NCERT Solutions Class 12

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 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 1)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 2)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 4)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 5)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 1)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 2)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 1)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 4)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 1)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 2)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 3)
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 Determinants : NCERT Solutions – Class 12 Maths (Ex 6)
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 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 2)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 3)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 4)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 6)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 8)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 9)

NCERT SolutionsChemistry
 Aldehydes, Ketones and Carboxylic Acids : NCERT Solutions – Class 12 Chemistry
 Alcohols, Phenols and Ethers : NCERT Solutions – Class 12 Chemistry
 Amines : NCERT Solutions – Class 12 Chemistry
 Biomolecules : NCERT Solutions – Class 12 Chemistry
 Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
 Chemistry in Everyday Life : NCERT Solutions – Class 12 Chemistry
 Coordination Compounds : NCERT Solutions – Class 12 Chemistry
 Electrochemistry : NCERT Solutions – Class 12 Chemistry
 General Principles and Processes of Isolation of Elements : NCERT Solutions – Class 12 Chemistry
 Haloalkanes and Haloarenes : NCERT Solutions – Class 12 Chemistry
 Polymers : NCERT Solutions – Class 12 Chemistry
 Surface Chemistry : NCERT Solutions – Class 12 Chemistry
 The dand fBlock Elements : NCERT Solutions – Class 12 Chemistry
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 The Solid State : NCERT Solutions – Class 12 Chemistry
 Solutions : NCERT Solutions – Class 12 Chemistry

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NCERT SolutionsPhysics
 Electrostatic Potential And Capacitance : NCERT Solutions – Class 12 Physics
 Electric Charges And Fields : NCERT Solutions – Class 12 Physics
 Semiconductor Electronics: Materials, Devices And Simple Circuits : NCERT Solutions – Class 12 Physics
 Ray Optics And Optical Instruments : NCERT Solutions – Class 12 Physics
 Nuclei : NCERT Solutions – Class 12 Physics
 Moving Charges And Magnetism : NCERT Solutions – Class 12 Physics
 Magnetism And Matter : NCERT Solutions – Class 12 Physics
 Electromagnetic Induction : NCERT Solutions – Class 12 Physics
 Dual Nature Of Radiation And Matter : NCERT Solutions – Class 12 Physics
 Current Electricity : NCERT Solutions – Class 12 Physics
 Communication Systems : NCERT Solutions – Class 12 Physics
 Atoms : NCERT Solutions – Class 12 Physics
 Alternating Current : NCERT Solutions – Class 12 Physics
Electrochemistry : NCERT Solutions – Class 12 Chemistry
3.1. 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
Answer
The following is the order in which the given metals displace each other from the solution of their salts.
Mg, Al, Zn, Fe, Cu
3.2. Given the standard electrode potentials,
K^{+}/K = −2.93V, Ag^{+}/Ag = 0.80V,
Hg^{2+}/Hg = 0.79V
Mg^{2+}/Mg = −2.37 V, Cr^{3+}/Cr = − 0.74V
Arrange these metals in their increasing order of reducing power.
Answer
The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K^{+}/K < Mg^{2+}/Mg < Cr^{3+}/Cr < Hg^{2+}/Hg < Ag^{+}/Ag.
Hence, the reducing power of the given metals increases in the following order:
Ag < Hg < Cr < Mg < K
3.3. Depict the galvanic cell in which the reaction Zn(s) + 2Ag^{+}(aq) → Zn^{2+}(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer
The galvanic cell in which the given reaction takes place is depicted as:
(i) Zn electrode (anode) is negatively charged.
(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.
(iii) The reaction taking place at the anode is given by,
The reaction taking place at the cathode is given by,
3.4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd^{2+}(aq) → 2Cr^{3+}(aq) + 3Cd
(ii) Fe^{2+}(aq) + Ag^{+}(aq) → Fe^{3+}(aq) + Ag(s)
Calculate the Δ_{r}G^{θ} and equilibrium constant of the reactions.
Answer
(i)
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
In the given equation,
n = 6
F = 96487 C mol^{−1}
= +0.34 V
Then, = −6 × 96487 C mol^{−1} × 0.34 V
= −196833.48 CV mol^{−1}
= −196833.48 J mol^{−1}
= −196.83 kJ mol^{−1}
Again,
= −RT ln K
= 34.496
K = antilog (34.496)
= 3.13 × 10^{34}
(ii)
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
Here, n = 1.
