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2.1

The Mole

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P. 3 / 66

A burrowing mammalwith fossorial forefeet

A small congenitalpigmented spot on the skin

An undercover agent,a counterspy,

a double agent

A breakwater

Mole

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V

B

ke

m 2

At fixed e, k, B and V

m can be determined.

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Q.5

Mass of one mole of atoms = 12.00 g mol1C126

= Mass of an Avogadros number of atomsC12

6

= Avogadros number 1.992648 10-23 g

Avogadros numberg101.992648

molg12.0023

1

= 6.022 1023 mol1

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2.1 The mole (SB p.18)

What is mole?

Item Unit used tocount

No. ofitems perunit

Shoes pairs 2

Eggs dozens 12

Paper reams 500

Particles inChemistry

moles 6.022 1023

for counting particles like atoms, ions, molecules

for

countingcommonobjects

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P. 8 / 66

1 mole ~

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

billion

quadrillion

trillionquintillion

sextillion

602.2 sextillions

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The fastest supercomputer can count 1.7591015atoms persecond.

Calculate the time taken for the superconductor to count1 mole of carbon-12 atoms.

s103.424101.759

106.022 815

23

10.85 years

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P. 10 / 66

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

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

23106.02particlesofno.molesofno.

massmolar

mass

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

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Relative atomic mass12.01

101.12

1.1213.003

101.12

100.0012.000

Q.6

Relativeisotopic mass Relativeintensity

12.000 100.00

13.003 1.12

C

12

C13

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Q.6


12.000 100.00

13.003 1.12

C

12

C13

Molar mass of carbon= 12.01 g mol1

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Q.6


12.000 100.00

13.003 1.12

C

12

C13

Relative isotopic mass is not exactlyequal to mass number of the isotope

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Number of moles of a substance

number of particles

6.022 1023mol1

mass

molar mass= =

Q.7

Number of moles of oxygen atoms

=

number of oxygen atoms

6.022 1023mol1

2 g

16 g mol1=

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Q.7

Number of moles of oxygen atoms

=number of oxygen atoms

6.022 1023mol1

2 g

16. g mol1=

Number of oxygen atoms = 2310022.6162

Number of atomsO17 %04.010022.6

16

2 23

= 3.011 1019

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Molar massis the same as the relative

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

molecular massin grams.

Molar massis the same as the formulamass in grams.

2.1 The mole (SB p.20)

Example 2-1A Example 2-1B Example 2-1C

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

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2.2

Molar Volume andAvogadros Law

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What is molar volume of gases?

Volume occupied by one mole ofmoleculesof a gas.

2.2 Molar volume and Avogadros law (SB p.24)

(S 2 )

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Depends on T & P

Two sets of conditions


2 2 M l l d A d l (SB 24)

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at 298 K & 1 atm

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


2 2 M l l d A d l (SB 24)

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at 273 K & 1 atm

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


22.4 dm3 22.4 dm322.4 dm3

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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 22.079NH3

Not constant~ 24 ~ 22.4

2 2 M l l d A d l (SB 24)

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24

Avogadros Law


Equal volumes of ALLgases at the sametemperature and pressure contain thesame number of moles of molecules.

At fixed T & P, V n

If n = 1, V = molar volume

2 2 M l l d A d l (SB 24)

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Avogadros Law

)mol(dmvolumemolar

)(dmgasofvolume

moleculesgasofmolesofno.

1-3

3

R.T.P.atmoldm24

)(dmgasofvolume

1-3

3


S.T.P.atmoldm22.4

)(dmgasofvolume1-3

3

V n

V = Vmn

2 2 Molar volume and Avogadros law (SB p 24)

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Interconversions involving numberof moles

Example 2-2A Example 2-2B Example 2-2C

Example 2-2D Check Point 2-2

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2.3Ideal GasEquation

2 3 Id l s ti (SB 27)

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Boyles law

2.3 Ideal gas equation (SB p.27)

At fixed n and T,

PV = constant or

P1V

n = number of moles of gas molecules

2 3 Ideal gas eq ation (SB p 28)

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Schematic diagrams explaining Boyles law

