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1. Arrange the following metals in the order in which they displace each
other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
•
Solution:
From electrochemical series the Eo values for reduction reactions of the above elements are given as
Eo Al3+/Al = -1.66 v
Eo Cu2+/Cu = + 0.34v
Eo Fe2+/Fe = -0.44v
Eo Mg2+/Mg = - 2.36v
Eo Zn2+/ Zn = - 0.76v
The reactivity order is given as
Mg > Al > Zn > Fe > Cu
Mg displaces all the other elements from their salt solution while Cu is least reactive.
2. Given the standard electrode potentials, arrange these metals in their
increasing order of reducing power.
K+/K = -2.93v, Ag+/Ag = 0.80v
Hg2+ / Hg = 0.79v
Mg2+/Mg = -2.37v
Cr3+/Cr = -0.74 v
•
Solution:
Higher value of Eo reduction values shows that the metal undergoes reduction easily. Therefore from
the Eo reduction values of the elements it is seen that
Ag < Hg < Cr < Mg < K
3. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+aq
Zn2+(aq) + 2Ag(s) takes place further show.
(i) which electrode is negatively charged
(ii) The carriers of the current in the cell
(iii) Individual reaction at each electrode.
Solution:
The galvanic cell is
Zn(s) | Zn2+(aq) || Ag+ | Ag(s)
in
•
Zn(s) / Zn2+(aq), serves as Anode as the oxidation takes place and is given the symbol (-ve sign)
The ions carry the current in the cell.
At anode the half cell reaction
Zn Zn2+ + 2e-
1
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At Cathode
2Ag + 2e Ag
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4. Calculate the standard cell potentials of galvanic cell in which the
following reaction takes place.
3+
(i) 12Cr(s) + 3Cd2+aq →€2Cr
→€
(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ∆ rGo and equilibrium constant of the reaction.
•
Solution:
(i) Eo Cr3+/Cr = - 0.74v
Eo Cd2+/Cd = -0.40 v
Eo cell = Eo cathode – Eo anode
= 0.40 + 0.74
+ Eocell = -0.40-[-0.74] = + 0.34v
Relationship between ∆ rGo and Eo cell is
∆ rGo = - n F Eo cell
n = 6 e-s
F = 96,500
Eocell = +0.34v
∆ rGo = -6 × 96,500 × 0.34 v
∆ rGo= -196860 J
∆ rGo= -196.86 J mol-1
∆ rGo= -2.303 RT log K
log K = log K =
=
log K = 34.5065
K = 3.21 × 1034
= 34.5065
in
(ii) Fe2+(aq) + AG+(aq) Fe3+(aq) + Ag(s)
Eo Fe3+ / Fe2+ = + 0.77v
Eo Ag+ / Ag = + 0.80v
Eo cell = 0.80 – 0.77
= 0.03 v
∆ rGo = -n F Eo cell
n=1
F = 96,500
Eo cell = 0.03v
∆ rGo = -1 × 96,500 × 0.03
= -2895 J
∆ rGo = -2.895 KJ mol-1
log K = =
2
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=
log K = 0.5073
K = Antilog 0.5073
K = 3.216.
=
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5. Write the Nernst equation and e.m.f of the following cells at 298 K
(i) Mg(s) | Mg2+(0.001M) || Cu2+(0.001M) | Cu(s)
(ii) Fe(s) | Fe2+(0.001M) || H+(1M)| H2(g)(1 bar), Pt(s)
(iii) Sn(s) | Sn2(0.050M) || H+(0.020M) | H2(g) (1bar) Pt(s)
(iv) Pt(s) | Br2(l) | Br-(0.010M) || H+(0.030M) | H2(g) (1 bar) Pt(s)
•
Solution:
Nernst equation brings the relationship between EMF of the cell under given set of conditions with
that of Eo cell . Under standard conditions.
