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CBSE XII CHEMISTRY
Board Paper – 2017 Solution
CBSE
Class XII Chemistry
Board Paper 2017 (Solution)
1. The formula of an oxo-anion of manganese (Mn) in which it shows the
oxidation state equal to its group number is
2
4
MnO
?
.
In
2
4
MnO
?
, the oxidation state of Mn is
Mn + (-2 × 4) = -1
Mn – 8 = -1
Mn = +7
The group number of manganese is 7.
2. The IUPAC name of the given compound is N-ethyl-N-methylethanamine.
3. The order of the given reaction is first order. The rate of reaction for the first
order reaction is independent of the initial concentration of the reaction.
4. Structure of 1-bromo-4-chlorobut-2-ene:
5. One similarity between physisorption and chemisorption is that the rate of
adsorption increases with an increase in the surface area of the adsorbent.
6. Completed reactions are
(i)
3 2 3
NH 3Cl (excess) NCl 3HCl ? ? ?? ? ?
(ii)
6 2 2 2
XeF 2H O XeO F 4HF ? ? ?? ? ?
OR
(i) When (NH4)2Cr2O7 is heated, it decomposes to produce Cr2O3, H2O and
N2.
4 2 2 7 2 3 2 2
(NH ) Cr O Cr O 4H O N
?
? ? ? ? ? ?
(ii) H2PO3 or orthophosphorous acid (or phosphorous acid) on heating
dissociates to give orthophosphoric acid or phosphoric acid and
phosphine.
3 3 3 4 3
4H PO 3H PO PH
?
? ? ? ? ?
Page 2
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
CBSE
Class XII Chemistry
Board Paper 2017 (Solution)
1. The formula of an oxo-anion of manganese (Mn) in which it shows the
oxidation state equal to its group number is
2
4
MnO
?
.
In
2
4
MnO
?
, the oxidation state of Mn is
Mn + (-2 × 4) = -1
Mn – 8 = -1
Mn = +7
The group number of manganese is 7.
2. The IUPAC name of the given compound is N-ethyl-N-methylethanamine.
3. The order of the given reaction is first order. The rate of reaction for the first
order reaction is independent of the initial concentration of the reaction.
4. Structure of 1-bromo-4-chlorobut-2-ene:
5. One similarity between physisorption and chemisorption is that the rate of
adsorption increases with an increase in the surface area of the adsorbent.
6. Completed reactions are
(i)
3 2 3
NH 3Cl (excess) NCl 3HCl ? ? ?? ? ?
(ii)
6 2 2 2
XeF 2H O XeO F 4HF ? ? ?? ? ?
OR
(i) When (NH4)2Cr2O7 is heated, it decomposes to produce Cr2O3, H2O and
N2.
4 2 2 7 2 3 2 2
(NH ) Cr O Cr O 4H O N
?
? ? ? ? ? ?
(ii) H2PO3 or orthophosphorous acid (or phosphorous acid) on heating
dissociates to give orthophosphoric acid or phosphoric acid and
phosphine.
3 3 3 4 3
4H PO 3H PO PH
?
? ? ? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
7.
(i) Colligative properties: Properties of solutions which depend on the
number of solute particles and are independent of their chemical identity
or their nature relative to the total number of particles present in the
solution are called colligative properties.
Examples: Lowering of vapour pressure, elevation of boiling point,
depression of freezing point, osmotic pressure
(ii) Molality: Molality (m) is the number of moles of the solute per kilogram
(kg) of the solvent. It is expressed as
Molesof soluteinkg
Molality(m)
Molesof solventinkg
?
Example: 1.00 mol kg
-1
(or 1.00 m) solution of KCl means that 1 mol
(74.5 g) of KCl is dissolved in 1 kg of water.
8.
(i) Structure of H2S2O7:
(ii) Structure of XeF6:
9. Degree of dissociation of a weak electrolyte is given by the relation:
m m
m
m
Where,
degreeof dissociationof theelectrolyte
molar conducivity
molar conductivity atinf inf initedilution
? ? ? ??
??
??
? ? ?
Page 3
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
CBSE
Class XII Chemistry
Board Paper 2017 (Solution)
1. The formula of an oxo-anion of manganese (Mn) in which it shows the
oxidation state equal to its group number is
2
4
MnO
?
.
In
2
4
MnO
?
, the oxidation state of Mn is
Mn + (-2 × 4) = -1
Mn – 8 = -1
Mn = +7
The group number of manganese is 7.
2. The IUPAC name of the given compound is N-ethyl-N-methylethanamine.
3. The order of the given reaction is first order. The rate of reaction for the first
order reaction is independent of the initial concentration of the reaction.
