PRE UNIVERSITY

CHEMISTRY

SEMESTER 1

� CHAPTER 1 : MATTER� CHAPTER 1 : MATTER

1.1 The Atom

1.1.1 Historical development of atomic theory� When atom was first discovered by John Dalton (1808), he claim that atom are the simplest unit in a substance.

� Later, physicist J.J. Thomson (1897) found out that atom are made up of even smaller particles with negative charge – electron, Dalton’s theory was being rejected. After that, subatom with a positive charge –� After that, subatom with a positive charge –proton was discovered by Rutherford in the center of the atom.

� At the same decade, Neils Bohr discover that electrons surrounding nucleus similar as planet surrounding the Sun, and electrons move about in a rich electron region called orbital.

� Few years later, Chadwick discovered that not only proton exist in the center of an atom but also a non-charge subatom – neutron.

Electron

Nucleus

Proton

Neutron

Particle Symbol Mass

(kg)

Relative

Mass

Charge

(C)

Relative Charge(kg) Mass (C) Charge

Proton p or 11H 1.67 x 10-27

1 a.m.u +1.6 x 10-19

+1

Neutron n or 10n 1.67 x 10-27

1 a.m.u 0 0

Electron e or 0-1e 9.11 x 10-31

_1_ amu 1834

-1.6 x 10-19

-1

The path of moving protons, electrons and neutrons is deflected by both

electric field and magnetic field

βα

electron proton

β

electron

α<

North neutron

neu

tron

+ –

α < β

α

pro

ton

<β

Path of proton, electron and neutron

in an electrical field

Path of proton, electron and neutron

in a magnetic field

Path of proton, electron and neutron

in an electrical field

Path of proton, electron and neutron

in a magnetic field

South

neu

tron

1.1.2 Structure of atoms

� In atom, proton & neutron are located at the centre of the atom, which is also known as nucleus while electrons are surrounding it. The mass of an atom are concentrated at the center of the atom, as the mass of electron is very small.

� In periodic table, the symbol of elements is usually written in such way

nucleon number charge

PA n+nucleon number charge

proton number no. of same element

� Ion is formed when electron is donated or received

PA

Z

n+

x

Positive charged ion (cation) Negative charged ion (anion)

Formed Formed when an atom donate electrons Formed when an atom receive electrons

Dot and

cross

diagra +

2-

+ 2 e-

diagra

ms

Ionic

equatio

n

Na � Na+ + e- O + 2 e- � O2-

No of

e-10 10

+

+ e-

oxide ion, O2-

Particles Number of protons Number of neutrons Number of electrons

2010 Ne

168 O2-

3517 Cl-

4020Ca2+

52 Cr3+

10 10 10

8 8 10

17 18 18

20 20 18

24 28 215224Cr3+

12251Sb3-

OH-

CO32-

21D3O

+

[168O-N-18

8O]-

24 28 21

51 71 54

9 8 10

30 30 32

11 11 10

23 25 24

1.2 Isotopes

� Isotopes are atoms which have the same proton number but different nucleon number

� In the previous table, ____________ are isotopes

� Other examples of isotopes

Element Isotopes No of proton No of neutron% of

abundance

Hydrogen

Protium, 11H 99.0

Deuterium, D : 2 H 0.99

HYDROGEN & DEUTERIUM

1 0

1 1Hydrogen Deuterium, D : 21H 0.99

Tritium, T : 31H 0.01

Oxygen

Oxygen-16 : 168O 98.9

Oxygen-17 : 178O 1.00

Oxygen-18 : 188O 0.01

Chlorine

Chlorine – 35 ; 3517Cl 75

Chlorine – 37 ; 3717Cl 25

1 1

1 2

8 8

8 9

8 10

17 18

17 20

Isotope 54Fe 56Fe 57Fe 58Fe

% composition 5.8 91.6 2.2 0.4

No. of protons 26 26 26 26

No. of neutron 28 30 31 32

The relative abundance (% composition) of the isotopes in the sample

of an element is not the same for all the isotopes present.