Ecell = Eocell +
log
(i) The cell reaction is
Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
n = 2 electrons
Ecell = Eocell +
log
Eocell = Eo Cathode - Eo Anode
Eo Mg2+ / Mg = -2.37 V
Eo Cu2+/ Cu = + 0.34 V
Eo cell = 0.34 + 2.37
= 2.71 v
Ecell = 2.71 +
log
= 2.71 + 0.0295 log 0.1
= 2.71 + 0.0295 × -1
= 2.71 - 0.0295
Ecell = 2.68v
Ecell = 0.44 +
in
(ii) The cell reaction is
Fe(s) + 2H+(aa) Fe2+(aq) + H2(g)
Eo Fe2+/Fe = -0.44v
Eo 2H+/H2 = 0
Eocell = + 0.44v
log
= 0.44 +
log
= 0.44 + 0.0295 log 1000
3
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Ecell
= 0.44 + 0.0295 × 3.0000
= 0.44 + 0.0885
= 0.5285v
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(iii) The cell reaction is
Sn(s) + 2H(aq)+ Sn2+(aq) + H2(g)
Eo Sn2+/Sn = -0.14v
Eo 2H+ / H2(g)= 0
Eocell = + 0.14 v
Ecell = Eocell +
log
= Eocell + 0.0295 log
= 0.14 + 0.0295 log
= 0.14 + 0.0295 log 0.0008
= 0.14 + 0.0295 ×
.9031
= 0.14 + 0.0295 × -2.0969
= 0.14 - 0.0295 × 2.0969
= 0.14 - 0.0618
Ecell = 0.0782 v
Cell reaction:
(Note carefully)
Nernst equation:
∴€
.
electrode.
in
Thus, oxidation will occur at the hydrogen electrode and reduction on the
(iv) The cell reaction is
Br2 + 2H+ 2Br- + H2
Eo Br2/ Br- = 1.09v
Eo 2H+ / H2 = 0
Eocell = 0 - 1.09 v
Eocell = -1.09v
Ecell = -1.09 +
4
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= - 1.09 +
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= -1.09 +
= -1.09 + (0.0295 log 9)
= - 1.09 + (0.0295 × 0.9542)
= -1.09 + 0.0281
E cell = -1.062 v.
6. In the button cell widely used in watches and other devices the
following reaction takes place.
Zn(s) + Ag2O(s) + H2O(l)
Zn2+(aq) + 2Ag(s) + OH-(aq)
Determine Eo and ∆ rGo for the reaction.
•
Solution:
EoZn2+/2n = - 0.76v
Eo Ag+/Ag = + 0.80v
Eocell = Eo cathode - Eo anode
= 0.80 + 0.76
Eocell = 1.56v
∆ rGo = -nF Eocell
n=2
F = 96,500 c
Eocell = 1.56 v
∆ rGo = 2 × 96,500 × 1.56
= 301080 J
∆ rGo = 301.080 KJ mol- .
7. Define conductivity and molar conductivity for the solution of an
electrolyte.Discuss their variation with concentration.
•
Solution:
Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1
sq.cm as the area of cross section.
in
Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one
mole of the electrolyte dissolved in Vcm3 of the solution when the electrodes are one cm apart and
the area of the electrodes is so large that the whole solution is contained between them. It is usually
represented by λm. From definition Λ m = K x V where K is specific conductivity and V is volume in
Cm3, containing 1g mole of electrotype.
Conductivity always decreases with the decrease in concentration both for weak and strong
electrolytes. This is due to the fact that the number of ions per unit volume that carry the current in
a solution decreases on dilution. The conductivity of a solution at any given concentration is the
conductance of unit volume of solution between two platinum electrodes with unit area of cross
section and at a distance of unit length. The relationship between conductivity and conductance of
the solution.
5
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Conductance = Conductivity x
C=K
both A and ℓ are unity in their appropriate units in m or cm.
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a. Molar conductivity λ m =κ /c where κ is the electrolytic conductivity and c is the concentration of
the solution in mol/cm3. For strong electrolytes the molar conductivity decreases slightly with
increasing concentration but for weak electrolytes the molar conductivity shows a steep increase as
concentration decreases.
b. Molar conductivity λ m = 1000 κ /c where κ is electrolytic conductivity and c is concentration.
Electrolytic conductivity =2.48×10-2 / Ω cm, concentration =0.20 mol/litre.
Therefore molar conductivity=1000 κ /c ohm-1mol-1cm2 = 1000×2.48×10-2 /0.2 =124 ohm-1mol1
cm2.
8. The conductivity of 0.20 M solution of K Cl at 298K is 0.0248 Scm-1.
Calculate its molar conductivity.
•
Solution:
λm =
K = 0.248 S cm-1
C = 0.20 M
= 124 S cm2 mol
λm =
–1
λm = 124 S cm2 mol-1 .
9. The resistance of a conductivity cell containing 0.001M KCl solution at
298K is 1500 Ω what is the cell constant if conductivity of 0.001 M KCl
solution at 298K is 0.146 × 10-3 cm-1?
