Section 1B
Section 1B
The Mole Concept & Chemical Calculation
/ Page 1
The Mole Concept & Chemical Calculation
Mole and Avogadro Number
Chemists developed a unit to represent the number of particles in a sample. This unit is
the mole (Latin, means 'pile').
of
The mole was defined as the amount of substances which contains
12
particles as
in 12 g of 6 C .
the same number
23
Experiments show that 12 g of carbon has 6.02 x 10 atoms. This number is known
as Avogadro number (honoring an Italian chemist, Amadeo Avogadro).
The Mass of One Mole of a Substance
The weight of 1 mole of substance (molar mass) always corresponds to its formula
express in gram units.
mass
In the definition for mole, the particles may be atoms, molecules, ions, electrons or other
particles.
no.
Example 1
of
mole
=
no. of pa r ticles
3
A
no.
6.02 x 1023
5 cm spoon can hold
23
1.67 x10
of
mole
=
Mass
Molar mass
molecules of water.
a)
What is the number of moles of water molecules in the spoon?
b)
What is the mass of water in the spoon?
Example 2 1 mole of helium atom weighs ______ g and contains ___________________ helium
atoms, i.e. ________ mole of helium atoms.
1 mole of nitrogen molecules weighs
moles nitrogen atoms
_________
g
and contains
____________
1 mole of AlCl3 weighs _______ g, contains _______ mole of aluminium ions and
______ moles of chloride ions.
_____________________________________________________________________________________
.
1
What is the mass of 2.5 moles of potassium atoms?
2
What is the number of nitrogen atoms in 0.5 mole of nitrogen gas, N 2?
3
The number of sodium atoms in 4.6 g of sodium metal is
4
Calculate the mass of
a] 3.0 mole of neon atoms
b]
0.10 mole of lithium atoms
Section 1B
The Mole Concept & Chemical Calculation
/ Page 2
The Molar Volume
The volume
occupied
by
one
mole
of
a
substance is called the molar volume.
For gases, experiments show that at the same temperature and pressure, molar volumes
of all gases are roughly equal.
Gas
Formula
Molecular mass
Hydrogen
Density
-3
(g dm )
0.083
Helium
0.166
Ammonia
0.706
Oxygen
1.333
Carbon
dioxide
Molar volume
3
(dm )
1.811
Chlorine
2.994
i.e.
The molar volume of any gas is approximately _______ at 25oC 。 and 1 atmospheric
pressure
(r.t.p.)
The molar volume of any gas is __________
at standard teperature and pressure (s.t.p.)
Avogadro's Law
Avogadro's law states that equal volumes of all gases at the same
pressure contains the same number of molecules.
For example, 24 dm
and
of any gases at r.t.p. contain 1 mole of molecules.
3
Example 1
.
3
temperature
How many moles of carbon dioxide is present in 5.58 dm of this gas at s.t.p. ?
5
What is the number of moles of molecules in 6 dm
6
1.4 g of a gas has a volume of 1.2 dm
of the gas.
no.
of
mole
=
3
3
of oxygen gas?
at r.t.p. Calculate the relative molecular mass
Vol ume
Mola r v ol ume
The Ideal Gas Equation
In 1662, Robert Boyle had discovered that the volume of a fixed mass of gas is inversely
proportional to its pressure, provided the temperature remains constant. This is known as
Boyle's Law, Which can be expressed as:
PV
=
constant
or
V

1
P
In1787, Charles showed that the volume of a fixed mass of gas is directly proportional to
its absolute temperature (0oC = 273K), provided the pressure remains constant. This is called
the Charles's Law which can be expressed as:
P
V
=
constant
or
V
 T
Section 1B The Mole Concept & Chemical Calculation
/ Page 3
In 1811, the Italian scientist Avogadro put forward his theory which stated that equal
volumes(V) of gases, under the same conditions of temperature and pressure, contain equal
numbers(n) of molecules.
i.e.
V  n
By considering the above 3 Laws, we can obtain the following result:
V