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1/P

2 3 Ideal gas equation (SB p 28)

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


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At fixed n and P,


Charles law

TV

T is the absolute temperature in Kelvin, K


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Schematic diagrams explaining Charles law


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A graph of volume against temperature for a gas atconstant pressure

0oCTemperature / oC

Volum

e

-273.15 oC


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A graph of volume against absolute temperaturefor a gas at constant pressure

/ K


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PV = nRT


Ideal gas equation

P

RnTV

R is the same for all gases

R is known as the universal gas constant

nV Avogadros law

P

1V Boyles law

Charles lawTV

Ideal gas equation


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Relationship between the ideal gas equation and theindividual gas laws

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At fixed n,

nRTPV a constant

......

3

33

2

22

1

11 T

VP

T

VP

T

VP

= a constant

Ideal gas behaviour is assumed in all gas

laws


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PV = nRT


Gas laws

nV Avogadros law

P

1V Boyles law

Charles lawTVIdeal gas equation


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What is the difference between a theory and a law?

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

A law describes what happens under a givenset of circumstances.

A theory attempts to explain why that

behaviour occurs.

Gas laws vs kinetic theory of gases

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Ideal gas behaviour

1. Gas particles are in a state of constantand random motion in all directions,

undergoing frequent collisions with oneanother and with walls of the container.

2. Gas particlesare treated as pointmasses, i.e. they do not occupy volume.

Four assumptions as stated in kinetic

theory of gases

Volume of a gas = capacity of the vessel

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Ideal gas behaviour

4. Collisions between gas particles are

perfectly elastic, i.e. kinetic energy isconserved.

3. There is no interaction among gasparticles.

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The ideal gas equation is obeyed byreal gases only at

(i) low pressure

(ii) high temperature

(less deviation from 24 dm3at R.T.P.)

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At high temperature,

gaseous molecules possess sufficientenergy to overcome intermolecular

interactions readily. (assumption 3)


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g q ( p )

(b) A reaction vessel is filled with a gas at 20

o

Cand 5 atm. Ifthe vessel can withstand a maximum internal pressure of10 atm, what is the highest temperatureit can be safelyheated to?

2

2

1

1

TVP

TVP

2T

atm10

20)K(273

atm5 T2= 586 K


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g q ( p )

(c) A balloon is filled with helium at 25

o

C. The pressureexerted and the volume of balloon are found to be 1.5 atmand 450 cm3respectively. How many molesof heliumhave been introduced into the balloon?

nRTPVK298molKdmatm0.082ndm0.450atm1.5 -1-133

n = 0.0276 mol Or

K298molKJ8.314nm10450Nm1013251.5 -1-13-6-2 n = 0.0276 mol


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g ( )

(d) 25.8 cm

3

sample of a gas has a pressure of 690 mmHganda temperature of 17 oC. What is the volumeof the gas ifthe pressure is changed to 1.85 atmand the temperature to345 K?

(1 atm = 760 mmHg)

2

22

1

11

T

VP

T

VP

K345Vatm1.85

K17)(273cm25.8 23mmHg760mmHg690

V2= 15.1 cm3


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

P = 1 atm or 101,325 Nm-2

(Pa)V = 22.4 dm3or 0.0224 m3

n = 1 mol

T = 273K

g q ( p )

Q.8

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


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g q ( p )

0.082

K273mol1

dm22.4atm1

nT

PVR

3

atm dm3 K1 mol1

K273mol1

m0.0224Nm101325R

32

= 8.314 Nm K1 mol1

= 8.314 J K1 mol1

Or,

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Q.9

PV = nRT

RTM

mPV

M

RT

V

mP

M

RT

PRTM

m = mass of the gas

M = molar mass of the gas

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2.4

Determinationof Molar Mass

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1. Mass Spectrometry

2. Density Measurement

P

RTM

Determination of Molar Mass

2.4 Determination of molar mass (SB p.32) m

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(Mass of syringe + liquid) before injection (m1)= 38.545 g

V

m


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(Mass of syringe + liquid) after injection (m2)= 38.260 g

