CBSE CLASS 12 CHEMISTRY NUMERICALS MADE EASY

CHEMISTRY NUMERICALS MADE EASY…
My dear students,
In CBSE XII class Chemistry, Numericals asked from Unit -1,2,3 & 4 it
carries Eight to Ten marks in the Board Examination. Most of the time in the Board
examination, one numerical problem for Three (3) or Two (2) mark question are usually
asked from each unit. The following important points to remember to carry numericals
and some solved numericals also given for your ready reference.
SOLID STATE
In the Numerical problem, usually calculating efficiency of packing in
FCC and BCC, Density of the Crystal, calculating the radius ratios and finding out the
Co-Ordination number abd also the type of the crystal to be identified may be asked.
Before starting, let us remember the following points:
1. NUMBER OF ATOMS PRESENT IN A UNIT CELL:
(a) If it is a simple cubic unit cell, only ONE (1) atom will be present.
(b) If it is a body centred cubic unit cell, only TWO (2) atoms will be present.
(c) If it is a face centred cubic unit cell, only FOUR (4) atoms will be present.
2. RELATION BETWEEN THE LENGTH OF THE UNIT CELL, RADIUS OF
THE ATOM AND NEAREST NEIGHBOUR DISTANCE:
(a). For a Simple cube : a = 2r = d
(b) For B.C.C. : a =
3
4r
=
3
2d
( c) For F.C.C : a = 2 2 r = 2 d
Let us identify each term in it :
a = Length of the Unit Cell, r = Radius of the Atom,
K.G.MALLIKARJUN, VICE PRINCIPAL, JNV, ANANTHAPUR(A.P.)
1

d = Nearest neighbour distance
HOW TO FIND OUT THE CO – ORDINATION NUMBER :
When r+
and r–
values were given, we can find out the Co – Ordination Number.
a). If −
+
r
r
ratio is between 0.225 – 0.414, the Co – ordination number is four (4)
and the shape is Tetrahedral
b) If −
+
r
r
ratio is between 0.414 – 0.732, the Co – ordination number is six(6)
and the shape is Octahedral.
c). If −
+
r
r
rafio is between 0.732 – 1.000, the Co –ordination number is Eight(8)
and the shape is B.C.C.
NOTE : -
Some time , Co –ordination number and one of the radius will be given and asked
to calculate the other radius. In this case, −
+
r
r
ration should be equated to the
MINIMUM value of the radius ratio. For example. The Co – ordination number of
crystal is sic(6). Then, you have to equate the radius ratio value to 0.414 and calculate the
other radius.
HOW TO CALCULATE THE EFFICIENCY OF PACKING:
(a) For a simple cubic
a = 2r
2

∴ Volume occupied by all the spheres in a unit cell = 1 x
3
4
πr3
( 1 sphere per unit
cell)
Volume of the unit cell = (2r)3
(volume = a3
)
Volume occupied by all the sphere in unit cell
Hence, packing efficiency = x
100
Volume of unit cell
=
3
3
)2(
100
3
4
1
r
r ×× π
=
67
22
×
= 52%
(b) In B.C.C
There are two atoms per unit cell.
Packing efficiency =
3
3
)
3
4
(
100
3
4
2
r
r ×× π
( aa
3
4
= )
= 2 x
3
4
x
444
33
xx
x 100 = 68 %
(c) In F.C.C.
Number of atoms per unit cell = 4
a =
2
4
r = 2 2 r
Packing efficiency =
3
3
)
2
4
(
100
3
4
4
r
r ×× π
= 74 %
3

HOW TO CALCULATE THE DENSITY OF A CRYSTAL:
d =
ANa
zM
3
d = Density of the crystal
z = Number of atoms in the Unit cell
M = Mass of the Unit cell
a = Length of the Edge of the Unit cell
NA = Avogadro’s Number.
NOTE : -
(a). Some times, the distance between the two(2) ions will be given. You can
calculate the length of the edge of the Unit cell (a) by multiplying the distance
between the Two(2) ions with 2.
2 x Distance between the two ions = Length of the edge of the Unit cell(a).
(b). To calculate the distance between the two ions, you divide the length of the
edge of the unit cell with 2.
2
a
will give the distance between the two ions.
4

SOLVED PROBLEMS
1.Calculate the efficiency of packing in case of metal crystal for :-(a) simple cubic
(b) Body centred cubic (c) face centred cubic (with the assumption that atoms are
touching each other)
(a) Simple cubic:-
Let cell edgelength be a and radius of an atom be r
a = 2r
Vol. of cube = a3
= 8r3
z = 1; Volume of one sphere = (4/3) πr3
Efficiency of packing= vol. of all spheres in one cube x 100/ Vol. of cube
= [(4/3) πr3
x 100 ]/ 8r3
= 52.3%
(b)Body centred cubic(BCC)
Body diagonal = 4r
(4r)2
=a2
+(√2a)2
=a2
+2a2
=3a2
16r2
=3a2
3
4r
a =
Vol. of cube =
33
64 3
3 r
a =
z=2; Vol. of two spheres = 2
3
4 3
×rπ
Efficiency of packing= vol. of two spheres x 100/ Vol. of cube
=
33
64
100
3
4
2
3
3
r
r ××× π
= 68%
(c) Face centred cubic (FCC)
Face diagonal = √2a = 4r (or)
2
4r
a =
Vol. of cube = a3
=
22
64 3
r
z = 4 ; Vol. of 4 spheres =
3
3
4
4 rπ×
5