1
P

V

PV
and

=
nT
P
nRT
V
 T

and
V
PV
 nT
 n
(The Ideal Gas Equation)
R
=
Universal gas constant
Determination of relative molecular masses
The Ideal Gas Equation provides the basis for the determination of the relative molecular mass
of a gas or low boiling liquid. By finding the volume of the gas (or vapour formed from a known
mass of a liquid), at a known temperature and pressure, the relative molecular mass of the
compound can be calculated.
The apparatus and experimental procedure used to determine the relative molecular
mass of a volatile substance (e.g. propanone) is given below:
steam i n
graduated gas syri nge
self-seali ng
rubber cap
hypodermi c
syri nge
thermometer
steam jacket
steam and
water out
Procedure:
1.
The apparatus is set up as shown, with a few cm3 of air being deliberately drawn into the
gas syringe. This allow for some space for the propanone to vaporise and expand smoothly
in the gas syringe.
2.
Pass steam through the jacket until the thermometer reading and the volume of air in the
syringe reach steady values. Record the temperature and air volume(V1).
3.
Rinse the hypodermic syringe with the liquid that is to be investigated. Recharge the
3
hypodermic syringe with about 1 cm of liquid, expel air from the needle by pushing in
the plunger gently. Dry the outside of the needle.
4.
Inject about 0.2 cm of liquid into the large syringe. The volume of liquid injected needs
not to be very accurately noted, since the mass of the liquid injected would be determined
by weighing the hypodermic syringe before and after injection. Between the weightings
handle the hypodermic syringe as little as possible to reduce loss of liquid by expansion or
evaporation.
5.
Record the volume(V2) of air plus vapour in the graduated syringe when the reading
become steady (this ensures complete vaporization of liquid and the vapour pressure inside
is equal to atmospheric pressure outside).
Pressure
=
1
atm
;
Volume
of
vaporized gas = V2 - V1
3
_____________________________________________________________________________________
Example 1
.
7
Section 1B The Mole Concept & Chemical Calculation
/ Page 4
3
Calculate the molar mass of B, given that 0.850 g of B occupied 59.5 cm
at r.t.p.
0.228 g
86.3 cm
of liquid was injected into a gas syringe. The volume of vapour formed was
3
at
r.t.p.
Calculate the molar mass of the substance.
Dalton's Law of Partial Pressure
Dalton's law of partial pressure states that the total pressure of a mixture of gases
which do not react together chemically is equal to the sum of the pressures that each gas would
exert if it occupied the space alone.
Ptotal
=
P1
+
P2
+
P3
+

+
Pn
Partial pressure of a gas is the pressure that the gas would exert if it alone occupied
the whole volume. Mathematically, it can be expressed as follow:
Ptotal

=
Partial
pressure
,
P1
=
The magnitude of the partial pressure of a gas is related to the number of mole of that gas
present in the container:
mole
fraction (1)
=



Partial
pressure , P1
=
_____________________________________________________________________________________
5
-2
Example 1: A mixture of gases at a pressure of 1.01 x 10 Nm
contains 25%
of oxygen. What is the partial pressure of oxygen in the mixture?
by volume
Section 1B
.
8
The Mole Concept & Chemical Calculation
A mixture of gases at a pressure
40% N2 ; 35% O2 ; 25% CO2.
4
-2
7.50 x 10 Nm
/ Page 5
has the volume composition
(a)
What is the partial pressure of each gas?
(b)
What will the partial pressures of nitrogen and oxygen be if the carbon dioxide is
removed by the introduction of some sodium hydroxide pellets?
Faraday and the mole
Quantity of Electricity
Quantity of electricity (Q)
=
in coulombs (C)
Example 1 :
current (I)
X
in amperes (A)
time (t)
in seconds (s)
A current of 0.25 A passing for 2 minutes. What is the quantity of electricity
passed ?
The Faraday Unit
1 Faraday
=
1 mole of electrons
=
96500 coulombs
Example 1: Find the quantity of electricity (in faradays) required to liberate
+
i)
1 mole of Na
ii)
.
9.
0.5 mole of Fe
2+
Find the quantity of electricity (in faradays) required to liberate
2+
i)
1 mole of Cu
ii)
2 moles of Al
iii)
2 moles of H2
3+
Section 1B
The Mole Concept & Chemical Calculation
/ Page 6
Faraday's Law of Electrolysis
Faraday's First Law
The mass (m) of a substance liberated at an electrode during electrolysis is directly
proportional to the quantity of elecltricity (Q) passed through the electrolyte.
i.e.
m

