calcium carbonate = Number of moles Molar mass

= 0.2 mol 100.1 g mol-1

= 20.02 g

Answer

91

Calculate the number of gold atoms in a 20 g gold pendant.


Back投影片 1

4Answer

Molar mass of gold = 197.0 g mol-1

Number of moles =

= 0.1015 mol

Number of gold atoms

= 0.1015 mol 6.02 1023 mol-1

= 6.11 1022

1mol g 197.0g 20

92

It is given that the molar mass of water is 18.0 g mol-1.

(a)What is the mass of 4 moles of water molecules?

(b)How many molecules are there?

(c)How many atoms are there?


Answer

93


(a) Mass of water = Number of moles Molar mass

= 4 mol 18.0 g mol-1

= 72.0 g

(b) There are 4 moles of water molecules.

Number of water molecules

= Number of moles Avogadro constant

= 4 mol 6.02 1023 mol-1

= 2.408 1024

94

Back


(c) 1 water molecule has 3 atoms (i.e. 2 hydrogen atoms and 1 oxygen atom).

1 mole of water molecules has 3 moles of atoms.

Thus, 4 moles of water molecules have 12 moles of atoms.

Number of atoms = 12 mol 6.02 1023 mol-1

= 7.224 1024

95

A magnesium chloride solution contains 10 g of magnesium chloride solid.

(a) Calculate the number of moles of magnesium chloride in the solution.


Answer

(a) The chemical formula of magnesium chloride is MgCl2.

Molar mass of MgCl2 = (24.3 + 35.5 2) g mol-1 = 95.3 g mol-1

Number of moles of MgCl2 =

= 0.105 mol

1mol g 95.3g 10

96

(b) Calculate the number of magnesium ions in the solution.


Answer

(b) 1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Cl- ions.

Therefore, 0.105 mol of MgCl2 contains 0.105 mol of Mg2+ ions.

Number of Mg2+ ions

= Number of moles of Mg2+ ions Avogadro constant

= 0.105 mol 6.02 1023 mol-1

= 6.321 1022

97

(c) Calculate the number of chloride ions in the solution.


Answer

(c) 0.105 mol of MgCl2 contains 0.21 mol of Cl- ions.

Number of Cl- ions

= Number of moles of Cl- ions Avogadro constant

= 0.21 mol 6.02 1023 mol-1

= 1.264 1023

98

(d) Calculate the total number of ions in the solution.


Answer(d) Total number of ions

= 6.321 1022 + 1.264 1023

= 1.896 1023

Back投影片 1

4

99

What is the mass of a carbon dioxide molecule?


AnswerThe chemical formula of carbon dioxide is CO2.

Molar mass of CO2 = (12.0 + 16.0 2) g mol-1 = 44.0 g mol-1

Number of moles = =

=

Mass of a CO2 molecule =

= 7.31 10-23 g

constant Avogadromolecules of Number

mass MolarMass

1-

2

mol g 44.0

molecule CO a of Mass1-23 mol 10 6.02

1

1-23

-1

mol 10 6.02mol g 44.0

Back

100


(a)Find the mass in grams of 0.01 mol of zinc sulphide.

(a) Mass = No. of moles Molar mass

Mass of ZnS = 0.01 mol (65.4 + 32.1) g mol-1

= 0.01 mol 97.5 g mol-1

= 0.975 g

Answer

101


(b) Find the number of ions in 5.61 g of calcium oxide.

Answer

(b) No. of moles of CaO =

= 0.1 mol

1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion.

No. of moles of ions = 0.1 mol 2

= 0.2 mol

No. of ions = 0.2 mol 6.02 1023 mol-1

= 1.204 1023

1mol g 16.0)(40.1g 5.61

102


(c) Find the number of atoms in 32.05 g of sulphur dioxide.

Answer(c) Number of moles of SO2 =

= 0.5 mol

1 SO2 molecule contains 1 S atom and 2 O atoms.

No. of moles of atoms = 0.5 mol 3

= 1.5 mol

No. of atoms = 1.5 mol 6.02 1023 mol-1

= 9.03 1023

1-mol g 2) 16.0 2.13(g 32.05

103


(d)There is 4.80 g of ammonium carbonate. Find the

(i) number of moles of the compound,

(ii) number of moles of ammonium ions,

(iii) number of moles of carbonate ions,

(iv) number of moles of hydrogen atoms, and

(v) number of hydrogen atoms. Answer

104


(d) Molar mass of (NH4)2CO3 = 96.0 g mol-1

(i) No. of moles of (NH4)2CO3 = = 0.05 mol

(ii) 1 mole (NH4)2CO3 gives 2 moles of NH4+ ions.