Efficiency of packing = 3
3
22
64
100
3
16
r
r ×π
= 74%
2. Niobium crystallizes in body centred cubic structure.. If Density is 8.55 g/cm3
.
Calculate atomic radius of niobium (At. Mass of Niobium=93).
Ans:-
ANa
Mz
3
×
=ρ
8.55= 233
10023.6
932
××
×
a
a3
= 23
10023.655.8
932
××
×
a=[3.6x10-23
]1/3
= 3.302 x 10-8
cm = 3.302 x 10-10
m
For BCC unit cell √3a = 4r : r = nm
ma
29.14
4
103.3732.1
4
3 10
=
××
=
−
3. Formula mass of NaCl is 58.45 g/mol and density of it’s pure form is 2.167 g/cm3
.
Average distance between adjascent sodium and chloride ions in the crystal is
2.814x10-8
cm.Calculate Avogadro constant.
M =5 8.45 g/mol
ρ = 2.167 g/cm3
a = 2 x distance between Na+
and Cl-
= 2 x 2.814x10-8
cm = 5.628 x10-8
cm
ANa
Mz
3
×
=ρ
NA =
( )
23
38
100523.6
167.210628.5
45.584
×=
××
×
−
4. Analysis shows that nickel oxide has formula Ni0.98 O1.00. What fractions of Nickel
exist as Ni2+
and Ni3+
ion?
Let no of Ni2+
be ‘x’ and the no. of Ni3+
be ‘y’.
6

x + y = 0.98 (1)
The total positive charge per formula unit = 2+, since charge of one oxide
ion is 2-. Therefore
2x + 3y = 2 (2)
Multiply (1) with 2 => 2x + 2y = 1.96 (3)
From (3) and (2) y = 0.04 & x = 0.94
No. of Ni3+
= 0.04, No of Ni2+
= 0.94
% of Ni3+
= %08.4
98.0
10004.0
=
×
% of Ni2+
= 95.92%
5. The compound CuCl has ZnS structure. Its density is 3.4 g/cm3
. What is the
length of the edge of the unit cell?
ANa
Mz
3
×
=ρ
ie, a3
= 23
1002.64.3
994
××
×
= 96.736 x 10-23
cm3
a = 5.75 x 10-8
cm
6. Thallium chloride, TlCl crystallizes in either a simple cubic lattice or a face
centred cubic lattice of Cl-
ion with Tl+
ion in the holes. If the density of the solid
is 9 gcm-3
and edge of the unit cell is 3.85 x 10-8
cm, what is the unit cell geometry?
ANa
Mz
3
×
=ρ ,
( ) 2338
10023.61085.3
240
9
×××
×
=
z
240
10023.61085.385.39 2324
×××××
=
−
z
z =1.289 ≈ 1
So TlCl is simple cubic.
7

7. Gold (atomic radis =0.144 nm). Crystallises in face centred Unit cell. What is the
length of a side of a cell?
For FCC, a = 2√2r = 2 x 1.414 x 0.144nm = 0.4072 nm
8. KF has NaCl structure. What is the distance between K+
and F-
in KF. If density
is 2.48 g/cm3.
ANa
Mz
3
×
=ρ ,
( ) 3/123
1002.648.2
584
××
×
=a = 5.374 x 10-8
cm
Distance b/w K+
& F-
= ½ a = ½ (5.374 x 10-8
cm) = 2.687x10-8
cm = 269 pm
9..A metal ( atomic mass = 50) has a body centred cubic crystal structure. The
density of the metal is 5.96 g /cm3
. Find the volume of its unit cell ( NA = 6.023 x
1023
atoms / mol.)
Ans:- Mass per unit cell = 23
10023.6
502
×
×
Density of unit cell = 5.96 g /cm3
Volume = ?
Density =
volume
Mass
Volume =
3233
23
108.2
96.5
1
10023.6
502
cmcm
Density
Mass −
×=×
×
×
=
8

SOLUTIONS
My Dear Students,
Let us see the various formulas involved in this chapter.
HOW TO CALCULATE THE CONCENTRATION OF A SOLUTION:
1. MOLARITY CAN BE CALCULATED AS :
No. of moles of the solute
M =
Volume of solution in liters
2. MOLALITY CAN BE CALCULATED AS :
No. of moles of the solute
m =
Volume of the solvent in Kilo grams.
3. NORMALITY CAN BE CALCULATED AS :
No. of gram equivalents of solute
N =
Volume of solution in liters
4. MOLE FRACTION OF EACH COMPONENT :
No. of moles of the components
XA (or) XB =
Total Number of moles
HOW TO CALCULATE THE NUMBER OF MOLES:
Weight of the component
Number of moles =
Molecular weight of the component
HOW TO CALCULATE THE WEIGHT OF THE SOLVENT:
Weight of the solvent = Weight of the solution - Weight of the solute
NOTE :
1. Students, while calculating Molality (m), you will have to convert the weight of
the solvent in Kgs.
9

2. While calculating the Molarity(M), you will have to convert the volume of the
solution in Liters.
3. While calculating the mole fraction, normality Component A is taken as Solvent
where as Component B is taken as Solute.
XA =
BA
A
nn
n
+
XB =
BA
B
nn
n
+
Where XA and XB are the mole fractions of the components of A and B
4. XA + XB = 1 . If you calculate XA, you can calculate XB = 1 - XA
HOW TO CALCULATE PARTIAL PRESSURE OF THE GAS (HENRY’S LAW):
Partial pressure of the gas in solution = KH x Mole fraction of the gas in solution
(or)
P = KH . X
KH = Henry’s law constant.
PROBLEMS ON COLLIGATIVE PROPERTIES:
1. Relative Lowering of Vapour pressure:
Xsolute =
solvent
o
solutionsolvent
o
P
PP −
(or)
WM
Mw
solute
solvent
×
×
=
solvent
o
solutionsolvent
o
P
PP −
(Xsolute =
WM
Mw
solute
solvent
×
×
)
Po
solvent = Vapour pressure of the pure Solvent
10

Psolution = Vapour pressure of the solution.
XB = Mole fraction of the solute
w = Weight of Solute
W = Weight of Solvent
Msolvent = Molecular mass of solvent
Msolute = Molecular mass of solute
NOTE: -
From the value of mole fraction (XB), you can find out the molecular weight
of the Non – Volatile solute(MB).
XB = BA
B
nn
n
+ where nB=
B
B
M
W
Where WB = Weight of the Non – Volatile solute
MB = Molecular weight of the Non –Volatile solute.
2. Problems on Elevation of boiling point:
Msolute =
WT
wK
b
b
×∆
××1000
Let us identify the each term in it :
Msolute = Molecular weight of the non-volatile solute
Kb = Molal Elevation constant
w = Weight of the non – volatile solute
W = Weight of the solvent
11

bT∆ = Elevation of boiling point.
NOTE : -
Students, it is observed that you are doing a mistake while calculating the
bT∆ value. Its value always should be in Kelvin only.
KTKTTb 12 −=∆
Where T1K = Boiling point of the pure Solvent, T2K = Boiling point of the solution.
3. Problem on Depression in Freezing Point:
Msolute = WT
wK
f
f
×∆
××1000
Msolute = Molecular weight of the non-volatile solute
Kf = Molal Elevation constant
w = Weight of the non – volatile solute
W = Weight of the solvent
fT∆ = Depression in Freezing point.
NOTE :-
12