Then,
= −1 × 96487 C mol^{−1} × 0.03 V
= −2894.61 J mol^{−1}
= −2.89 kJ mol^{−1}
Again,
= 0.5073
K = antilog (0.5073)
= 3.2 (approximately)
3.5. Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s)  Mg^{2+}(0.001M)  Cu^{2+}(0.0001 M)  Cu(s)
(ii) Fe(s)  Fe^{2+}(0.001M)  H^{+}(1M)H_{2}(g)(1bar)  Pt(s)
(iii) Sn(s)  Sn^{2+}(0.050 M)  H^{+}(0.020 M)  H_{2}(g) (1 bar)  Pt(s)
(iv) Pt(s)  Br_{2}(l)  Br^{−}(0.010 M)  H^{+}(0.030 M)  H_{2}(g) (1 bar)  Pt(s).
Answer
(i) For the given reaction, the Nernst equation can be given as:
= 2.7 − 0.02955
= 2.67 V (approximately)
(ii) For the given reaction, the Nernst equation can be given as:
= 0.52865 V
= 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:
= 0.14 − 0.0295 × log125
= 0.14 − 0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:
3.6. In the button cells widely used in watches and other devices the following reaction takes place:
Zn(s) + Ag_{2}O(s) + H_{2}O(l) → Zn^{2+}(aq) + 2Ag(s) + 2OH^{−}(aq)
Determine and for the reaction.
Answer
= 1.104 V
We know that,
= −2 × 96487 × 1.04
= −213043.296 J
= −213.04 kJ
3.7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer
Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of crosssection 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbolκ. If ρ is resistivity, then we can write:
The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of crosssection and at a distance of unit length.
i.e.,
(Since a = 1, l = 1)
Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of crosssection A and distance of unit length.
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
The variation of withfor strong and weak electrolytes is shown in the following plot:
3.8.The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm^{−1}. Calculate its molar conductivity.
Answer
Given,
κ = 0.0248 S cm^{−1}
c = 0.20 M
Molar conductivity,
= 124 Scm^{2}mol^{−1}
3.9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^{−3} S cm^{−1}.
Answer
Given,
Conductivity, κ = 0.146 × 10^{−3} S cm^{−1}
Resistance, R = 1500 Ω
Cell constant = κ × R
= 0.146 × 10^{−3} × 1500
= 0.219 cm^{−1}
3.10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Concentration/M 0.001 0.010 0.020 0.050 0.100
10^{2} × κ/S m^{−1} 1.237 11.85 23.15 55.53 106.74
Calculate for all concentrations and draw a plot between and c½. Find the value of.
Answer
Given,
κ = 1.237 × 10^{−2} S m^{−1}, c = 0.001 M
Then, κ = 1.237 × 10^{−4} S cm^{−1},^{ }c½ = 0.0316 M^{1/2}
= 123.7 S cm^{2} mol^{−1}
Given,
κ = 11.85 × 10^{−2} S m^{−1}, c = 0.010M
Then, κ = 11.85 × 10^{−4} S cm^{−1}, c½ = 0.1 M^{1/2}
= 118.5 S cm^{2} mol^{−1}
Given,
κ = 23.15 × 10^{−2} S m^{−1}, c = 0.020 M
Then, κ = 23.15 × 10^{−4} S cm^{−1}, c^{1/2} = 0.1414 M^{1/2}
= 115.8 S cm^{2} mol^{−1}
Given,
κ = 55.53 × 10^{−2} S m^{−1}, c = 0.050 M
Then, κ = 55.53 × 10^{−4} S cm^{−1}, c^{1/2} = 0.2236 M^{1/2}
= 111.1 1 S cm^{2} mol^{−1}
Given,
κ = 106.74 × 10^{−2} S m^{−1}, c = 0.100 M
Then, κ = 106.74 × 10^{−4} S cm^{−1}, c^{1/2} = 0.3162 M^{1/2}
= 106.74 S cm^{2} mol^{−1}
Now, we have the following data:

0.0316
0.1
0.1414
0.2236
0.3162
123.7
118.5
115.8
111.1
106.74
Since the line interruptsat 124.0 S cm^{2} mol^{−1}, = 124.0 S cm^{2} mol^{−1}.