V

m


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Mass of liquid injected (m1 m2)= 38.545 g 38.260 g = 0.285 g

V

m


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Volume of air in syringe before injection (V1)= 10.5 cm3

V

m


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

= 146.6 cm3

V

m


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Volume of vapour in syringe (V2V1)= 146.6 cm3- 10.5 cm3= 136.1 cm3

V

m

2.4 Determination of molar mass (SB p.32)

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Once m and V of the vapour are known,

V

m density( )can be determined


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Temperature = 273 + 65 = 338 K


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Pressure = 1 atm

1 atm


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Molar mass

PRT

VVmm

PRT

Vm

PRT

12

12

atm1

K)(338molKdmatm0.082

dm10136.1

g0.285 113

33

= 58.0 g mol1

Relative molecular mass = 58.0

Q.10

2.5 Daltons law of partial pressures (SB p.35)

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R = 8.314 J K1mol1= 0.082 atm dm3 K1mol11m3= 103dm3= 106cm3

1 atm = 760 mmHg = 101325 Nm

2

= 101325 Pa

Unit conversions : -


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(a) 0.204 g of phosphorus vapour occupies a volume of

81.0 cm3at 327 oC and 1.00 atm. Determine the molar massof phosphorus.

P

RTM

atm1.00

K327)(273molKdmatm0.082 -1-13dm0.0810

g0.2043

= 124 g mol-1

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(c) A sample of 0.037 g magnesium reacted with hydrochloric

acid to give 38.2 cm3of hydrogen gas measured at 25 oCand 740 mmHg. Use this information to calculate therelative atomic mass of magnesium.

RT

PVn

K25)(273molKdmatm0.082

0.0382dm1-1-3

3mmHg760mmHg740

= 1.52103mol


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acid to give 38.2 cm3of hydrogen gas measured at 25 oCand 740 mmHg. Use this information to calculate therelative atomic mass of magnesium.

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)1.52103mol1.52103mol

Mgofmassmolar

g0.037

Mgofmassmolar

massmol101.52 3

1-3-

molg24.3mol101.52

g0.037Mgofmassmolar

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Exp rim nt 1

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At fixed T & n,

PV = constant

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

empty

Experiment 1

Tap opened

Gas A

Experiment 2

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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

Experiment 3

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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

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Gas B

12 atm

Tap opened

Partial pressure of a constituent gas in amixture is the pressure that the gas would

exert if it were present aloneunder thesame conditions

PA= 5 atm

PB= 8 atm

D lt L f P ti l P2.5 Daltons law of partial pressures (SB p.35)

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In a mixture of gases which do not react chemically,

the total pressureof the mixture is the sum of thepartial pressuresof the component gases (the sumof the pressure that each gas would exert if it werepresent aloneunder the same conditions).

PT = PA + PB + PC

Daltons Law of Partial Pressures


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Consider a mixture of gases A, B and C at fixed T & V.

nA, nBand nCare the numbers of moles of each gas.

The total number of moles of gases in the mixture

nT= nA+ nB+ nC Multiply by the constant RT/V

If gases A, B and C obey ideal gas behaviour

Ptotal= PA+ PB+ PC

V

RTn

V

RT)nn(n TCBA

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

Derivation from ideal gas equation

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Partial Pressures and Mole Fractions

Consider a mixture of two gases A and B in acontainer of capacity V at temperature T

RTnVP AA RTnVP BB RTnVP TT

RTnRTn

VPVP

T

A

T

A

RTn

RTn

VP

VP

T

B

T

B

ABA

A

T

A Xnn

nPP

BBA

B

T

B

Xnn

n

P

P

PA = PTXA PB = PTXB 1XX BA

Mole fractionsof A & B

Consider a mixture of gases A B C D

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

1...XXXX DCBA PA = PTXA

PB = PTXBPC = PTXC

PD

= PT

XD

Q.11

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Q.11

At 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, P1V1 = P2V2

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

By Daltons law of partial pressures

22 ONTPPP = 0.05 Pa + 0.2 Pa = 0.25 Pa

Q.12

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Q

At 40oC, only N2exists as a gas in the mixture

For a given amount of N2at fixed V, P T

2

1

2

1

T

T

P

P

K40)(273

K200)(273

atm1.50

P1

2N1Patm3.05P

At 200oC

propaneNT PPatm4.50P 2

atm1.45atm3.05)-(4.50P-PP2NTpropane

At fixed T & V

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At fixed T & V,

0.322atm4.50

atm1.45

P

PX

T

propanepropane

propaneTpropane XPP

Q 13(a)