Students it is observed frequently that you are doing mistake while
calculating fT∆ value. Its value always should be in Kelvin only. Some times, fT∆
value is given in the negative sign. But, you should take only the positive value only.
f
o
f
o
f TTT −=∆
Where To
f = Freezing point of pure solvent, Tf = Freezing point of the solution.
4. Problems on Osmotic Pressure :
MRT=Π
(or)
RT
V
n
=Π (or) Msolute = V
wRT
Π
Π = Osmatic pressure in atmospheres only.
R = Gas constant = 0.0821 Ltrs. Atms.
V = Volume of the solution in Litres only
T = Temperature in Kelvin only.
n = No. of moles of the Non – Volatile solute.
M = Concentration of the solution in molarity.
NOTE :-
1. Students, the common mistake committed by most of you is substituting
the ‘R’ value. It is observed that the value of ‘R’ is taken as 8.314 instead
of 0.0821.
13

2. As the value of ‘R is in Ltrs Atms, care should be taken that the volume
of the solution should be in Litres and also the value of Osmotic Pressure
in Atmospheres inly.
3. Some times, Osmatic pressure value will be given in mm mercury. Hence,
it is to be converted into atmospheres. We know that 760mm Hg = 1
atmosphere.
4. Isotonic solutions were given to find out the molecular weight of the non
– volatile solute, Then equate both the Osimatic pressure values and use
this formula:
n 1 = n 2
Where n1 , n2 are the number of moles of the two solutes.
HOW TO CALCULATE THE VAN’T HOFF FACTOR (i) , DEGREE OF
ASSOCIATION AND DEGREE OF DISSOCIATION.
Normal Moleculat Mass
i =
Observed Molecular mass
NOTE :-
In case of dissociation the value of i < 1 and in case of association the value of i > 1.
To calculate degree of dissociation and association:
1.For Dissociation:
i =
1
1 α+
(or) i = 1 + α
Let us Know :
i = Van’t Hoff factor, α = Degree of dissociation
14

% of Dissociation = Degree of dissociation x 100
2.For Association:
i = 1 -
2
α
Let us Know : i = Van’t Hoff factor, α = Degree of association
% of Association = Degree of Association x 100
SOLVED PROBLEMS
1. 18 grams of glucose ( molar mass 180 gram /mol) is present in 500 cm3
of its
aq. Solution. What is the molarity of the solution? What additional data is
required if the molarity of the solution is also required to be calculated.
Ans ; - The amount of glucose present in one litre of solution
= 18grams x 3
3
500
1000
cm
cm
= 36 grams
Number of moles of glucose present in one litre of solution
= 36 grams x grams
mol
180
1
= 0.2 mol.
2. Calculate the molarity and molality of a 15% solution (by weight) of H2SO4
of density 1.020 g / cm3
.
Ans ; - Molarity of 15% H2SO4
Weight of H2SO4 in 100 grams of solution = 15 grams
Volume of 100 grams of solution =
02.1
100 = 98 ml
( Density of H2SO4 = 1.02 g /cm3
)
Moles of H2SO4 present in 98 ml of solution =
98
15
Molarity of solution =
98
15 x
98
1000 = 1.56M
Molality of 15% H2SO4
Weight of H2SO4 in 100 grams of solution = 15 grams
Weight of water in 100 grams of solution = 85 grams
Moles of H2SO4 present in 85 grams of water =
98
15
Hence Molality =
98
15 x
85
1000 = 1.8m.
15

3. If N2 gas is bubbled through water at 293K, how many millmoles of N2 gas
would dissolve in 1 liter of water. Assume that N2 exerts a partial pressure of
0.987 bar. Given that Henry’s law constant for N2 at 293K is 76.48 K. bar.
Ans ; - The solubility of gas is related to its mole fraction in the aq. Solution. The
mole fraction of the gas in the solution is calculated by applying Henry,s law.
Thus.
xN2 =
H
N
K
p 2
=
bar
bar
76480
987.0
= 1.29x 10-5
As a litre water contains 55.5mol of it (unit 8, class – XI) therefore, if n
represents number of moles of N2 in solution.
XN2 =
5.55+n
n
=
5.55
n
= 1.29 x 10-5
Thus n = 1.29 x 10-5
x 55.5 mol. = 7.16 x 10-4
mol.
= 7.16 x 10-4
mol x
1
1000m
mol
= 0.716 m mol.
4. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A
non-volatile non electrolyte solid weighing 0.5grams is added to 39.0 grams of
benzene(molar mass 78 gram / mol.). The vapour pressure of the solution
then is o.845 bar. What is the molecular mass of the solid substance?
Ans :- p0
Solvent = 0.850 bar pSolution = 0.845 bar
MSolvent = 78 gram / mol. w = 0.5 gram and W = 39 grams.
solvent
solutionsolvent
p
pp
0
0
−
=
WM
Mw
solute
solvent
×
×
bar
barbar
850.0
845.0850.0 −
=
gramsM
mol
gramgrams
solute 39
785.0
×
×
Msolute =
39005.0
850.0785.0
×
××
bar
barmolgramgrams
= 170 grams / mol
5. A solution of 12.5 grams of urea in 170 grams water gave a boling elevation
of 0.63K calculate the molecular mass of urea taking Kb = 0.52 K / m.
Ans: - ∆Tb = kb . m (or) Msolute =
WT
wk
b
b
×∆
××1000
16