Q
Faraday's Second Law
The quantity of electricity (Q) required to produce a mole of a substance from its ions is
proportional to the charge on the ions (Mn+).
Q

n
Example 1:
A current of 1.5 A was passing through a solution of silver nitrate for 1.5 hours.
What was the mass of silver produced ?
Example 2: How many hours will it take for a current of
from a solution of copper sulphate ?
.
1.5 A
to deposit
0.80 g
of copper
10. A metal M has a relative atomic mass of 88. When a current of 0.5A was passed
for 16 minutes 5 seconds, 0.22 g of M was deposited on the cathode.
i)
Calculate the number of faradays of electricity needed to liberate 1 mole of
atoms.
ii)
What is the charge carried by an ion of
iii)
Write an ionic equation for the reaction at the cathode.
M
M?
11. Whan an aqueous solution of sodium hydroxide is electrolysed using platinum
electrodes, the hydroxide ions are discharged at the anode to give oxygen according to
the following half-equation:
4 OH- (aq)

2 H2O (l) + O2 (g) + 4 eIf a current of 2.5 A is passed for 15 minutes,
i)
how many moles of OH- ions are discharged ?
ii)
how many moles of oxygen (O2) are produced ?
iii)
what is the volume of this oxygen at r.t.p. ?
3
24 dm )
(molar volume of gases at r.t.p. =
Section 1B
The Mole Concept & Chemical Calculation
/ Page 7
12. A metal spoon is electroplated with silver in silver nitrate solution, using a current of
0.5 A for one hour.
(relative atomic mass:
Ag = 108.0 ; 1 Faraday = 96500 C)
i)
Draw a labelled diagram showing how this experiment can be carried out in the
laboratory.
ii)
Write an equation for the reaction which occurs at the metal spoon.
iii)
Calculate the mass of silver deposited on the spoon.
Chemical Formulae
Chemical formulae can sometimes be predicted from the theoretical argument based on the
atomic structure, but basically they are determined experimentally.
In order to find the formula of a compound, we should know
a]
b]
what elements are present ; and
relative number of each atoms/ions present in the compound.
There are several types of formulae which are commonly used. These are:
1.
Empirical formula - It is the formula which shows the simplest integral
ratio of the relative number of atoms or ions present.
(whole number)
2.
Molecular formula - It is the formula which shows the actual number of each kind of
atom in one molecule of the compound. (This type of formula is only applicable for covalent
compounds because only covalent compounds are exist in discrete molecules)
Determination of formulae of simple compounds from percentage
composition by mass
In order to find the formula of a compound, it is first necessary to find the ratio
mass of each element in the compound.
by
After finding the masses of each element, these masses can be converted into moles. The
ratio of these mole values gives the empirical formula.
Example 1: To find
the formula
of
copper
copper oxide
oxide
excess town gas
burning with a
small flame
Town gas
HEAT
In performing the above experiment, the following points should be noted:
i]
It is necessary to pass the town gas for a few seconds before igniting it at the outlet. This is
to ensure that all the air inside the tube has been displaced. The gas coming out of the
hole is thus pure town gas, which will burn quietly and safely on ignition.
ii]
iii]
Section 1B The Mole Concept & Chemical Calculation
/ Page 8
Excess town gas has to be burnt away, otherwise it will escape into the laboratory. Town
gas is poisonous and may catch fire easily.
The solid residue after complete reaction should be allowed to cool with town gas still
passing over it. If the town gas supply is stopped at once after the reaction, air will be
drawn into the glass tube as it cools, then the hot copper would easily combine with
oxygen in the incoming air to form back copper oxide. To prevent this, copper is allowed to
cool in an atmosphere of town gas.
Specimen results:
mass
mass
mass
mass
mass
mass
of
of
of
of
of
of