Same Different

�Proton number

�No. of electron (neutral atom)

�Electronic configuration

�Chemical properties

�No. of neutron in nucleus

�Density

�Mass

�Rate of diffusion

� If compounds are formed from different isotopes, the molecular mass of that particular compound is the summation of all the isotopic mass involved

� Example

� C16O2 and C18O2

1 C + 2 (16O) = 44 1 C + 2 (18O) = 48

� H2O and T2O

2 H + 1 O = 18 2 T + 1 O = 22

� N35Cl3 and N37Cl31 N + 3 (35Cl) = 119 1 N + 3 (37Cl) = 125

�79Br–79Br and 79Br–81Br

79 + 79 = 158 79 + 81 = 160

1.2.4 Uses of Radioisotopes

� Widely used as tracers in biological processes. For example tracing the uptake of phosphorus by plant using 32P

� Carbon-14 is used in carbon dating, which is used to determine the age of archeological artifacts

� Gamma radiation from 60Co is used in radiotherapy to destroy malignant tissues in cancer patients.

� Energy released by nuclear fission is used to generate electricity in nuclear plants.

To sterilise food or surgical instruments.� To sterilise food or surgical instruments.

� Used in leak management. Underground leakage, especially in water or fuel pipeline leakage. Sudden increase in radioactivity mean that there’s a leakage.

1.3 Relative Mass

� Mass of an atom is approximately equal to the sum of the mass of all the sub-atomic particles present. Example

� In this method, the mass of the atom is compared to the mass of another atom which is used as reference. Initially, hydrogen was used as standard because it is the lightest. Subsequently, the oxygen atom was used to replace hydrogen as standard atom was used to replace hydrogen as standard due to a few reason.

� In 1961, carbon-12 was chosen as the standard for comparing relative atomic masses because it is easily available and its solid in room temp. It is known as 12C scale. On this scale, an atom of 12C has the mass exactly 12 atomic mass unit (a.m.u).

1.3.1 Relative Isotopic Mass (RIM)

� The relative isotopic mass of an isotope is the mass of 1 atom of the isotope relative to 1/12 times the mass of one atom of 12C.

RIM = mass of 1 atom of the isotope

1/12 x mass of 1 atom of C-12

� The relative isotopic mass of an isotope is approximately equal to its nucleon number. For example, example,

Isotope Relative isotopic mass40Ca 40.08019F 18.999127I 126.910

� Thus for most calculation involving atomic mass, the nucleon number can be use as a substitute for the actual RIM.

1.3.2 Relative Atomic Mass (RAM)� Most elements consist of a mixture of isotopes with different abundance as mention in slide 7. Therefore, the relative abundance of the abundance of the isotopes has to be taken into consideration when calculating the average mass of an atom of the element.

� The relative atomic mass (Ar) of an element is defined as the average mass of 1 atom of the element relative to 1/12 times the mass of 1 atom of 12C.12C.

RAM = average mass of 1 atom of the element1/12 x mass of 1 atom of C-12

Example : The element oxygen consist of 3 isotopes, 16O, 17O and 18O in the ratio of 99.76 : 0.04 : 0.20. Calculate RAM of oxygen.

RAM=(16 x 99.76) + (17 x 0.04) + (18 x 0.20) = 16.004499.76 + 0.04 + 0.20

1.3.3 Relative Molecular Mass (RMM)

� RMM of a molecular substance is the mass of 1 molecule of the substance relative to 1/12 times the mass of 1 atom of 12C

RMM = average mass of 1 molecule of substance1/12 x mass of 1 atom C-12

� ~ is equal to the sum of the relative mass of all the atoms shown in the molecular formula.

� For ionic compound the terms relative formula mass (RFM) is used because it do not exist as discrete molecules but consist of an infinite array of ions. RFM is defined as the mass of one of an infinite array of ions. RFM is defined as the mass of one formula unit of the compound relative to 1/12 times the mass of 1 atom of carbon-12.

1.4 Introduction to Mass Spectroscopy (MS)

� Relative mass of atom can be determine by using analytical instrument. One of the instrument that is frequently used nowadays is MASS SPECTROSCOPY

� Mass Spec can be used to determine :

� Relative isotopic mass

� Relative abundance of the isotopes

� Relative atomic mass� Relative atomic mass

� Relative molecular mass

� Structural formula of compound

� Figure on the next slide shows a simplified diagram of a mass spec.

Mass Spectrometer

This machine is used to find out the relative atomic mass of an element. There are 4 main stages in the process:

1. Ionisation - after a vapourised sample is put into the mass spectrometer, it is ionised - electrons are removed -usually one electron is removed but sometimes two. The positively charged species then go into the accelerating chamber.