•
Solution:
Conductance =
R = 1500 ohms K = 0.146 × 10-3 cm-1
=
= 0.000666 ohms-1
Specific conductance (k) = Conductance × cell constant
Specific conductance of 0.0001 M KCl = 0.146 × 10-3 S cm-1
in
C=
C=K×
Cell constant =
=
Cell constant = 0.219 cm-1.
6
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10. The conductivity of Sodium chloride at 298 K has been determined at
different concentration and the results are given below:
Concentration / M 0.001 0.010 0.020 0.050 0.100 102 × K/Sm-1 1.233
11.85 23.15 55.53 106.74. Calculate Λ for all concentrations and draw a
plot between Λ and C1/2 . Find the value of Λ o
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Solution:
The relationship between Λ€(molar conductance) and specific molar – Conductivity Λ =
(unit conversion factor)
Conc. (M)
0.0316
0.100
0.141
0.224
0.316
in
11. Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1 λ ∞ .
Calculate its molar conductivity and if λ ∞ for acetic acid is 390.5 S cm2mol-1.
What is the dissociation constant?
•
Solution:
=
K = 7.896 × 10-5 Cm-1
C = 0.00241
λ
m
λ
m
=
7
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λ m = 32.76 S cm2 mol-1
λ m = Acetic acid = 32.76
λ ∞ m Acetic acid = 390.5
α the degree of dissociation =
= 0.0839
α = 8.4%
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[CH3COOH] = C (1-α )
[ CH3COO-] = Cα
[ H3O+] = Cα
[CH3COOH] = 0.0024 × [1 – 0.0839] = 0.00219
[CH3COO-] = 0.0024 × 0.0839 = 0.00020
[H2O+] = 0.0024 × 0.0839 = 0.00020
Kα =
=
Kα = 1.8 × 10-5 .
12. How much charge is required for the following reaction of
(i)1 mole of Al3+ to Al
(ii)1 mole of cu2+ to Cu
(iii)1 mole of Mno
•
to Mn2+
Solution:
in
8
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13. How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from Molten Ca Cl2
(ii) 40.0 g of Al from molten Al2 O3
•
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Solution:
According to Faraday’s first law the amount of the substance deposited is indirectly proportional to
the quantity of electricity passed through it.
To deposit one gram atom or mole of an element nF coulombs are required where n is the charge on
the ion
Atomic mass of Ca = 40 g
To produce 40 gms 2F is required. ∴ 20 gms is produced by 1F.
14. How much electricity is require in coulomb for the oxidation of"1
mole of H2O to O2 1 mole of FeO to Fe2O3
•
Solution:
4 Faraday is required to oxidise 2 moles of water. According to the
balanced equation and 2
Faraday is required to oxidise 1 mole of water.
FeO to Fe2O3
The balanced chemical reaction is
in
According to the balanced chemical reaction 2 moles of Fe2+ ions is oxidised to 2 moles of Fe3+ by
losing two electrons. Therefore to oxidise one Fe2+ to Fe3+ one e- is lost. Hence the charge carried by
mole of electrons is one Faraday.
15. A solution of Ni (NO3)2 is electrolysed between platinum electrodes
using a current of 5 amperes for 20 minutes. What mass Ni is deposited at
the cathode?
9
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•
Solution:
To deposit one mole of Ni two moles of electrons are needed, therefore
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2 x 96,500 C deposits 58.5 gms of Ni
5 x 20 x 60C deposits =
Amount of Ni deposited = 1.838 g.
16. Three electrolytic cells ABC containing solution of ZnSO4, AgNO3 and
Cu SO4 respectively are connected in series. A steady current of 1.5
amperes was passed through them until 1.45 g of silver deposited. At the
cathode cell B. How long did the current flow? What mass of copper and of
Zinc were deposited.
•
Solution:
For the deposition of Ag reaction is
Thus 1 mol = 107.8 g Ag is deposited by
1F= 96,500 C
107.8 gm of Ag is deposited by 96,500 C
1.45g of Ag is deposited by =
According Faraday’s first Law = Q = I t
1298 C = 1.5 x t
t=
= 1298 C
= 865.22 sec
time in minutes =
= 14.4 min
time = 14.4 min
The amount of Cu deposited according to the equation is
in
10
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17. Using the standard electrode potential given in table 5 predict if the
reaction between the following is feasible.