No. of moles of NH4+ ions = 0.05 mol 2 = 0.1 mol

(iii) 1 mole (NH4)2CO3 gives 1 mole of CO32- ions.

No. of moles CO32- ions = 0.05 mol

(iv) 1 (NH4)2CO3 formula unit contains 8 H atoms.

No. of moles of H atoms = 0.05 mol 8

= 0.4 mol

(v) No. of H atoms = 0.4 mol 6.02 1023 mol-1 = 2.408 1023

1mol g 96.0g 4.80

Back

105

What is the difference between a theory and a law?

Back


A law tells what happens under a given set of circumstances while a theory attempts to explain why that behaviour occurs.

Answer

106

Find the volume occupied by 3.55 g of chlorine gas at room temperature and pressure.

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)Answer

Molar mass of chlorine gas (Cl2) = (35.5 2) g mol-1 = 71.0 g mol-1

Number of moles of Cl2 =

= 0.05 mol

Volume of Cl2 = Number of moles of Cl2 Molar volume

= 0.05 mol 24.0 dm3 mol-1

= 1.2 dm3

1mol g 71.0g 3.55


Back

107

Find the number of molecules in 4.48 cm3 of carbon dioxide gas at standard temperature and pressure.

(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1; Avogadro constant = 6.02 1023 mol-1)

Answer

Molar volume of carbon dioxide at S.T.P. = 22.4 dm3 mol-1

= 22400 cm3 mol-1

Number of moles of CO2 =

= 2 10-4 mol

Number of CO2 molecules = 2 10-4 mol 6.02 1023 mol-1

= 1.204 1020

13

3

mol cm 22400cm 4.48


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108

The molar volume of nitrogen gas is found to be 24.0 dm3 mol-1 at room temperature and pressure. Find the density of nitrogen gas.

Answer


Back

Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1 = 28.0 g mol-1

Density = =

Density of N2

=

=1.167 g dm-3

VolumeMass

volume Molarmass Molar

1-3

-1

mol dm 24.0mol g 28.0

109

1.6 g of a gas occupies 1.2 dm3 at room temperature and pressure. What is the relative molecular mass of the gas?

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)Answer


Number of moles of the gas =

= 0.05 mol

Molar mass of the gas =

= 32 g mol-1

Relative molecular mass of the gas = 32 (no unit)

13

3

mol dm 24.0dm 1.2

mol 0.05g 1.6

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110


(a)Find the volume occupied by 0.6 g of hydrogen gas at room temperature and pressure.

(Molar volume of gas at R.T.P. = 24.0 dm3 mol-

1) Answer

(a) No. of moles of H2 = = 0.3 mol

Volume = No. of moles Molar volume

= 0.3 mol 24.0 dm3 mol-1

= 7.2 dm3

1mol g 2)(1.0g 0.6

111


(b) Calculate the number of molecules in 4.48 dm3 of hydrogen gas at standard temperature and pressure.

(Molar volume of gas at S.T.P. = 22.4 dm3 mol-

1)Answer

(b) No. of moles of H2 = = 0.2 mol

No. of H2 molecules = 0.2 mol 6.02 1023 mol-1

= 1.204 1023

13

3

mol dm 2.42dm 4.48

112


(c)The molar volume of oxygen gas is 22.4 dm3 mol-1 at standard temperature and pressure. Find the density of oxygen gas in g cm-3 at S.T.P. Answer

(c) Density = =

Molar mass of O2 = (16.0 2) g mol-1 = 32.0 g mol-1

Molar volume of O2 = 22.4 dm3 mol-1 = 22400 cm3 mol-1

Density =

= 1.43 10-3 g cm-3

VolumeMass

volume Molarmass Molar

13

-1

mol cm 22400mol g 32.0

113


(d) What mass of oxygen has the same number of moles as that in 3.2 g of sulphur dioxide?