One electron is removed : A (g) � A+ (g) + e–

2nd electron is removed : A+ (g) � A2+ (g) + e–

If molecule involved e.g. : A–B–C

A–B–C � (A–B–C)+ + e-

or A+ + (B–C)+ + 2 e- / C+ + (A–B)+ + 2 e-

or A+ + B+ + C+ + 3 e–

2. Acceleration - the ions are subjected to a negatively charged electric plate in order to accelerate the ion

2. Acceleration - the ions are subjected to a negatively charged electric plate in order to accelerate the ion.

3. Deflection - the heavy ions (ones with the larger atomic mass) are deflected less than the the lighter ions. Therefore the ions are separated according to their atomic masses and travel a different path in the mass spectrometer (shown by the dotted lines in the diagram below).56Fe+ ; 59Fe+

56Fe+ ; 56Fe2+

4. Detection - only ions of a certain mass actually end up at this point (the ones taking the green path). To make sure that all of the ions are detected, you have to vary the strength of the accelerating field. The detector records each species as a peak on a trace.

� Ratio of 79Br to 81Br is 50 : 50 or 1 : 1, so RAM of Br = 80

� For peak m/e 158, it is due to the existence of (79Br–79Br)+ ; so the probability of peak P(79Br–79Br) = (1/2)(1/2) = ¼

� For peak m/e 160, it is due to the existence of (79Br–81Br)+ ; or, it may also be (81Br–79Br)+ , so the probability of peak is

P(79Br–81Br) = (1/2)(1/2) + P(81Br–79Br) = (1/2)(1/2) = ½

� For peak m/e 162, it is due to the existence of (81Br–81Br)+ ; so the probability of peak P(81Br–81Br) = (1/2)(1/2) = ¼

RAM = 206(15) + 207(72) + 208(8) + 209(5)___15 + 72 + 8 + 5

= 207.03

O+

N+

O+

NO+

N2O+

NO2+

N2O2+

N2O3+

N2O4+

CH2+

CH3+

C2H2+

C2H3+

C3H5+

C3H6+

C4H8+

CH3+

O+

C2H3+

C2H5+

COH+

C3H6+

C2H3O+

C3H6O+

1.5 Mole

� One mole of substance is the amount of substance that contains the same number of particles as the number of atom in exactly 12 g of the C-12 isotope.

� The number of particles in one mole of any substance is a constant known as the Avogadro constant (L)

Avogadro Constant = 6.023 x 1023 mol-1

� 1 mole of any substance is the same as the relative atomic/ molecular/formula mass of that substance expressed in gram

e.g. : 3 mole of boron

= 3 x 6.023 x 1023 atoms of boron

1.5.1 Moles of Gases

� In reactions involving gases, the volume of the gases that take part in the reaction is usually more important than the mass of the gases involved.

� The relationship between the amount of gas (in moles) and the volume of gas is given in Avogadro’s Law

� Avogadro’s Law state that under the same conditions of temperature and pressure, an equal volume of gases contains equal number of moles where ;

volume α number of moles

For example, under room condition, 5 cm3 of Cl will contain � For example, under room condition, 5 cm3 of Cl2 will contain the same number of mole/molecules as5 cm3 NH3

� At standard temperature and pressure, (s.t.p ; 273 K and 101 kPa), 1 mole of all gases will occupy a volume of 22.4 dm3 . This volume is known as molar volume (Vm). Under room condition, 1 mole of gas occupies 24.4 dm3.

1.5.2 Moles and Solutions

� The concentration of a solution is usually expressed as the mass solute per 1.0 dm3 of solution (g dm-3) or mole of solute in 1.0 dm3 of solution (mol dm-3).

� Concentration in unit mol dm-3 is also known as molarity, M.

� The relationship between molarity and concentration is given by expression :

Molarity (M) = concentration (g dm-3)relative molecular mass of solution (g mol-1)

The number of moles of solute present in a given volume of � The number of moles of solute present in a given volume of solution (of molarity M) is

mole = molarity x volume = MV

1.6 Empirical Formula and Molecular Formula

� Empirical Formula of a compound shows the simplest whole number ratio for atom of all the different elements present in one molecule of the compound

� Molecular Formula of a compound shows the actual number of atoms of different elements in one molecule of compound

Example

Compound Molecular Formula Empirical FormulaCompound Molecular Formula Empirical Formula

Ethene C2H4 CH2

Phosphorous (V) oxide P4O10 P2O5

Hydrogen peroxide H2O2 HO

Ethanoic acid CH3COOH CH2O

1. A saturated hydrocarbon (hydrocarbon which only have carbon and hydrogen in it) contains 82.66% of carbon.

i. What is its empirical formula?