W = 170 grams w = 12.5 grams, ∆Tb = 0.63, kb = 0.52K/m
Msolute =
63.0170
5.12100052.0
×
××
= 60.7 grams / mol.
6. An aq. Solution of a weak mono basic acid containing 0.1 gram in 21.7 gram
of water freezes at 273.813 K. If the value of kf for water is 1.86 degree / mol.
What is the molecular mass of the mono basic acid.
Ans:- ∆Tf = kf . m (or) Msolute = WT
wk
f
b
×∆
××1000
W = 21.7 grams w = 0.1 grams, ∆Tf = 273.813 – 273 = 0.813K/m,
Kf= 1.86K/m
Msolute =
813.07.21
1.0100086.1
×
××
= 10.54 grams / mol.
7. The osmatic pressure of blood is 8.21 atms. At 370
. How much of glucose
should be used per litre for an intravenous injection that is at the same
osmatic pressure on blood.
Ans:- We know MSolute =
v
RTwsolute
π
.
π = 8.21 atms. T = 37 + 273 = 310K v = 1 litre
Msolute = 180 wsolute = ?
wsolute =
310082.0
21.8180
×
×
= 58.06 grams.
8. One litre aqueous solution of sucrose ( molar mass = 342 grams/
mol)weighing 1015 grams is found to record an osmatic pressure of 4.82
atms. At 293K. What is the molarity of the sccrose solution( R = 0.0821
atom / mol./K)
Ans :- weight of solution = 1015 grams
We know π =
v
n RT
π = 4.82 atms. R= 0.0821 atms/mol./K , T = 293K
4.82 =
v
n x 0.0821 x 293
4.82 = molarity x 0.021 x 293 (
v
n = molarity)
17

Molarity = 0.2 molar.
Mass of solute in one liter of solution = 0.2 x 342 = 68.4 grams
Mass of water = (1015 – 68.4) grams = 946.6 grams = 0.9466 kg
Hence molarity = 0.2 / 0.9466 = 0.211 kg . mol.
9. Two elements ‘A’ and ‘B’ form compounds having molecular formulae AB2
and AB4, when dissolved in 20grams of C6H6, 1 grams of AB2 lowers the
freezing point by 2.3K, while 1.0 grams of AB4 lowers it by 1.3K. The molar
depression constant for benzene is 5.1kg/mol. calculate atomic mass of A and
B.
Ans :- Using the relation ∆Tf = kf . m
For compound AB2, we get
2.3K = 5.1 K kg/mol x
1
1
M x kg02.0
1
( M1 = Molecular mass of AB2)
M1 =
02.03.2
1.5
× = 110.8 / mol --------(1)
For compound AB4, we get
1.3K = 5.1K.kg/mol x
2
1
M x kg02.0
1 )
( M2 = Molecular mass of AB4)
M2 =
02.03.1
1.5
× = 196.15 / mol -------- (2)
From (1) and (2)
A + 2B = 110.8 ---------(3)
A + 4B = 196.15 ---------(4)
Solving (3) and (4)
- 2B = - 85.35
B = 42.675
A = 25.43
10. At 300K, 36 grams of glucose present per litre in its solution has an osmatic
pressure 4.98 bars. If the osmotic pressure of solution is 1. 52 bar at the same
temperature, what would be its concentration.
Ans:- Using the relation π = CRT , we get
π1 = C1RT and π2 = C2RT
18

2
1
π
π
=
2
1
C
C
52.1
98.4 =
180
36 x
2
1
C
C2 = 0.061 mol./lit
Concentration of solution will be 0.061 mol./ lit.
ELECTROCHEMISTRY
My dear student,
This chapter involved very few formulas, Hence, you can score
more marks in this chapter.
How to calculate Molar Conductivity ( λm)
Molar Conductivity ( λm) =
M
1000×κ
Let us identify each term in it :-
κ (kappa) = Specific Conductivity
M = Concentration of the solution in Molarity.
How to calculate Equivalent Conductivity
Equivalent Conductivity =
N
1000×κ
Let us identify each term in it :-
κ (kappa) = Specific Conductivity
19

N = Concentration of the solution in Normality
How to Calculate Degree of Dissociation
Degree of Dissociation ∞
≈
m
m
λ
λ
λ
Let us identify each term in it
mλ = Molar Conductivity
∞
mλ = Molar Conductivity at infinity dilution.
How to calculate cell constant :
Cell Constant =
A
l
Let us identify each term in it:
l = Distance between the electrodes
A = Area of the electrode
Cell Constant can also be calculated as:
Cell Constant = Specific conductivity X Resistance
How to calculate Ionization constant (K):
K =
α
α
−1
2
C
20

K = Ionisation Constant
α = Degree of dissociation
How to calculate cell potential :
Eo
Cell = Eo
cathode - Eo
Anode
NOTE:
Students, you are not able to Identify which is cathode and which is anode by
looking at Electrode potential Values. Let us see, how to decide them on the basis of
Electrode potential values.
1. If Eo
Values for both the Electrodes were in the NEGATIVE SIGN, the LEAST
VALUE is taken as ANODE . For example Eo
of Metal X = - 3.05 V and Eo
of
Metal Y = - 2.92V. Here X is taken as ANODE and Y is taken as CATHODE.
2. Similarly, if E0
values for both Electrodes were in the POSITIVE SIGN, We have
to follow the same as in case of the above.
3. If both are in different signs, the NEGATIVE values is taken as ANODE AND
positive IS TAKEN AS cathode. For example E0
of Metal X = - 2.37V, E0
= of
Metal Y = + 0.34V. Here X is taken as ANODE and Y is taken is CATHODE.
PROBLEMS ON NERNST EQUATION:
This equation is useful to find out the electrode potential of an Electrode in
which concentration of Metal ions is not 1M.
Ecell = E0
cell +
n
0592.0
log
( )
ellOxidationc
lductioncelRe
21

n = Number of moles of electrons Transferred
(Reduction Cell) = Concentration of the reduction half cell
(Oxidation Cell) = Concentration of the Oxidation half cell
E0
cell = Standard cell potential
NOTE:
1. Some times the number of Electrons lost by one Electrode may not be tha
same that of the other Electrode gaining.
For example :
Cr / Cr+3
(0.1) // Fe+2
(0.01) / Fe
Here chromium is loosing three(30 electrons where as Iron is gaining only
two(2) electrons. Hence, Redox equation is to be balanced.
2 x ( Cr - 3e-
→ Cr+3
) ⇒ 2Cr - 6e-
→ 2Cr+3
3 x (Fe +2
+ 2e-
→ Fe ) ⇒ 3Fe+2
+6e-
→ 3 Fe
2Cr + 3 Fe+2
→ 2Cr+3
+ 3 Fe
Hence , the concentration terms of oxidation and reduction half cells are to be
raised to the respective powers. That is:
Ecell = E0
cell +
6
0592.0
log
[ ]
[ ]2
3
1.0
01.0
Similarly whenever Silver Electrode is connected to other electrodes, you
also should be very careful about the value of ‘n’ and also in raising the concentration
terms.
How to calculate the single electrode potential
22