(a) Fe3+ (aq) and I- (aq)
(b) Ag+ (aq) and Cu(s)
(c) Fe3+ (aq) and Br- (aq)
(d) Ag (s) and Fe3+(aq)
(e) Br2(aq) and Fe2+(aq)
•
Solution:
In a given redox reaction, the species undergoing reduction should have relatively higher E0 Red value
than the species which undergoes oxidation. If this condition is fulfilled the reaction will be feasible.
If the E0cell of the reaction is positive, the reaction will be feasible.
a. E cell :- E0 Fe+3 / Fe+2 - E0 I-/I2 =0. 77V- 0.54V = 0.23V. So the reaction taking place in the cell is
feasible.
2Fe+3 + 2I- →€2Fe2+ + I2
b. E cell :- E0 Ag+/ Ag - E0 Cu+2/Cu =0.80V- 0.34V = 0.46V. The reaction taking place in the cell will
be feasible.
Cu + 2Ag+ ® Cu
2+
+ 2 Ag
c. E cell :- E0 Fe+3 / Fe+2 - E0 Br2/Br- = 0.77V-1.08V =-0.31V. The reaction taking place in the cell is
not feasible.
Fe+3 cannot oxidise Br- to Br2.
d. E cell :- E0 Fe+3 / Fe+2 - E0 Ag+/ Ag = 0.77V-0.80V=-0.03V .The reaction taking place in the cell is
not feasible.
Fe3+ cannot oxidise Ag to Ag+.
e. E cell :- E0 Br2/Br- -E0 Fe+3 / Fe+2 = 1.08V- 0.77V =0.31V. The reaction taking place in the cell is
feasible.
Br2 + 2Fe+2 → 2Br- + 2Fe
3+
•
Solution:
in
18. Predict the products of electrolysis in each of the following
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of Ag NO3 with P+ electrodes
(iii) A dilute solution of H2SO4 with Pt electrodes.
(iv) An aqueous solution of CuCl2 with Pt electrodes.
(i) AgNO3 undergoes ionisation in the presence of Ag electrodes.
AgNO3 Ag+ + NO3At Cathode Ag+ + e- Ag
At Anode Ag Ag+
(ii) AgNO3 undergoes ionisation in the presence of pt electrodes forming
AgNO3 Ag+ + NO3-
11
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At cathode Ag+ + e- Ag Eo = 1.98V
H2O also undergoes reduction as follows:
2H2O + 2e- 2OH- + H2 Eo – 0.83 V
As the Eo value of Ag+/Ag is greater Silver is deposited at Cathode
(reduction) At anode
H2O undergoes oxidation as well as NO3- ion represented as follows:
2H2O 4H+ + 4e- + O2Eo = + 0.40v
NO3- NO + O2 Fo = very low -eTherefore H2O undergoes oxidation in preference to NO3- . Therefore O2 is liberated at anode
Therefore the products are Ag at Cathode O2 at anode
(iii) Dilute H2SO4 undergoes ionization according to the equation
H2SO4 2H+ + SO42At Cathode
2H+ + 2e- H2
At Anode
2H2O 4H+ + 4e- + O2 Eo = +0.40 v
(iv) An aq solution of CuCl2 in the presence of Pt electrodes undergoes ionization as follows
CuCl2 → Cu2+ + 2ClAt Cathode
Cu2 + 2e- → Cu. Eo + 0.34v
2H2O + 2e- → H2 + 2OH- Eo ?0.83v
At anode
2H2O → 4H+ + 4e- + O2 Eo + 0.40v
2Cl- → Cl2 + 2e- Eo = -1.36v
Eo oxidation of 2Cl-/Cl2 is very low.
But due to over potential chlorine is liberated instead of oxygen
19. How would you determine the standard electrode potential of the
system Mg2+|Mg?
•
Solution:
in
The electrode is connected as anode to standard hydrogen electrode and a cell is formed. The
potential of the cell is read on the voltmeter . It is the oxidation electrode potential of the given
electrode as it is connected at the anode. The electrode potential is given with negative sign.
20. Can you store copper sulphate solutions in a zinc pot?
•
Solution:
When copper sulphate is stored in a nickel container, nickel forms the anode and copper forms the
cathode. The cell formed will be spontaneous with a positive cell potential and so reaction will take
12
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place which means nickel will go into solution and copper will be deposited. This will form holes in the
container and so you cannot store it.