Element

C H

Mass 82.66 17.34

Mol 82.6612

= 6.89 mol

17.341

=17.34 mol

Ratio 6.89/6.89= 1

17.34/6.89= 2.5

Empirical formulaC2H5

ii. What is its molecular formula if given the relative molecular mass of the hydrocarbon is 58.0

= 1 = 2.5

Empirical formula = C2H5

(C2H5)n = 58(12(2) + 5(1))n = 58n = 2

Molecular formula = (C2H5)2= C4H10

2. Ester is the main chemical substance applied in perfume. In an analysis of determining the molecular formulae of the ester, it is say that this ester contains 54.5% of carbon, 9.10% of hydrogen and 36.4% of oxygen. Given the molecular mass of the ester is 88.0, determine the molecular formulae of this ester.

Element C H O

Mass 54.5 9.10 36.4

Mol54.512

=4.542

9.101

=9.10

36.416

= 2.275Mol 12

=4.5421

=9.1016

= 2.275

Ratio 4.542 / 2.275= 2

9.10 / 2.275= 4

2.275 / 2.275= 1

Empirical formula = C2H4O(C2H4O)n = 88

(12(2) + 4(1) + 16(1))n = 88n = 2

Molecular formula = (C2H4O)2= C4H8O2

3 An organic acid has the following composition by mass: C, 40.0%; H, 6.7%; O, 53.3%. Its mass spectrum shows major peaks (including the molecular ion) at the following m/e (mass) values: 15, 17, 43, 45, 60.

(a) Calculate the empirical formula of the acid, and use the mass spectrum to suggest its molecular formula and its structural formula. [3]

Element C H O

Mass 40.0 6.7 53.3

Mol40.0 / 12

= 3.336.7 / 1= 6.7

53.3 / 16= 3.33

Empirical = CH2O(CH2O)n = 60(12(1) + 1(2) + 16(1))n = 60

n = 2

(b) By suggesting their molecular formulae, identify the various species responsible for the peaks in the mass spectrum. [3]

m/e = 15 � CH3+

m/e = 17 � OH+

m/e = 43 � CH3CO+

m/e = 45 � COOH+

m/e = 60 � CH3COOH+

= 3.33 = 6.7 = 3.33

Ratio 3.33 / 3.33= 1

6.7 / 3.33= 2

3.33 / 3.33= 1

n = 2Molecular = (CH2O)2

= C2H4O2

1.6 Stoichiometry

� The term Stoichiometric comes from a balanced chemical equation where the amount of mole required for reactants to form a certain amount of mole of products.

� The moles of reactants required to form how many moles of products are referred from the chemical equation

Example 1 : In the reaction of

NaOH (aq) + HCl (aq) � NaCl (aq) + H2O (l)

� Stoichiometrically : 1 mole of sodium hydroxide is required to form 1 mole of sodium chloride

1 mol NaOH ≡ 1 mol NaCl

Example 2 : In the reaction of

2 KOH (aq) + H2SO4 (aq) � K2SO4 (aq) + 2 H2O (l)

� Stoichiometrically : 2 mole of potassium hydroxide is required to form 1 mole of potassium sulphate

2 mol KOH ≡ 1 mol of K2SO4

� Students are required to understand thus balanced the chemical equation before knowing the stoichiometry between the reactants and products.

1. The composition of an organic compound is 76.6 % C, 6.38 % H and 17.02 % O. Its relative molecular mass is 94. What are the empirical and molecular formulas of the compound?

Element C H O

Mass 76.60 6.38 17.02

Mol

76.6012

= 6.38

6.381

= 6.38

17.0216

= 1.06= 6.38 = 6.38 = 1.06

Ratio 6.38 / 1.06= 6

6.38 / 1.06= 6

1.06 / 1.06= 1

Empirical formula = C6H6O(C6H6O)n = 94

[(12(6) + 1(6) + 16(1)]n = 94n = 1

Molecular formula = (C6H6O)1= C6H6O

2. What volume of oxygen (at STP) is required to burn exactly

(a) 100 cm3 methane, CH4?