4. Structure of 1-bromo-4-chlorobut-2-ene:
5. One similarity between physisorption and chemisorption is that the rate of
adsorption increases with an increase in the surface area of the adsorbent.
6. Completed reactions are
(i)
3 2 3
NH 3Cl (excess) NCl 3HCl ? ? ?? ? ?
(ii)
6 2 2 2
XeF 2H O XeO F 4HF ? ? ?? ? ?
OR
(i) When (NH4)2Cr2O7 is heated, it decomposes to produce Cr2O3, H2O and
N2.
4 2 2 7 2 3 2 2
(NH ) Cr O Cr O 4H O N
?
? ? ? ? ? ?
(ii) H2PO3 or orthophosphorous acid (or phosphorous acid) on heating
dissociates to give orthophosphoric acid or phosphoric acid and
phosphine.
3 3 3 4 3
4H PO 3H PO PH
?
? ? ? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
7.
(i) Colligative properties: Properties of solutions which depend on the
number of solute particles and are independent of their chemical identity
or their nature relative to the total number of particles present in the
solution are called colligative properties.
Examples: Lowering of vapour pressure, elevation of boiling point,
depression of freezing point, osmotic pressure
(ii) Molality: Molality (m) is the number of moles of the solute per kilogram
(kg) of the solvent. It is expressed as
Molesof soluteinkg
Molality(m)
Molesof solventinkg
?
Example: 1.00 mol kg
-1
(or 1.00 m) solution of KCl means that 1 mol
(74.5 g) of KCl is dissolved in 1 kg of water.
8.
(i) Structure of H2S2O7:
(ii) Structure of XeF6:
9. Degree of dissociation of a weak electrolyte is given by the relation:
m m
m
m
Where,
degreeof dissociationof theelectrolyte
molar conducivity
molar conductivity atinf inf initedilution
? ? ? ??
??
??
? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
Given
21
21
3
21
m
(H ) 349.6Scm mol
(CH COO ) 40.9Scm mol
Molar conducvtivity( ) 39.05Scm mol
??
??
?
? ? ?
? ? ?
??
Hence, limiting molar conductivity of acetic acid is
m3
21
acetic acid (H ) (CH COO )
40.9 349.8
390.7cm mol
??
?
? ? ? ? ? ? ? ?
??
?
Degree of dissociation of acetic acid,
m m
39.05
390.7
0.0999
0.1
? ? ? ??
?
?
?
Degree of dissociation of acetic acid is 0.1.
10.
(i) Wolff–Kishner reduction
(ii) Etard reaction
Page 4
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
CBSE
Class XII Chemistry
Board Paper 2017 (Solution)
1. The formula of an oxo-anion of manganese (Mn) in which it shows the
oxidation state equal to its group number is
2
4
MnO
?
.
In
2
4
MnO
?
, the oxidation state of Mn is
Mn + (-2 × 4) = -1
Mn – 8 = -1
Mn = +7
The group number of manganese is 7.
2. The IUPAC name of the given compound is N-ethyl-N-methylethanamine.
3. The order of the given reaction is first order. The rate of reaction for the first
order reaction is independent of the initial concentration of the reaction.
4. Structure of 1-bromo-4-chlorobut-2-ene:
5. One similarity between physisorption and chemisorption is that the rate of
adsorption increases with an increase in the surface area of the adsorbent.
6. Completed reactions are
(i)
3 2 3
NH 3Cl (excess) NCl 3HCl ? ? ?? ? ?
(ii)
6 2 2 2
XeF 2H O XeO F 4HF ? ? ?? ? ?
OR
(i) When (NH4)2Cr2O7 is heated, it decomposes to produce Cr2O3, H2O and
N2.
4 2 2 7 2 3 2 2
(NH ) Cr O Cr O 4H O N
?
? ? ? ? ? ?
(ii) H2PO3 or orthophosphorous acid (or phosphorous acid) on heating
dissociates to give orthophosphoric acid or phosphoric acid and
phosphine.
3 3 3 4 3
4H PO 3H PO PH
?
? ? ? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
7.
(i) Colligative properties: Properties of solutions which depend on the
number of solute particles and are independent of their chemical identity
or their nature relative to the total number of particles present in the
solution are called colligative properties.
Examples: Lowering of vapour pressure, elevation of boiling point,
depression of freezing point, osmotic pressure
(ii) Molality: Molality (m) is the number of moles of the solute per kilogram
(kg) of the solvent. It is expressed as
Molesof soluteinkg
Molality(m)
Molesof solventinkg
?