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Q.13(a)

24

NHNHT Nm109.8PPPP 223

At fixed P & T, V n

T

NH

T

NHV

V

n

n33

)(20%)Nm10(9.8XPP 24NHTNH 33

=1.96 104Nm2

3NHX = 20%

Q 13(a)

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Q.13(a)

55%V

V

n

nX

T

H

T

HH 222

)(55%)Nm10(9.8XPP 24HTH 22

=5.39 104Nm2

Q 13(a)

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Q.13(a)

25%V

V

n

nX

T

N

T

NN 222

)(25%)Nm10(9.8XPP 24NTN 22

=2.45 104Nm2

Q 13(b) NH i d

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Q.13(b)

223 NHNHT PPPP

22 NHPP

= 5.39 104

Nm2

+ 2.45 104

Nm2

= 7.84 104Nm2

NH3is removed

but PTchanges

Note : remain unchanged,22 NH

P&P


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(c) The valve between a 6 dm3vessel containing gas A at a

pressure of 7 atm and an 8 dm3vessel containing gas B ata pressure of 9 atm is opened. Assuming that thetemperature of the system remains constant and there isno reaction between the gases, what is the final pressure

of the system?


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2211 VPVP

atm3dm8)(6

dm6atm7

V

VPPAgasofpressurePartial

3

3

2

112

atm5.1dm8)(6dm8atm9

VVPPBgasofpressurePartial

3

3

2

112

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


(d) 2 0 g of helium 3 0 g of nitrogen and 4 0 g of argon are

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(d) 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon areintroduced into a 15 dm3vessel at 100 oC.

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

mol0.50molg4.0

g2.0

massmolar

massHeofmolesofno.

1-

mol0.11molg28.0g3.0

massmolarmassNofmolesofno. 1-2

mol0.10molg39.9

g4.0

massmolar

massArofmolesofno.

1-

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

0.700.71

0.50XHe 0.150.71

0.11X

2N 0.14

0.71

0.10XAr


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(d) 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon areintroduced into a 15 dm3vessel at 100 oC.

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

atm1.45

dm15

K373molKdmatm0.082mol0.71

V

RTnP

3

-1-13T

T

atm1.00.71

0.50atm1.45XPP HeTHe

atm0.220.71

0.11

atm1.45XPP 22 NTN

atm0.200.71

0.10atm1.45XPP ArTAr

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The END

2.1The mole (SB p.20)

Back

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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


Back

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

Back

14

AnswerMolar mass of gold = 197.0 g mol-1

Number of moles =

= 0.1015 mol

Number of gold atoms

= 0.1015 mol 6.02 1023mol-1

= 6.11 1022

1

molg197.0

g20


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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


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(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 1023mol-1

= 2.408 1024

Back


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

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 1023mol-1

= 7.224 1024


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A magnesium chloride solution contains 10 g of magnesium

chloride solid.

(a) Calculate the number of moles of magnesium chloridein 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

1molg95.3

g10


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96

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

Answer

(b) 1 mole of MgCl2contains 1 mole of Mg2+ions and 2 moles of Cl-

ions.

Therefore, 0.105 mol of MgCl2contains 0.105 mol of Mg2+ions.

Number of Mg2+ions

= Number of moles of Mg2+ions Avogadro constant

= 0.105 mol 6.02 1023mol-1

= 6.321 1022

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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

14


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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 CO2molecule =

= 7.31 10-23g

constantAvogadro

moleculesofNumber

massMolar

Mass

1-

2

molg44.0moleculeCOaofMass

1-23 mol106.021

1-23

-1

mol106.02

molg44.0


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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


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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 1023mol-1

= 1.204 1023

1molg16.0)(40.1

g5.61


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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 SO2molecule 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-molg2)16.02.13(

g32.05


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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


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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)2CO3gives 2 moles of NH4+ions.