Ecell = E0
cell +
6
0592.0
log +n
M
1
E0
cell = Standard cell Potential
n = Number of moles of electrons transferred
[ ]+n
M = Concentration of Electrolyte.
How to calculate the Equilibrium Constant:
E0
cell =
n
0592.0
log Kc
E0
n = Number of moles of electrons Transferred
K = Equilibrium constant
How to calculate the Free energy change (or) Maximum work done by the cell
0
cellnFEG −=∆
∆G = Free energy change
n = Number of moles of electrons transferred
F = 96,500 Coloumbes
E0
How to calculate mass of Metal deposited at Cathode:
Charge = Current X Time (or) Q = It
23

NOTE:- One mole of the electron is required 1 Faraday(96,500 C) for the reduction of 1
mole of Metal .
How to calculate Efficiency of a cell
Efficiency of a cell (η) =
H
G
∆
∆
=
Totalwork
Usefulwork
SOLVED PROBLEMS
1. How much charge is required for the following reduction of
Ans:-
(i) Al3+
+ 3e-
 Al
Three mole electrons are required for the reduction of 1 mole of Al ions. We
know
that the charge on one electron= 1.6021 x 10-19
C.
Charge on 3 mole = 3x6.023x1023
x 1.6021x10-19
=3F
(ii) Cu2+
+ 2e-
 Cu
Similarly 2F
(iii) MnO4
-
 Mn2+
MnO4
-
+ 8H+
+ 5e-
 Mn2+
+ 4H2O
Similarly 5F
2. How much electricity in terms of Faraday is required to produce.
Ans:-
(i) CaCl2  Ca
Ca2+
+ 2e-
 Ca (40g)
Therefore one F is required to produce 20g Ca.
24

ii) Al2O3  Al
Al3+
+ 3eAl
ie, to get one mole Al(27g) 3F is required. So in order to get 40g of Al.
27
3
x 40 = 4.44 F is required
3. How much electricity is required in coulomb for the oxidation of
Ans:-
(i) H2O  O2
H2O  ½ O2 + 2H+
+ 2e-
2F is required.
(ii) FeO  Fe2O3
2FeO + H20  Fe2O3 + 2H+
+ 2e-
1F is required
4. A solution of Ni(NO3)2 is clectrolysed between platinum electrodes using a current
of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode.?
Ans:-t = 1200s
Charge = current x time
= 5A x 1200s = 6000C
Ni2+
+ 2e-
 Ni
We require 2F or 2x96487 C to deposit 1mol or 59 g of Ni
For 6000C the mass of Ni deposited
= 1
1
964872
600059
−
−
×
×
cmol
cgmol
=1.834 g
5. Three electrolytic cells A,B,C containing solution of ZnSO4, AgNO3, and CuSO4
respectively are connected in series. A steady current of 1.5 amperes was passed
through them until 1.45 grams of silver deposited at the cathode of cell B. How
long did the current flow? What mass of copper and of zinc were deposited?
Ans:-
25

By Faradays second law,
E1/m1 = E2/1.45g = E3/m3
Mass of Cu (or) m1 = 31.77 x
108
45.1 g
= 0.426g
Mass of Zn (or) m3 = 0.439g
108g Ag is deposited by passing 96500 C
1g Ag is deposited by passing
108
96500
1.45 g Ag is deposited by passing
108
96500 x 1.45
Q=It
t =
5.1108
45.196500
×
×
= 863.7s =14.39 minutes
6. Using the standard electrode potential given in the table 5.4(page no.95 in Text
book) predict if the reaction between the following is feasible.
a) Fe3+
(aq) and I-
(aq)
E0
Fe3+
/Fe2+
= 0.77 V
E0
I2/I-
= 0.54 V
The reaction is feasible. Fe3+
is 0.77 V which is more than that of I2
(0.54V).
2Fe 3+
+ 2I -
 2Fe 2+
+ I2
b) Ag+
(aq) and Cu(s)
E0
Ag+
/Ag = 0.80 V
E0
Cu2+
/Cu = 0.34 V
Reaction b/n Ag+
and Cu is feasible because E0
of Ag, 0.8V is more than
that of Cu, 0.34 V.
2Ag+
(aq) + Cu(s)  Cu 2+
(aq)  2Ag(s)
c) Fe3+
(aq) and Br-
(aq)
E0
Fe3+
/Fe2+
= 0.77 V
26

E0
Br/Br-
= 1.09V
The reaction b/w Fe3+
and Br-
is not feasible since E0
of Fe3+
is less than
that of Br-
.
d) Ag(s) and Fe3+
(aq)
E0
Ag+
/Ag 0.80 V
E0
Fe3+
/Fe2+
 0.77V
The reaction b/w Ag and Fe3+
is not feasible since E0
of Fe3+
is less than
Ag.
e) Br (aq) and Fe2+
(aq)
E0
Br/Br-
=1.09 V
E0
Fe2+
/Fe=-0.44 V
The reaction is feasible
7. Calculated the standard cell potentials of galvanic cell in which the following
reactions take place.
(i) 2Cr(S) + 3Cd+2
(aq) → 2Cr+3
(aq) + 3 Cd,
(ii) Fe+2
((aq) + Ag+
(aq) → Fe+3
(aq) + Ag(s)
i) Cathode
3Cd2+
+6e 3Cd E0
= -0.40V
Anode
2Cr 2Cr3+
+ 6e E0
= -0.74V
ECell = Ecathode - Eanode
= -0.40 - (-0.74)
= 0.34V
We know ∆G0
= -nFE0
∆G0
= -6 x 96487 x 0.34 = -196.86 kJ mol-1
E0
cell = 0.0592 x logk / n => K =3.039x1034
27