(b) 200 cm3 ethanol, C2H5OH?

CH4 + 2 O2 � CO2 + 2 H2O

Since 1 CH4 = 2 O2

Volume of oxygen gas = 2 (100)= 200 cm3

C2H5OH + 3 O2 � 2 CO2 + 3 H2O

Since 1 C2H5OH = 3 O2

Volume of oxygen gas = 3 (200)

(c) 2.2 dm3 propanone, C3H6O?

(d) 3.0 dm3 octane, C8H18?

Volume of oxygen gas = 3 (200)= 600 cm3

C3H6O + 4 O2 � 3 CO2 + 3 H2O

Since 1 C3H6O = 4 O2

Volume of oxygen gas = 4 (2.2)= 8.8 dm3

C8H18 + 25/2 O2 � 8CO2 + 9H2O

Since 1 CH4 = 25 / 2 O2

Volume of oxygen gas = 25/2 (3.0)= 37.5 dm3

3. When 1.25 g of a mixture of ethane C2H6 and propene, C3H6 was burned in excess oxygen, 3.78 g of CO2 was obtained. What is the percentage by mass of C2H6 in the mixture?

Since gas is mixture of C2H6 and C3H6

If the total mass of gas = 1.25 gThen, let’s assume mass of C2H6 = x ; while C3H6 = 1.25 – x The mole of each gas are

Given the mass of CO2 formed from mixture is 3.78 g

30

xHCofmol 62 =

42

x25.1HCofmol 63

−=

Given the mass of CO2 formed from mixture is 3.78 g

mol of CO2 = 3.78 / 44 ; mol = 0.0859 mol

The amount of CO2 released from each gas is known from equationC2H6 + 7/2 O2 � 2 CO2 + 3 H2O (2 mol of CO2 is given by 1 C2H6)C3H6 + 9/2 O2 � 3 CO2 + 3 H2O (3 mol of CO2 is given by 1 C3H6)

Hence the total mol of CO2,

x = 0.69 g% C2H6 = 0.69 / 1.25 x 100%

= 55.2 %

0859.042

x25.13

30

x2 =

−+

4. 20.0 cm3 of a gaseous hydrocarbon X mixed with 150 cm3

of oxygen and bum completely. When the mixture is cooled the total volume of gas is 110 cm3. When the gaseous mixture is passed through concentrated potassium hydroxide solution, 30.0 cm3 of gas remains. Determine the molecular formula of X.Given the chemical equation of combustion for hydrocarbon

CxHy + (x + y/4) O2 � x CO2 + y/2 H2O

Initial 20 cm3 150 cm3

After 0 cm3 30 cm3 80 cm3 ?

The volume of O2 used for water = 150 – 110 = 40 cm3

According to Avogadro’s Law

Since 20 cm3 of CXHY formed 80 cm3 of CO2

So, the mol of CO2 = 80 / 20 = 4 ; Hence x = 4

As for H, since the 20 cm3 of CXHY react with 40 cm3 oxygen for water

So, mol of O for water = 40 / 20 = 2

Since y / 4 = 2 ; so y = 8

As a conclusion, the formula of hydrocarbon is C4H8

5 Q is more electropositive than Y. When 1.92 g of metal Q is added to an aqueous solution containing Y2+ ions, 12.4 g of metal Y is obtained. In this reaction, Q3+ ions are produced.

� (a) Write an ionic equation to represent the reaction above.

2 Q + 3 Y2+ � 3 Y + 2 Q3+

� (b) What is the relative atomic mass of Q if the relative atomic mass of Y is 207?

Mol of Y = mass / RAM Mol of Y = mass / RAM Y = 12.42 / 207Y = 0.060 mol

Since from equation 3 Y = 2 QMol of Q = 0.060 x 2 / 3

Q = 0.040 molRAM of Q = 1.92 / 0.040RAM of Q = 48

6. 1.0 dm3 sample of air containing carbon dioxide is passed through aqueous calcium hydroxide. If 0.080 g of calcium carbonate is formed, determine

(a) the number of moles of CO2 present.

The equation between CO2 and Ca(OH)2CO2 + Ca(OH)2 � CaCO3 + H2OSince 1 mol of CO2 = 1 mol of CaCO3

= 0.080 g / [40 + 12 + 3 (16)]= 8.0 x 10-4 mol

(b) the percentage by volume of CO2 in the sample at STP.