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

(iii) 1 mole (NH4)2CO3gives 1 mole of CO32-ions.

No. of moles CO32-ions = 0.05 mol

(iv) 1 (NH4)2CO3formula 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 1023mol-1= 2.408 1023

1molg96.0g4.80


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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


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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 dm3mol-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 Cl2Molar volume

= 0.05 mol 24.0 dm

3

mol

-1

= 1.2 dm3

1molg71.0g3.55

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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 dm3mol-1; Avogadroconstant = 6.02 1023mol-1)

Answer

Molar volume of carbon dioxide at S.T.P. = 22.4 dm3mol-1

= 22400 cm3mol-1

Number of moles of CO2=

= 2 10-4mol

Number of CO2molecules = 2 10-4mol 6.02 1023mol-1

= 1.204 1020

13

3

molcm22400

cm4.48

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108

The molar volume of nitrogen gas is found to be

24.0 dm3mol-1at room temperature and pressure. Find thedensity 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

volumeMolarmassMolar

1-3

-1

moldm24.0

molg28.0


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109

1.6 g of a gas occupies 1.2 dm3at room temperature and

pressure. What is the relative molecular mass of the gas?

(Molar volume of gas at R.T.P. = 24.0 dm3mol-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

moldm24.0dm1.2

mol0.05

g1.6

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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 dm3mol-1)

Answer

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

Volume = No. of moles Molar volume

= 0.3 mol 24.0 dm3mol-1

= 7.2 dm3

1molg2)(1.0

g0.6


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111

(b) Calculate the number of molecules in 4.48 dm3of

hydrogen gas at standard temperature and pressure.

(Molar volume of gas at S.T.P. = 22.4 dm3mol-1)

Answer

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

No. of H2molecules = 0.2 mol 6.02 1023mol-1

= 1.204 1023

13

3

moldm2.42

dm4.48


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112

(c) The molar volume of oxygen gas is 22.4 dm3mol-1at

standard temperature and pressure. Find the density ofoxygen gas in g cm-3at 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 dm3mol-1= 22400 cm3mol-1

Density =

= 1.43 10-3g cm-3

Volume

Mass

volumeMolar

massMolar

13

-1

molcm22400

molg32.0


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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

1

molg2)16.0(32.1

g3.2

Back

2.3 Ideal gas equation (SB p.30) Back

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114

A 500 cm3sample of a gas in a sealed container at 700 mmHg

and 25 oC is heater to 100 oC. What is the final pressure of thegas?

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.

T

PV

1

11

T

VP

2

22

TVP

K25)(273

cm500mmHg700 3

K)001(273

cm500P 32


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115

A reaction vessel of 500 cm3

is filled with oxygen gas at 25o

Cand the final pressure exerted on it is 101 325 Nm-2. Howmany moles of oxygen gas are there?

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

PV = nRT

101325 Nm-2500 10-6m3= n 8.314 J K-1mol-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 pressureof 50 atm. If 2 moles of nitrogen gas are pumped into thevessel, what is the highest temperature it can be safelyheated to?

AnswerApplying the equation,

T = = = 1523.4 K

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

1-1-

3-3-2

molKJ8.314mol2

m105Nm10132550

nR

PV


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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 of10 atm, what is the highest temperature it can be safelyheated to?

2

2

1

1

TVPTVP

2T

atm10

20)K(273

atm5

T2= 586 K


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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 atmand 450 cm3respectively. How many moles of heliumhave been introduced into the balloon?

nRTPV

K298molKdmatm0.082ndm0.450atm1.5 -1-133

n = 0.0276 mol Or

K298molKJ8.314nm10450Nm1013251.5

-1-13-6-2

n = 0.0276 mol


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(d) 25.8 cm3sample of a gas has a pressure of 690 mmHg and

a temperature of 17 oC. What is the volume of the gas ifthe pressure is changed to 1.85 atm and the temperature to345 K?