Answer

(d) No. of moles of SO2 =

No. of moles of O2 = 0.05 mol

Mass = No. of moles Molar mass

Mass of O2 = 0.05 mol (16.0 2) g mol-1

= 1.6 g

1mol g 2)16.0(32.1g 3.2

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114


A 500 cm3 sample of a gas in a sealed container at 700 mmHg and 25 oC is heater to 100 oC. What is the final pressure of the gas?

Answer

As the number of moles of the gas is fixed, should be a constant.

=

=

P2 = 876.17 mmHg

The final pressure of the gas at 100 oC is 876.17 mmHg.

Note: All temperature values used in gas laws are on the Kelvin scale.

TPV

1

11

T

VP

2

22

T

VP

K 25)(273cm 500 mmHg 700 3

K )001(273

cm 500 P 3

2

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115


A reaction vessel of 500 cm3 is filled with oxygen gas at 25 oC and the final pressure exerted on it is 101 325 Nm-2. How many moles of oxygen gas are there?

(Ideal gas constant = 8.314 J K-1 mol-1)Answer

PV = nRT

101325 Nm-2 500 10-6 m3 = n 8.314 J K-1 mol-1 (273 + 25) K

n = 0.02 mol

There is 0.02 mol of oxygen gas in the reaction vessel.

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116


A 5 dm3 vessel can withstand a maximum internal pressure of 50 atm. If 2 moles of nitrogen gas are pumped into the vessel, what is the highest temperature it can be safely heated to? Answer

Back

Applying the equation,

T = = = 1523.4 K

The highest temperature it can be safely heated to is 1250.4 oC.

1-1-

3-3-2

mol K J 8.314 mol 2m 105 Nm 10132550

nRPV

117


(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to?

2

2

1

1

TVP

TVP

2Tatm 10

20)K(273atm 5

T2 = 586 K

118


(c) A balloon is filled with helium at 25 oC. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How many moles of helium have been introduced into the balloon?

nRTPV

K 298 mol K dm atm 0.082 n dm 0.450 atm 1.5 -1-133

n = 0.0276 molOrK 298 mol K J 8.314 n m 10 450Nm 101325 1.5 -1-13-6-2

n = 0.0276 mol

119


(d)25.8 cm3 sample of a gas has a pressure of 690 mmHg and a temperature of 17 oC. What is the volume of the gas if the pressure is changed to 1.85 atm and the temperature to 345 K?

(1 atm = 760 mmHg)

2

22

1

11

TVP

TVP

K347 V atm 1.85

K 17)(273

cm 25.82

3mmHg 760mmHg 690

V2 = 15.1 cm3

120


A sample of gas occupying a volume of 50 cm3 at 1 atm and 25 oC is found to have a mass of 0.0286 g. Find the molar mass of the gas.

(Ideal gas constant = 8.314 J K-1 mol-1; 1 atm = 101325 Nm-2) Answer

Back

M = 13.99 g mol-1

Therefore, the molar mass of the gas is 13.99 g mol-1.

RTMm

PV

K 25) (273 mol K J 8.314M

g 0.0286m 10 50 Nm 101325 11362-

121


The density of a gas at 450 oC and 380 mmHg is 0.0337 g dm-3. What is its molar mass? (1 atm = 760 mmHg = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)

Answer

Back

The unit of density of the gas has to be converted to g m-3 for the calculation.

0.0337 g dm-3 = 0.0337 103 g m-3 = 33.7 g m-3

PM = RT

M =

= = 4.0 g mol-1

Therefore, the molar mass of the gas is 4.0 g mol-1.

PRT

2

-1-1-3

Nm 101325760380

K 450) (273 mol K J 8.314 m g 33.7

122

(a)0.204 g of phosphorus vapour occupies a volume of 81.0 cm3 at 327 oC and 1 atm. Determine the molar mass of phosphorus.


Answer


(a) PV = RT

101325 Nm-2 81.0 10-6 m3

= 8.314 J K-1 mol-1 (273 + 327) K

M = 123.99 g mol-1

The molar mass of phosphorus is 123.99 g mol-1.

Mm

Mg 0.204

123

(b) A sample of gas has a mass of 12.0 g and occupies a volume of 4.16 dm3 measured at 97 oC and 1.62 atm. Calculate the molar mass of the gas.


Answer


(b) PV = RT

1.62 101325 Nm-2 4.16 10-3 m3

= 8.314 J K-1 mol-1 (273 + 97) K

M = 54.06 g mol-1

The molar mass of the gas is 54.06 g mol-1.