ii) Cathode => Ag+
+e Ag, E = 0.80V
Anode => Fe2+
 Fe3+
+ e, E = 0.77 V
Ecell =Ecathode-Eanode = 0.80-0.77 = 0.03V
We know ∆G0
=-nFE0
= -1 x 96487 x 0.03 = -2.89461 kJ mol-1
We know at equilibrium
E0
cell = 0.592 x log K/n ; n = 1 => K = 3.211
8. Write the Nerest equation and e.m.f. of the following cells at 298K
(i) Mg(s) /Mg2+
(0.001M) // Cu2+
(0.0001M) /Cu(s)
Mg Mg2+
(0.001M)
Cu2+
+ 2e  Cu
Mg + Cu2+
 Mg2+
+Cu
K=
[ ][ ]
[ ][ ]+
+
2
2
CuMg
CuMg
K=0.001M/0.0001M = 10
Ecell =E0
cell +
10
1
log
2
0592.0
E0
cell = 0.34 – (-2.36) = 2.70V
Ecell = 2.70 - 0.0592/2 x 1 = 2.6704 V
(ii) Fe(s)/Fe2+
(0.001M)// H+
/H2(g) (1 bar)/ Pt(s)
Fe Fe2+
+ 2e
2H+
+2eH2
Fe+2H+
 Fe2+
+ H2
K=
[ ][ ]
M
M
H
HFe
1
001.02
2
=+
+
=0.001
28

Ecell = 0.44 + 1000log
2
0592.0
= 0.5288 V
(iii)Sn(s)/Sn2+
(0.05M)//H+
(0.02M)/H2(g)
SnSn2+
+ 2e
2H+
+ 2e  H2
Sn + 2H+
 Sn2+
+ H2
K=
[ ][ ]
[ ][ ] ( )
125
02.0
050.0
22
2
2
==
+
+
HSn
HSn
Ecell = 0.14 +
125
1
log
2
0592.0
Ecell = 0.078V
iv) Pt(s)/Br2(l)/Br(0.01M)//H+
(0.03M)/H2(g)/H2(g)/bar/Pt(s)
Br2  2Br-
+ 2e
2H+
+ 2e  H2
Br2+2H+
 2Br-
+ H2
K =
[ ] [ ]
[ ][ ]
( )
( ) 9
1
03.0
01.0
2
2
2
2
2
2
==
+
M
M
HBr
HBr
Ecell =E0
cell + 0.0592/2 log 1/k
Ecell = -1.09 + 0.0592/2 log 9 = -1.0618V
9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K
is 1500Ω. What is the cell constant if conductivity of 0.001M KCl solution at
298 K is 0.146 x 10-3
S cm-1
.
Ans : - Conductivity = 0.146 x 10-3
S cm-1
R=1500Ω
29

We know R = ρ l/a ; l / a =cell constant
ρ = 1/(0.146x10-3
Scm-1
)
l / a= 1500 x 0.146x10-3
= 0.2190cm-1
10.Conductivity of 0.00241M acetic acid is 7.896 x 10-5
S cm-1
. Calculate its molar
conductivity and if λ0
for acetic acid is 390.5 S cm2
mol-1
, what is its
dissociation constant.
Ans:-
Conductivity = 7.896x10-5
cm-1
molar conduvctivity= molarity
tyconductivi
=
M00241.0
10896.7 5−
×
=32.76 Scm2
mol-1
α= 00838.0
5.390
76.32
0
==
λ
λ
K =
α
α
−1
2
C
=
( ) 8467.1
0838.01
0838.000241.0
2
=
−
×
x 10-5
11. In the button cell widely used in watches and other devices the following reation
takes place
Zn (s) + Ag2O (s) + H2O (l)  Zn 2+
(aq) + 2Ag (s) + 2 OH-
(aq)
Determine Eo
and ∆G0
for the reaction.
Ans :-
E0
cell = 0.344V-(-0.76V) = 1.104V
∆G0
= -nFE0
= - 2 x 96500Cmol-1
x 1.104V = - 2.131 x 105
J mol-1
12. Depict the galvanic cell in which the reaction Zn(s) + 2 Ag+ (aq) → Zn+2
(aq) +
2 Ag( s) takes place. Further show (i) Which of the electrode is negatively
charged (ii) The carries of the current in the cell (iii) Individual reaction at each
electrode.
Ans:-
(i) Zn electrode is negatively charged
ii) Ag+
and Zn2+
ions are the carriers of current in the cell.
30

iii) At anode (oxidation)
Zn  Zn2+
+ 2e-
. Here electrons are generated. It is rich in electrons. Thus
it is negative electrode.
At Cathode(reduction)
2Ag+
+ 2e  2 Ag
CHEMICAL KINETICS
My Dear Students,
Even this chapter contains very less formulas. Hence you can score full
marks in this chapter.
1. How to write the Rate expression for a given equation :
Examples:-
1. 2NO(g) + O2 (g) → 2NO2 (g)
Rate Expression = -
( )
dt
NOd
2
1
= -
( )
dt
Od 2
= +
( )
dt
NOd 2
2
1
2. 5Br-
(aq.) + BrO3
-
(aq.) + 6H+
(aq.) → 3Br2 (aq.) + 3H2O (l)
Rate Expression = -
( )
dt
Brd −
5
1
= -
( )
dt
BrOd
−
3
= -
( ) ( )
dt
Brd
dt
Hd 2
3
1
6
1
+=
+
(Note : - In aq. Solution changes in concentration of water is very small and
hence we do not use change in concentration of water for expressing the rate)
2. Units of Rate constant for different order can be calculated as :
31

For the First order Reaction:
Rate = k [Concentration]1
( )
dt
ionconcentratd
= k [ concentration]1
ond
litermoles
sec
= k [ moles / liter]1
k =
litermolesond
litermoles 1
sec
×
k =
1
sec
sec
1 −
= ond
ond
Similarly we can calculated the unitsfor the other orders also.
NOTE:- For :- . Zero Order :- k = mol L-1
s-1
Second Order : - k = L mol-1
s-1
nth
Order : - k = (mol / L)(1-n)
s-1
3. When one Fast reaction and one Slow reation is given, how to calculate the
order of a reaction?
Examples:-
1. In the reaction 3ClO-
→ ClO3
-
+ 2 Cl-
Various steps are:
ClO-
+ ClO-
→ ClO2
-
+ Cl-
(Slow step)
ClO2
-
+ ClO-
→ ClO3
-
+ Cl-
(Fast step)
Rate determine step is always SLOW STEP
∴ RATE = k1 [ClO-
] [ClO-
]
= k1 [ClO-
]2
ORDER = 2
32