test tube
test tube + copper oxide
test tube + copper
copper oxide
copper in oxide
oxygen in oxide
(Given: atomic mass of copper
=
=
=
=
21.320 g
22.950 g
22.622 g
______ g
= ______ g
= ______ g
=
63.5
;
no. of mole of copper in copper oxide =
of
Copper(Cu) : Oxygen(O)
= 16)
=
no. of mole of oxygen in copper oxide
ratio
atomic mass of oxygen
=
=
= ___ : ___
Therefore, the formula of copper oxide is ________________ .
Example 2:
To
determine
the
empirical
crucible
formula
lid
of
magnesium oxide
pipe-clay triangle
magnesium
heat
Specimen
mass
mass
mass
mass
mass
mass
of
of
of
of
of
of
results:
crucible + lid
crucible + lid + magnesium
crucible + lid + magnesium oxide
magnesium oxide
magnesium
oxygen present in the oxide
Mg
= 20.635 g
= 21.555 g
= 22.165 g
= _____ g
= _____ g
= _____ g
O
no. of mole
ratio
Therefore, the formula of magnesium oxide is _____________
Example 3: A compound has the empirical formula CxHy . On analysis, 1.000 g of the
compound is found to contain 0.857 g of carbon. Find values of x and y .
C
H
masses
no. of mole
=
=
ratio
 Empirical formula is ______.
Therefore ,
x = ___ and
y = ___.
Example 4:
Section 1B The Mole Concept & Chemical Calculation
/ Page 9
A compound X contains 80 % carbon and 20 % hydrogen. Find its empirical
formula. Find also its molecular formula if the vapour density of X is 15 .
Assume
we
have
_____ g
of
compound
Carbon
X.
Hydrogen
masses
no. of mole
=
=
ratio
 Empirical formula is ______.
Vapour density
=
densit y of t he gas
densit y of h yd r ogen
=
=
molar mass of t he gas
molar v ol ume
molar mass of H gas
2
molar v ol ume
=
molec ular mass of the gas
2
mass o f t he gas
vol ume
mass of H gas
2
vol u me
mola r gas of t he gas
mola r mass of H 2 gas
=
Molecular mass of X
=
2 x V.D. =
Let ______ be the molecular formula of X.
Molecular
formula
of
X
=
is _________.
13 A sample of a hydrated compound was analysed and found to contain 2.10 g of
cobalt, 1.14 g of sulphur, 2.28 g of oxygen and 4.50 g of water. Calculate its
empirical formula.
(CoSO4 . 7 H2O)
Co
S
O
H2O
masses
no. of mole
=
=
=
=
ratio
 Empirical formula is ___________.
(no molecular formula for ionic compound)
14 10.00 g of hydrated barium chloride is heated until all the water is driven off. The
mass of anhydrous compound is 8.53 g . Determine the value of x in BaCl 2 . x
H2O.
(2)
15 Determine the formula of a mineral with the following mass composition: Na = 12.1
% ; Al = 14.2 % ; Si = 22.1 % ; O = 42.1 % ; H 2O = 9.48 %.
(Na2Al2Si3O10 . 2 H2O)
Section 1B
The Mole Concept & Chemical Calculation
/ Page 10
Calculation based on equations
If the balanced equation for one reaction is known, it is possible to calculate the masses
of the products if the masses of the reactants are given, or vice versa. The steps in this type of
calculation are as follow:
a]
write down the balanced equation for the reaction
b]
convert the amounts of given substances into mole quantities (this may not be necessary
in case of gases reactions)
c]
calculate the mole quantities of the required substances using the stoichiometric
coefficient given by the equation
d]
change the mole quantities back into mass or volume as required by the question
Example 1: Calculate the volume of hydrogen evolved at r.t.p. when 1.3 g of sodium reacts
with excess water.
Step 1
Equation
Step
2
into
Step
3
moles
Step
4
into
Example 2:
no. of
moles
by
equation
volume
Calculate the mass of magnesium oxide formed when
are burnt with excess oxygen.
2.43 g
of magnesium
Calculation involving gaseous volume
In this type of calculation, we must first know a very important fact - the Avogadro's Law. This
law states that equal volumes of all gases contain the same number of molecules (if
temperature and pressure are the same). Therefore, by applying Avogadro's law, mole ratio of
gases can be converted into volume ratio and vice versa.
Example 1: Methane, CH4, burns in oxygen to give carbon dioxide and water. If 10 dm 3 of
methane burns, calculate :