Since mol of CO2 = 8.0 x 10-4 mol

So volume of CO2 in sample = 8.0 x 10-4 (22.4 dm3)V = 0.018 dm3

% by V of CO2 = 0.018 dm3 / 1.0 dm3 x 100%= 1.8 %

7. When 6.70 g of iron is burned in 3.64 g oxygen, Fe2O3 is formed. [Ar O = 16; Fe = 55.8]

(a) Write a balanced equation for the action.

4 Fe + 3 O2 � 2 Fe2O3

(b) What mass of Fe2O3 will be produceMol of Fe = mass / RAM ; mol of Fe = 6.70 / 55.8

mol of Fe = 0.120 molSince 4 Fe = 2 Fe2O3

Mol of Fe2O3 = 0.120 / 2 = 0.060 molMass of Fe2O3 = 0.060 x [2(55.8) + 3(16)]

(c) What mass of oxygen will be left over at the end of the reaction?

Mass of Fe2O3 = 0.060 x [2(55.8) + 3(16)]= 9. 58 g

From equation above, since 4 Fe = 3 O2

Mol of O2 = 0.120 x 3 / 4 = 0.090 molMass of O2 = 0.090 x [2(16)]

= 2.88 gMass of O2 unreacted = 3.64 – 2.88

= 0.96 g

8. Calculate the concentration in mol dm-3 of the resulting solution when 300 cm3 of 0.40 mol dm-3 Na2SO4 is mixed with 200 cm3 of 1.2 mol dm-3 Na2SO4. What are the molar concentration of Na+ and SO4

2- ions in the resulting solution?

When mixing both Na2SO4

Mol of Na2SO4 in A mol of Na2SO4 in B

mol = 0.12 mol mol = 0.24 molTotal mol = 0.12 + 0.24 = 0.36 mol

1000

)300)(40.0(mol;

1000

MVmol ==

1000

)200)(2.1(mol;

1000

MVmol ==

Total mol = 0.12 + 0.24 = 0.36 molWhen mixed, total volume, V = 300 + 200

V = 500 cm3

Concentration after mixture

M = 0.72 mol dm-3

Since Na2SO4 � 2 Na+ + SO42-

So, [Na+] = 2 (0.72) = 1.44 mol dm-3

[SO42-] = 1 (0.72) = 0.72 mol dm-3

500

)1000)(36.0(M;

V

1000molM

tot

=×

=

9. Manganate (VII) ions react with oxalate ions according to the reaction below: ;

2 MnO4- + 5 C2O4

2- + 16 H+ � 8 H2O + 2 Mn2+ + 10 CO2

What is the volume of 0.200 mol dm-3 KMnO4- required to

completely oxidize 28.5 cm3 of 0.500 mol dm-3 Na2C2O4? [Ar H = 1.0; C = 12.0; O = 16.0; K = 39.1]

5

2

)50.28)(500.0(

V)200.0(;

b

a

VM

VM a

bb

aa ==

V = 28.50 cm3Va = 28.50 cm3

10. Brass is an alloy of copper, containiq 90.0% copper and 10.0 % zinc by mass. When nitric acid, HNO3 is added to tin alloy, the following reactions occur:

Cu + 4 H+ + 2 NO3- � Cu2+ + 2 NO2 + 2 H2O

4 Zn + 10 H+ + NO3- � 4 Zn2+ + NH4

+ + 3 H2O

(a) What volume of 2.00 M nitric acid is required to react completely with 10.0 g of brass?

In 10.0 g of brass ; 9.0 g is Cu and 1.0 g ZnMol of Cu = 9.0 / 63.5 = 0.1417 molMol of Zn = 1.0 / 65.4 = 0.0153 molMol of Zn = 1.0 / 65.4 = 0.0153 molBased on the mol of H+ in both equation ;

Total mol of HNO3 = 4 (0.1417) + 10/4 (0.0153)= 0.605 mol

VHNO3 = mol x 1000 / M= 0.605 x 1000 / 2.00= 302.5 cm3

(b) What volume of NO2 gas will be produced at 25°C and

1.01 x 105 Pa? [Ar Cu = 63.5: Zn = 65.4]

Since NO2 is only produced from reaction with CuSo, mol of NO2 = 2 (0.1417)

= 0.2834 molUnder room condition,

V = mol x Vm= 0.2834 x 22.4 dm3

= 6.91 dm3= 6.91 dm3