(1 atm = 760 mmHg)

2

22

1

11

T

VP

T

VP

K347Vatm1.85

K17)(273cm25.8 2

3mmHg760mmHg690

V2= 15.1 cm3

2.4 Determination of molar mass (SB p.33) Back

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120

A sample of gas occupying a volume of 50 cm3at 1 atm and

25 oC is found to have a mass of 0.0286 g. Find the molar massof the gas.

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

Answer

M = 13.99 g mol-1

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

RTM

mPV

K25)(273molKJ8.314

M

g0.0286m1050Nm101325 11362-

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(a) 0.204 g of phosphorus vapour occupies a volume of

81.0 cm3

at 327o

C and 1 atm. Determine the molar mass ofphosphorus.

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

Answer(a) PV = RT

101325 Nm-281.0 10-6m3

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

M = 123.99 g mol-1

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

M

m

M

g0.204


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123

(b) A sample of gas has a mass of 12.0 g and occupies a

volume of 4.16 dm3

measured at 97o

C and 1.62 atm.Calculate the molar mass of the gas.


Answer(b) PV = RT1.62 101325 Nm-24.16 10-3m3

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

M = 54.06 g mol-1

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

M

m

M

g12.0


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acid to give 38.2 cm3

of hydrogen gas measured at 25o

Cand 740 mmHg. Use this information to calculate therelative atomic mass of magnesium.

(1 atm = 760 mmHg = 101325 Nm-2;

ideal gas constant = 8.314 J K-1

mol-1

) Answer


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

PV = nRT

101325 Nm-238.2 10-6m3

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

n = 1.52 10

-3

molNo. of moles of H2produced = 1.52 10

-3mol

No. of moles of Mg reacted = No. of moles of H2produced

= 1.52 10-3mol

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

The relative atomic mass of Mg is 24.34.molesofNo.

Massmol101.52g0.0373-

760

740

Back2.5 Daltons law of partial pressures (SB p.36)

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126

Air is composed of 80 % nitrogen and 20 % oxygen by

volume. What are the partial pressures of nitrogen andoxygen in air at a pressure of 1 atm and a temperature of25 oC?

Answer

Mole fraction of N2=

Mole fraction of O2=

Partial pressure of N2=

= 81060 Nm-2

Partial pressure of O2=

= 20265 Nm-2

10080

100

20

2Nm101325100

80

2Nm101325100

20


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127

The valve between a 5 dm3vessel containing gas A at a

pressure of 15 atm and a 10 dm3

vessel containing gas B at apressure of 12 atm is opened.

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

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

Answer


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

By Boyles law, P1V1= P2V2

Partial pressure of gas A (PA)

=

= 5 atm

Partial pressure of gas B (PB)

=

= 8 atm

By Daltons law of partial pressures, Ptotal= PA+ PB

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

3

3

dm15

dm5atm15

3

3

dm15

dm10atm12


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129

(b) Mole fraction of gas A =

=

= 0.385

Mole fraction of gas B =

=

= 0.615

total

A

PP

atm13

atm5

total

B

P

P

atm13

atm8


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0.25 mol of nitrogen and 0.30 mol of oxygen are introduced

into a vessel of 12 dm3

at 50o

C. Calculate the partialpressures of nitrogen and oxygen and hence the totalpressure exerted by the gases.


Answer


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Let the partial pressure of nitrogen be PA.

Using the ideal gas equation PV = nRT,

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

PA= 55946 Nm-2(or 0.552 atm)

Let the partial pressure of oxygen be PB.


PB12 10-3m3 = 0.30 mol 8.314 J K-1mol-1(273 + 50) K

PB= 67136 Nm-2(or 0.663 atm)


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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 is1.215 atm.


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4.0 g of oxygen and 6.0 g of nitrogen are introduced into a

5 dm3

vessel at 27o

C.(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-1mol-1)

Answer


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(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

1gmol32.0g4.0

1gmol28.0

g6.0

mol0.339

mol0.125

mol0.339

mol0.214


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


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

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

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Ch02 (EE).ppt