2. In the reaction NO (g) + O2 (g) → 2NO2 (g)
Various steps are :
NO (g) + O2(g) → NO3 (g) (Fast step)
NO3 (g) + NO (g) → 2NO2 (g) (slow step)
Rate = k [NO3] [NO]
But in the actual reaction NO and O2 are the reactants, Hence NO3
intermediate product.
NO3 → NO + O2
Rate = [NO] [O2] [NO]
= [NO]2
[O2]
ORDER = 3
4. Order of a reaction is also calculated by :
a. Integrated Method
b. By initial rate Method
c. Half – life period method
Integrated Method
In this method, we have to make use of the general equation pertaining to
First Order reaction. The equation can be written as
K =
R
R
t
0
log
303.2
(or) k =
2
1
12
log
303.2
R
R
tt −
Let us identify the terms in it : -
33

k = Rate constant, R0 = Initial concentration, R = Concentration left after
time ‘t’, t = time , R1 and R2 are the concentration of the reactants at time t1
and t2
If the k value is the same for different values given in the problem, then it is a
First order constant.
By Initial rate method
In this method, a table comprises of concentration values of the reactants in
different experiments and also the rate constant values will be given. If you look at it, it is
clear that the concentration of one of the teactants will be kept constant and the other will
be changed. To begin the problem, first of all you have to write the rate equation for the
given chemical reaction.
Let us take some examples:-
Example:1:-
Consider the following data for the reaction : 2A + B → C + D
Determine the order of the reaction with respect of ‘A’ and with respect to
‘B’ and the overall order of the reaction? What are the units of the rate
constant?
Run
[A]
Mol. L-1
[B]
Mol. L-1
Rate of reaction(mol. L-1
sec.-
1
)
01 0.1 0.1 6.0x10-3
02 0.3 0.2 7.2x10-2
03 0.3 0.4 2.88 x 10-1
04 0.4 0.1 2.4 x 10-2
34

Solution:- Rate law = k [A]p
[B]q
By looking at the table, you can find that, in experiment 1 and 4, the
concentration of reactants B is the same. It is clear that, in experiment 2 and 3 the
concentration of reactants A is the same.
Rate Law is : Rate = k [A]p
[B]q
Where ‘p’ and ‘q’ values can be calculated by dividing the given experimental values.
Experiment 3 / 2 :
( )
( )q
q
litmol
litmol
..2.0
..4.0
102.7
1088.2
2
1
=
×
×
−
−
4 = (2)q
22
= 2q
q = 2
Experiment 4 /1: Similarly If you find ‘p’ value, we get, p = 1
Order of ‘A’ = 1, Order of “B’ = 2
Rate : k [A]p
[B]q
, Rate = k [A]1
[B]2
Then order of the reaction = 3
Example: 2:- If the initial rate is given and the concentration are not given for some
Experiments at that time by using following method we can calculate the concentrations
35

The reaction between A and B is first order with respect to A and zero order
with respect to B. fill in the blank in the following table.
Experiment [A]/M [B]/M initial rate / M min-1
1 0.1 0.1 2 x 10-2
2 -- 0.2 4 x 10-2
3 0.4 0.4 --
4 -- 0.2 2 x 10-2
Rate = k [A] 1
[B] 0
= k [A] 1
From (1) 2 x 10-2
= k x 0.11
à (1)
K = (2 x 10-2
/ 0.1) = 2 x 10-1
min-1
Form (2) 4 x 10-2
= k [A]
[A] = 4 x 10-2
/ 2 x 10-1
) = 0.2 M.
Form (3) rate = k x 0.41
= 2 x 10-1
x 0.41
= 8 x 10-2
M min-1
From (4) 2 x 10-2
=k [A] 1
2 x 10-2
= 2 x 10-1
[A]
[A] = (2 x 10-2
/ 2 x 10-1
) = 0.1M.
Half life method
t1/2 =
k
693.0
t1/2 = Half life time, k = Rate constant
For Zero order reaction t1/2 α [R0]
First order reaction t1/2 is independent of [R0]
Second order reaction t1/2 α
[ ]0
1
R
n th
order reaction t1/2 α
[ ] 1
0
1
−n
R
[R0] = initial concentration
5. How to calculate the ActivationEnergy:
36

log 




 −
=
21
12
1
2
303.2 TT
TT
R
E
k
k a
k1 and k2 are the rate constant at the temperature T1T2
Ea = Activation Energy, R = Gas Constant.
Note :-
r
a
f
ar EEH −=∆ −
ΔrH-
= Enthalpy of the reaction
Ea
f
= Activation energy of the forward reaction
Ea
r
= Activation energy of reverse reaction.
For endothermic reaction ( ΔHθ
> 0 ) hence Ea
r
< Ea
f
and for exothermic
reaction ( ΔHθ
< 0 ) hence Ea
r
> Ea
f
6. How to calculate slope of the line:
R
E
slope a
303.2
−
=
Where Ea = Activation Energy, R = Gas Constant.
CHEMICAL KINETICS
1. Form the rate expression for the following reactions determine their order of
reaction and dimensions of the rate constants.
a) 3NO(g) à N2O(g) + NO2 (g) Rate = K [NO]2
b) H2O2 (aq) + 3 I –
(aq) + 2H+
à 2H2O (l) +I-
3 Rate =- k [H2O] [I-
]
c) CH3 CHO(g) à CH4(g) + CO(g) Rate = k [CH3 CHO]3/2
d) CHCl3 (g) + Cl2 (g) à CCl4 (g) + HCl(g): Rate = [CHCl3] [Cl2] 1/2
37