21. Consult the table of standard electrode potentials and suggest three
substances that can oxidise ferrous ions under suitable conditions.
Solution:
Chlorine, bromine and oxygen are the three substances that can oxidize ferrous ions under suitable
conditions.
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22. Calculate the emf of the cell in which the following reaction takes
place:
Ni(s) + 2Ag+ (0.002 M) Ni2+ (0.160 M) + 2Ag(s)
Given that E0(cell) = 1.05 V
•
Solution:
E0(cell) = 1.05V
[Ag+] = 0.002M
[Ni2+] = 0.16M
= 1.05 - 0.0295 x 4.6021
=1.05 – 0.1358
=0.9142.
in
23. The cell in which the following reaction occurs:
2Fe3+ (aq) + 2I- (aq) 2Fe2+ (aq) + I2 (s) has E0(cell) = 0.236 V at 298 K. Calculate the standard
Gibbs energy and the equilibrium constant of the cell reaction.
•
Solution:
13
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Kc = 108
∆G0 = - RT ln Kc = - 2.303 RT log Kc
= - 2.303 x 8.314 x 298 x log 108
= -45646.8J
24. Suggest a way to determine the٨0m value of water.
•
Solution:
value of water can be determined by using Kohlrausch law.
25. The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S
cm2 mol–1. Calculate its degree of dissociation and dissociation constant.
Given λ0(H+)= 349.6 S cm2 mol–1 and λ0(HCOO–) =54.6 S cm2 mol–1.
•
Solution:
٨0m (HCOOH) = λ0(H+) + λ0(HCOO-)
= 349.6 + 54.6
= 404.5 S cm2mol-1
in
= 9.562 x 10-3.
26. If a current of 0.5 ampere flows through a metallic wire for 2 hours,
then how many electrons would flow through the wire?
•
Solution:
14
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Charge = current x time = 0.5 A x 2 x 60 x 60 s = 3600 C
Number of electrons that flow =3600€×€6.023€×€1023/ 96500
=2.25€×1022
Note- 96500 coulumbs is the charge of one mole of electrons
27. Suggest a list of metals that are extracted electrolytically.
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Solution:
The metals like sodium, potassium, calcium and aluminium which belong to the high reactivity series
can be extracted electrolytically.
28. Write the chemistry of recharging the lead storage battery,
highlighting all the materials that are involved during recharging.
•
Solution:
A lead storage battery can be charge by passing electric current of a suitable voltage in the opposite
direction. The electrode reaction gets reversed. As a result, the flow of electrons gets reversed and
lead is deposited on anode and PbO2 on the cathode. The density of sulphuric acid also increases.
The reaction may be written as:
2 PbSO4(s) + 2H2O
charge
> Pb(s) + PbO2(s) + 2H2SO4
It may be noted that storage battery acts as voltaic cell as well as electrolytic cell. For example,
when it is used to start the engine of the automobile, it acts as a voltaic cell and produces electric
energy. During recharging, it acts as an electrolytic cell.
29. Suggest two materials other than hydrogen that can be used as fuels
in fuel cells.
•
Solution:
Methanol, potassium hydroxide and phosphoric acid are some materials that can be used as fuels in
fuel cells.
30. Explain how rusting of iron is envisaged as setting up of an
electrochemical cell.
•
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Solution:
Virtually all corrosion reactions are electrochemical in nature and in the same way, the corrosion of
iron can occur in an aqueous electrolyte.
At anodic sites on the surface the iron goes into solution as ferrous ions, this constituting the anodic
reaction. As iron atoms undergo oxidation to ions they release electrons whose negative charge
would quickly build upin the metal and prevent further anodic reaction, or corrosion. Thus this
dissolution will only continue if the electrons released can pass to a site on the metal surface where a
cathodic reaction is possible. At a cathodic site the electrons react with some reducible component
of the electrolyte and are themselves removed from the metal. The rates of the anodic and cathodic
reactions must be equivalent according to Faraday’s Laws, being determined by the total flow of
electrons from anodes to cathodes which is called the "corrosion current", Icor. Since the corrosion
current must also flow through the electrolyte by ionic conduction the conductivity of the electrolyte
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
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"mixed electrode" since simultaneous anodic and cathodic reactions are proceeding on its surface.
The mixed electrode is a complete electrochemical cell on one metal surface.
in
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:16 [email protected] web site www.mathematic.in