3.11. Conductivity of 0.00241 M acetic acid is 7.896 × 10^{−5} S cm^{−1}. Calculate its molar conductivity and if for acetic acid is 390.5 S cm^{2} mol^{−1}, what is its dissociation constant?
Answer
Given, κ = 7.896 × 10^{−5} S m^{−1}
c = 0.00241 mol L^{−1}
Then, molar conductivity,
= 32.76S cm^{2} mol^{−1}
Again, = 390.5 S cm^{2} mol^{−1}
Now,
= 0.084
Dissociation constant,
= 1.86 × 10^{−5} mol L^{−1}
3.12. How much charge is required for the following reductions:
(i) 1 mol of Al^{3+} to Al.
(ii) 1 mol of Cu^{2+ }to Cu.
(iii) 1 mol of to Mn^{2+}.
Answer
(i)
Required charge = 3 F
= 3 × 96487 C
= 289461 C
(ii)
Required charge = 2 F
= 2 × 96487 C
= 192974 C
(iii)
i.e.,
Required charge = 5 F
= 5 × 96487 C
= 482435 C
3.13. How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl_{2}.
(ii) 40.0 g of Al from molten Al_{2}O_{3}.
Answer
(i) According to the question,
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium
= 1 F
(ii) According to the question,
Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al
= 4.44 F
3.14. How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H_{2}O to O_{2}.
(ii) 1 mol of FeO to Fe_{2}O_{3}.
Answer
(i) According to the question,
Now, we can write:
Electricity required for the oxidation of 1 mol of H_{2}O to O_{2} = 2 F
= 2 × 96487 C
= 192974 C
(ii) According to the question,
Electricity required for the oxidation of 1 mol of FeO to Fe_{2}O_{3} = 1 F
= 96487 C
3.15. A solution of Ni(NO_{3})_{2} is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer
Given,
Current = 5A
Time = 20 × 60 = 1200 s
Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
3.16. Three electrolytic cells A,B,C containing solutions of ZnSO_{4}, AgNO_{3} and CuSO_{4}, 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 of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Answer
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by =
= 1295.43 C
Given,
Current = 1.5 A
Time
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit
= 0.439 g of Zn
3.17. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe^{3+}(aq) and I^{−}(aq)
(ii) Ag^{+} (aq) and Cu(s)
(iii) Fe^{3+} (aq) and Br^{− }(aq)
(iv) Ag(s) and Fe^{3+} (aq)
(v) Br_{2 }(aq) and Fe^{2+ }(aq).
Answer
Sincefor the overall reaction is positive, the reaction between Fe^{3+}_{(}_{aq}_{)} and I^{−}_{(}_{aq}_{)} is feasible.
Since for the overall reaction is positive, the reaction between Ag^{+}_{ (}_{aq}_{)} and Cu_{(}_{s}_{)} is feasible.
Since for the overall reaction is negative, the reaction between Fe^{3+}_{(}_{aq}_{)} and Br^{−}_{(}_{aq}_{)} is not feasible.
Since E for the overall reaction is negative, the reaction between Ag_{ (}_{s}_{)} and Fe^{3+}_{(}_{aq}_{)} is not feasible.
Since for the overall reaction is positive, the reaction between Br_{2(}_{aq}_{)} and Fe^{2+}_{(}_{aq}_{)} is feasible.
3.18. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO_{3} with silver electrodes.
(ii) An aqueous solution of AgNO_{3}with platinum electrodes.
(iii) A dilute solution of H_{2}SO_{4}with platinum electrodes.
(iv) An aqueous solution of CuCl_{2} with platinum electrodes.
Answer
(i) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag^{+}.
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
Since Pt electrodes are inert, the anode is not attacked by ions. Therefore, OH^{−} or ions can be oxidized at the anode. But OH^{−} ions having a lower discharge potential and get preference and decompose to liberate O_{2}.
(iii) At the cathode, the following reduction reaction occurs to produce H_{2} gas.
At the anode, the following processes are possible.
For dilute sulphuric acid, reaction (i) is preferred to produce O_{2} gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
At anode:
The following oxidation reactions are possible at the anode.
At the anode, the reaction with a lower value of is preferred. But due to the overpotential of oxygen, Cl^{−} gets oxidized at the anode to produce Cl_{2} gas