e) C2 H5 Cl(g) à C2H4(g) = HCl(g) : Rate = k [C2H5Cl]
Ans:-
a) Order = 2 Unit of rate constant is s-1
mol-1
L [M-1
s-1
]
b) Order = 2; Unit of rate constant is s-1
mol-1
L or M-1
s-1
c) Order = 3/2 ; Unit of rate constant is s-1
; mol-1/2
L1/2
d) Order = 3/2 ; Unit of rate constant is s-1
mol-1/2
L1/2
e) Order = 1 Unit of rate constant is s-1
2. For the reaction
2A + B + C à A2 B+C
The rate = k[A] [B]2
with k = 2 x 10-6
M-2
S-1
, calculate initial rate of the
reaction when [A] = 0.1M [B] = 0.2M and [c] = 0.8 M. If the rate of reverse
reaction is negligible, then calculate the rate of reaction after [A] is reduced to
0.06M.
Ans:-
Rate = k [A] [B] 2
= 2 x 10-6
M-2
s-1
x 0.1 M x (0.2M)2
= 8 x 10-9
5Ms-1
.
When [A] is reduced to0.06m?
[B] = 0.2 – 0.02 = 0.18 M
∴ Rate = 2 x 10-6
x 0.06 x (0.18)2
= 38.88 x 10-10
= 3.888 x 10-9
Mol L-1
s-1
.
3. The rate of decomposition of NH3 on platinum surface is zero order. What are the
rate of production of N2 and H2 if k = 2.5 x 10-4
M-1
s-1
.
Ans:-
2NH3 à N2 + 3H2
Rate of decomposition of NH3 = k[NH3]0
= k = 2.5 x 10-4
M s-1
-
[ ] [ ] [ ]
dt
Hd
dt
Nd
dt
NHd 223
3
1
2
1
==
38

Rate of formation of N2 =
[ ]
dt
NHd
22
1 3
= ½ x 2.5 x 10-4
M s-1
= 1.25x10-4
Rate of formation of H2 =
[ ]
dt
NHd 3
2
3
=
14
105.2
2
3 −−
×× Ms
= 3.75 x 10-4
M s-1
4. In a pseudo first order hydrolysis of ester in water following results were
obtained.
T/s 0 30 60 90
[Ester]/M 0.55 0.31 0.17 0.085
i) Calculate the average rate of reaction between the time intervals 30 to 60
seconds.
ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
i) Average rate =
( ) ( )
3060
31.017.0
12
12
−
−−
=
−
−−
tt
CC
=
30
14.0
= 4.6 x 10-3
Mol s-1
ii) at t = 30 min. k =
R
R
t
0
log
303.2
=
31.0
55.0
log
30
303.2
=
12
min1091.1249.0
30
303.2 −−
×=×
At t = 60 min. k =
17.0
55.0
log
60
303.2
= 51.0
60
303.2
× = 1.95 x 10-2
min-1
Since the value of k [H2O] is constant so it is a pseudo first order reation.
5. In a reaction between A and B, the initial rate of reaction was measured for
different initial concentration of A and B as given below.
39

[A]/M 0.2 0.2 0.4
[B]/M 0.3 0.1 0.05
r0 / Ms-1
5.07 x 10-5
5.07 x 10-5
7.6 x 10-5
What is the order of reaction with respect to A and B?
-
[ ]
dt
Ad
= rate = k [A]x
[B]y
5.07 x 10-5
= k x 0 .2x
x 0 .3y
à (1)
5.07 x 10-5
= k x 0 .2x
x 0.1y
à (2)
7.6 x 10-5
= k x 0 .4x
x 0.05y
à (3)
Dividing (1) by (2)
1 = (0.3./0.1)y
= 3y
y = 0.
Dividing equation (2) by (3)
( )x
x
2
2.0
4.0
1007.5
106.14
5
5
=





=
×
×
−
−
2.87 = 2x
log2.87= xlog2
x = 1.5
Order with respect of [A] is 1.5
Thus order or reaction w.r.t. A is 1.5 and w.r.t. B is zero. The overall order is 1.5.
6. Reaction between NO2 and F2 to give NO2 F takes place by the following
mechanism.
NO2 (g) + F2(g) slow > NO2 F(g) + F(g)
NO2(g) + F(g) fast > NO2F(g)
Write the rate expression for the reaction.
The rate-determining step is the slow step. So rate law is
Rate = k[NO2] [F2].
7. The rate constant for a first order reaction 60 s-1
. How much time will it take to
reduce the initial concentration of the reactant to its 1/16th
value?
40

k = 60 s-1
, Initial concentration = R0
Concentration at time t =
16
1
R0
t =
tR
R0
log
60
303.2
=
16
log
60
303.2
0
0
R
R
= 2041.1
60
303.2
16log
60
303.2
×=
= 4.62 x 10-2
s.
8. The rate of most of the reactions double when their temperature is raised from
298k to 308k. Calculate their activation energy.
log
1
2
k
k
= 





−
21
11
303.2 TTR
Ea
Ea =
12
2
21 1
log303.2
TT
k
k
TRT
−
×
=
298308
2log298308314.8303.2
−
××××
= 52.897 kJ mol -1
.
9. The half-life for radioactive decay of 14
C is 5730 y. an archeological artefact
contained wood had only 80% of the 14
C found in a living tree. Estimate the age
of sample.
λ =
2/1
693.0
t
= years5750
693.0
t =
tN
N0
log
303.2
λ
=
80
100
log
693.0
5730303.2 years×
= 1845 years.
41

10. During nuclear explosion, one of the product is 90
Sr with half life 28.1 Y. if 1µg
of 90
Sr was absorbed in the bones of a newly born baby instead of calcium, how
much of it will remain after 10 years and 60 years if it is not lost metabolically?
λ =
yearst 1.28
693.0693.0
2/1
=
t =
tN
N0
log
303.2
λ
10 years =
tN
gµ1
log
693.0
1.28303.2 ×
1071.0
2.28303.2
693.0101
log =
×
×
=
tN
tN
1
= 1.279
Nt =
279.1
1
= 0.7818 μg
When t = 60 years
6427.0
2.28303.2
693.0601
log =
×
×
=
tN
tN
1
= 4.392
Nt=
392.4
1
= 0.228 μg.
42

CBSE CLASS 12 CHEMISTRY NUMERICALS MADE EASY

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