[a]
the volume of oxygen required
;
[b] the volume of the products formed (measured at room temperature and
pressure).
Equation :
mole by
equation
volume by
Avogadro‘s law
Volume
Volume
Volume
Example 2:
of
of
of
oxygen
products (
products (
=
)=
________
________
) = ________
Calculate the volume of oxygen required for the complete combustion of 100
3
cm of ethane, C2H6 . What is the volume of carbon dioxide and water formed,
assuming
all the volumes are measured at room temperature and pressure?
Section 1B
The Mole Concept & Chemical Calculation
/ Page 11
Molarity of a Solution
Molarity (M) is the number of moles of solute dissolved in 1 dm3
Molarity of a solution(M)
of solution.
n umber of mole of sol ute (n)
vol ume of sol ution in dm3(V)
=
Number of moles of solute present = Molarity x Volume of solution in dm
i.e.
n =
M x
3
V
_____________________________________________________________________________________
3
Example 1 How many grams of sodium hydroxide are needed to prepare 3 dm of
Example 2
How many
dm
3
of
0.05 M
NaCl
can be prepared from
11.7 g
2M
of
NaOH?
NaCl ?
_____________________________________________________________________________________
16 Calculate the molarity of each of the following solutions:
3
a) 4 g of NaOH in 1 dm
b)
9.8 g
of H2SO4
in
500 cm
3
17 Calculate the mass of solute in each of the following solutions:
3
a) 250 cm of 2M NaOH
b)
100 cm
3
of
0.5 M
HCl
_____________________________________________________________________________________
Calculations in Acid-Base titrations
3
Example 1
20.0 cm of a solution of barium hydroxide, Ba(OH) 2 , of unknown concentration
is placed in a conical flask and titrated with a solution of hydrochloric acid, HCl ,
3
which has a concentration of 0.06 M . The volume of acid required is 25.0 cm .
Calculate the concentration of the barium hydroxide solution.
Step
1
Equation
Step
2
into
Step
3
moles
Step
4
into
moles
by
equation
concentration (M)
Section 1B
Example 2
2.65 g
The Mole Concept & Chemical Calculation
/ Page 12
of sodium carbonate (Na 2CO3) were dissolved in water and made up to 250
3
3
3
cm
solution. 25.0 cm
of this required 20.0 cm
of a hydrochloric acid
solution for complete reaction. Find the molarity of the hydrochloric acid.
Step
1
Equation
Step
2
into
Step
3
moles
Step
4
into
moles
by
equation
molarity (M)
3
Example 3. 50.0 cm of 0.50 M nitric acid were added to 1.28 g of a sample of calcium
carbonate containing sand as impurity. The excess acid was found to be
3
neutralized by 20.0 cm
of 0.40 M sodium hydroxide solution. Find the
percentage purity of the calcium carbonate.
Step
1
Equation
1
Equation
2
Step
2
into
Step
3
moles
Step
4
into
.
moles
by
equation
mass (g)
18 On diluting a sample of ethanoic acid
3
(CH3COOH)
five times,
it
was
found
3
that 25 cm of the dilute solution required 30.0 cm
of 0.10 M sodium
hydroxide solution for complete neutralization. What was the concentration of original
-3
acid in mol dm ?
(0.60 M)
Step
1
Step
2
Step
3
Step
4
19 5.00 g of an impure sample of potassium carbonate (K 2CO3) required
2.00 M hydrochloric acid for complete reaction.
What is the percentage purity of the carbonate?
Step
1
Step
2
Step
3
Step
4
34.2 cm
3
of
(94.4 %)
Section 1B
20 8.0 g
The Mole Concept & Chemical Calculation
/ Page 13
of a sample of hydrated sodium carbonate , Na2CO3 . n H2O , are dissolved in
3
water and the solution is made up to 250 cm . Using methyl orange as indicator, 25
3
3
cm of this solution required 28.0 cm
of 0.10 M sulphuric acid for complete
reaction. Calculate n , the number of molecules of water of crystallization, in the
sample of sodium carbonate crystals.
(10)
Step
1
Step
2
Step
3
Step
4
_____________________________________________________________________________________
Calculations in Redox Titration
Example 1
Acidified potassium permanganate solution oxidizes oxalic acid to carbon dioxide
and water according to the following equation:
2 MnO4-(aq)
+
5 H2C2O4(aq)
+
+
6 H (aq)