Mm

Mg 12.0

124


(1 atm = 760 mmHg = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)

Answer


125


(c) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

PV = nRT

101325 Nm-2 38.2 10-6 m3

= n 8.314 J K-1 mol-1 (273 + 25) K

n = 1.52 10-3 mol

No. of moles of H2 produced = 1.52 10-3 mol

No. of moles of Mg reacted = No. of moles of H2 produced

= 1.52 10-3 mol

Molar mass of Mg = = = 24.34 g mol-1

The relative atomic mass of Mg is 24.34.

moles of No.Mass

mol 10 1.52g 0.037

3-

760740

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126

Air is composed of 80 % nitrogen and 20 % oxygen by volume. What are the partial pressures of nitrogen and oxygen in air at a pressure of 1 atm and a temperature of 25 oC?

Answer

Back2.5 Dalton’s law of partial pressures (SB

p.36)

Mole fraction of N2 =

Mole fraction of O2 =

Partial pressure of N2 =

= 81060 Nm-2

Partial pressure of O2 =

= 20265 Nm-2

10080

10020

2Nm 10132510080

2Nm 10132510020

127

The valve between a 5 dm3 vessel containing gas A at a pressure of 15 atm and a 10 dm3 vessel containing gas B at a pressure of 12 atm is opened.

(a) Assuming that the temperature of the system remains constant, what is the final pressure of the system?

(b) What are the mole fractions of gas A and gas B?

Answer


128


(a) Total volume of the system = (5 + 10) dm3 = 15 dm

By Boyle’s law, P1V1 = P2V2

Partial pressure of gas A (PA)

=

= 5 atm

Partial pressure of gas B (PB)

=

= 8 atm

By Dalton’s law of partial pressures, Ptotal = PA + PB

Final pressure of the system = (5 + 8) atm = 13 atm

3

3

dm 15dm 5 atm 15

3

3

dm 15dm 10 atm 12

129


(b) Mole fraction of gas A =

=

= 0.385

Mole fraction of gas B =

=

= 0.615

total

A

P

P

atm 13atm 5

total

B

P

P

atm 13atm 8

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130

0.25 mol of nitrogen and 0.30 mol of oxygen are introduced into a vessel of 12 dm3 at 50 oC. Calculate the partial pressures of nitrogen and oxygen and hence the total pressure exerted by the gases.

(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1) Answer


131


Let the partial pressure of nitrogen be PA.

Using the ideal gas equation PV = nRT,

PA 12 10-3 m3 = 0.25 mol 8.314 J K-1 mol-1 (273 + 50) K

PA = 55946 Nm-2 (or 0.552 atm)

Let the partial pressure of oxygen be PB.


PB 12 10-3 m3 = 0.30 mol 8.314 J K-1 mol-1 (273 + 50) K

PB = 67136 Nm-2 (or 0.663 atm)

132


Total pressure of gases

= (55946 + 67136) Nm-2

= 123082 Nm-2

Or

Total pressure of gases

= (0.552 + 0.663) atm

= 1.215 atm

Hence, the partial pressures of nitrogen and oxygen are 0.552 atm and 0.663 atm respectively, and the total pressure exerted by the gases is 1.215 atm.

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133

4.0 g of oxygen and 6.0 g of nitrogen are introduced into a 5 dm3 vessel at 27 oC.

(a) What are the mole fraction of oxygen and nitrogen in the gas mixture?

(b) What is the final pressure of the system?

(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1) Answer


134


(a) Number of moles of oxygen =

= 0.125 mol

Number of moles of nitrogen =

= 0.214 mol

Total number of moles of gases = (0.125 + 0.214) mol

= 0.339 mol

Mole fraction of oxygen = = 0.369

Mole fraction of nitrogen = = 0.631

1gmol 32.0g 4.0

1gmol 28.0g 6.0

mol 0.339mol 0.125

mol 0.339mol 0.214

135


(b) Let P be the final pressure of the system.


P 5 10-3 m3 = 0.339 mol 8.314 J K-1 mol-1 (273 + 27) K

P = 169 107 Nm-2 (or 1.67 atm)

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Documents

1 The Mole Concept 2.1The Mole 2.2Molar Volume and Avogadro’s Law 2.3Ideal Gas Equation 2.4 Determination of Molar Mass 2.5Dalton’s Law of Partial Pressures