2+
2 Mn (aq)
+ 8 H2O(l)
3
+
10 CO2(g)
3
25 cm
of 0.20 M oxalic acid solution required 20.0 cm
of potassium
permanganate solution for complete reaction. Find the molarity of the potassium
permanganate solution.
Step
2
Step
3
Step
4
.
21 Potassium permengate solution oxidizes iron(II) salt to iron(III) salt according to the
following ionic equation:
5 Fe
2+
(aq)
+
-
MnO4 (aq)
+
+
8 H (aq)

5 Fe
3+
(aq)
+
Mn
2+
(aq)
+ 4 H2O (l)
3
1.817 g
of potassium permanganate are dissolved in water to give 250.0 cm of
aqueous solution. 2.185 g of an iron(II) salt are dissolved in water and the solution
3
3
3
made up to 100 cm . 40.0 cm of this iron(II) salt solution require 25.0 cm of
the permanganate solution for complete reaction.
(a)
What is the molarity of the potassium permanganate solution?
(0.046 M)
(b)
What is the molarity of the
(0.1438 M)
Fe
2+
ions in the iron(II) salt solution?
(c)
Section 1B The Mole Concept & Chemical Calculation
2+
Given that 1 mole of the iron(II) salt contains 1 mole of Fe
is the molar mass of the iron(II) salt ?
(152.0 g)
(d)
What is the percentage by mass of iron in the salt ?
/ Page 14
ions, what
(36.84%)
3
22 25.0 cm of a solution of potassium permanganate(KMnO4) were added to excess of
potassium iodide(KI) solution and the iodine(I 2) liberated was titrated with sodium
3
thiosulphate(Na2S2O3) solution. 25.2 cm of 0.12 M thiosulphate solution were
used. Calculate the molarity of the permanganate solution.
(0.0242 M)
Equation
1:
Equation
2:
no. of
mole
of
S2O32-
no. of
mole
of
I2
formed
=
no. of
mole
of
I
reacted
=
no. of
mole
of
MnO4-
Molarity
of
-
MnO4-
used =
used =
solution =
Past Paper
96IIA
2(a)
(ii)
95IIA
3(a) (i)
(ii)
N2 at 0.20 Pa pressure was mixed with 2.0 dm3
of O2 at 0.40 Pa in a 4.0 dm 3 container. Assuming that both N2 and
O2 behave ideally, calculate the pressure of the gaseous mixture at 298 K.
(2 marks)
At
298 K, 1.0 dm3 of
Derive an expression for the molar mass M of an ideal gas, in terms of its
density d, pressure P, and absolute temperature T.
At 98.6 kPa and 300 K, the density of a sample of dry air is 1.146 g dm 3. Assuming that dry air contains only nitrogen and oxygen and behaves
ideally, calculate the composition of the sample.
(Gas constant, R = 8.31 J K- 1 mol- 1)
94IA
(5 marks)
1(d)
An aqueous solution of a titanium (Ti) salt was electrolysed by passing a
current of 5.00 A for 2.50 hours. As a result, 5.60 g of metallic Ti were
deposited at the cathode. Deduce the charge on the Ti ion in the solution.
(Given:
1 faraday = 96500 C mol - 1)
(3 marks)
Section 1B
91IA
The Mole Concept & Chemical Calculation
/ Page 15
2(c)
A container holds a gaseous mixture of nitrogen and propane. The
pressure in the container at 200
is 4.5 atm. At - 40oC the propane
completely condenses and the pressure drops to 1.5 atm. Calculate the mole
fraction of propane in the original gaseous mixture.
(3 marks)
Section 1B
Making
96IIA
The Mole Concept & Chemical Calculation
Scheme
2(a) (ii) PV
= nRT
0.2(1) = n N 2 RT
and
0.4(2) =
n O2 RT
Ptotal (4)= ( n N 2 + n O2 )RT
= 0.2 + 0.8
=1
Ptotal
= 14 = 0.25 Pa
95IIA
/ Page 16
3(a) (i)
(2 marks)
PV = nRT
1
2
m
where m is the mass of dry air
M
m
PV =
(RT)
M
m
M =
(RT)
PV
d
=
(RT) where d is the density of dry air
P
n
=
(ii) Molar
=
=
mass
of
dry
1
2
1
air
1.146
x 8.31 x 300
98.6
28.98
1
Let mole fraction of N2 be x.
28.02 x + 32.0 (1 - x)
=
28.98
1
x
= 0.759
Composition
1
of
dry
air
75.9%
N2
and
24.1%
O2.
(5 marks)
94IA
1(d)
i.e. 1
Tin+ + nemole of Ti
requires
No.
of
moles
of
Ti
liberated =
No.
of
moles
of
e-
passed
n =
91IA
 Ti
produced
2(c) P1V1
=
P2V1 = nt(1 - x)RT2
where
nt is
propane.
(2)  (1)
0.4663
0.1169
n
=
=
0.4663
=
3.989 =
x
e -.
1
1
4
1
(1)
number
of
P2
P1
1.5 473
(1 - x) = (
)(
) = 0.6767
4.5 233
(1 - x)
of
(3 marks)
ntRT1
total
moles
5.60
= 1.169 x 10-1
47.90
5.00 x 2.50 x 60 x 60
96500
(2)
moles,
x
is
mole
fraction
T2
=
T1
=
0.3233
2
1
(3 marks)
of