Example: 1.00 mol kg
-1
(or 1.00 m) solution of KCl means that 1 mol
(74.5 g) of KCl is dissolved in 1 kg of water.
8.
(i) Structure of H2S2O7:
(ii) Structure of XeF6:
9. Degree of dissociation of a weak electrolyte is given by the relation:
m m
m
m
Where,
degreeof dissociationof theelectrolyte
molar conducivity
molar conductivity atinf inf initedilution
? ? ? ??
??
??
? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
Given
21
21
3
21
m
(H ) 349.6Scm mol
(CH COO ) 40.9Scm mol
Molar conducvtivity( ) 39.05Scm mol
??
??
?
? ? ?
? ? ?
??
Hence, limiting molar conductivity of acetic acid is
m3
21
acetic acid (H ) (CH COO )
40.9 349.8
390.7cm mol
??
?
? ? ? ? ? ? ? ?
??
?
Degree of dissociation of acetic acid,
m m
39.05
390.7
0.0999
0.1
? ? ? ??
?
?
?
Degree of dissociation of acetic acid is 0.1.
10.
(i) Wolff–Kishner reduction
(ii) Etard reaction
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
11.
12 22 11
-1
f
Solution:
Molarmassof sugar(C H O )=12x12+22x1+11x16
=342gmol
?T = 2 7 3 . 1 5 -2 6 9 . 1 5
=4K
10% solutionof sucrosemeans10gof canesugar ispresent
=100-10=90gof water
10
No.of molesof cane sugar=
342
=029mol
molalityof the solu
-1
ff
f
-1
6 12 6
-1
0.029x1000
tion=
90
=0.324molkg
Weknow,
?T = K x m
4
K=
0.324
=12.334Kkgmol
Molarmassof glucose(C H O )=6x12+12x1+6x16
=180gmol
10% solutionof glucose means10gof canesugar ispresent
=100-10
=90gof water
Number of moles
-1
ff
10
of glucose= mol=0.055mol
180
0.055x1000
Molalityof the solution= =0.616molkg
90
Usingthe relation,
?T = K x m
=12.33x0.616
=7.6
Therefore,the freezing pointof10%glucose solution=273.15-7.6
=265.5K
Page 5
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
CBSE
Class XII Chemistry
Board Paper 2017 (Solution)
1. The formula of an oxo-anion of manganese (Mn) in which it shows the
oxidation state equal to its group number is
2
4
MnO
?
.
In
2
4
MnO
?
, the oxidation state of Mn is
Mn + (-2 × 4) = -1
Mn – 8 = -1
Mn = +7
The group number of manganese is 7.
2. The IUPAC name of the given compound is N-ethyl-N-methylethanamine.
3. The order of the given reaction is first order. The rate of reaction for the first
order reaction is independent of the initial concentration of the reaction.
4. Structure of 1-bromo-4-chlorobut-2-ene:
5. One similarity between physisorption and chemisorption is that the rate of
adsorption increases with an increase in the surface area of the adsorbent.
6. Completed reactions are
(i)
3 2 3
NH 3Cl (excess) NCl 3HCl ? ? ?? ? ?
(ii)
6 2 2 2
XeF 2H O XeO F 4HF ? ? ?? ? ?
OR
(i) When (NH4)2Cr2O7 is heated, it decomposes to produce Cr2O3, H2O and
N2.
4 2 2 7 2 3 2 2
(NH ) Cr O Cr O 4H O N
?
? ? ? ? ? ?
(ii) H2PO3 or orthophosphorous acid (or phosphorous acid) on heating
dissociates to give orthophosphoric acid or phosphoric acid and
phosphine.
3 3 3 4 3
4H PO 3H PO PH
?
? ? ? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
7.
(i) Colligative properties: Properties of solutions which depend on the
number of solute particles and are independent of their chemical identity
or their nature relative to the total number of particles present in the
solution are called colligative properties.
Examples: Lowering of vapour pressure, elevation of boiling point,
depression of freezing point, osmotic pressure
(ii) Molality: Molality (m) is the number of moles of the solute per kilogram
(kg) of the solvent. It is expressed as
Molesof soluteinkg
Molality(m)
Molesof solventinkg
?
Example: 1.00 mol kg
-1
(or 1.00 m) solution of KCl means that 1 mol
(74.5 g) of KCl is dissolved in 1 kg of water.
8.
(i) Structure of H2S2O7:
(ii) Structure of XeF6:
9. Degree of dissociation of a weak electrolyte is given by the relation:
m m
m
m
Where,
degreeof dissociationof theelectrolyte
molar conducivity
molar conductivity atinf inf initedilution
? ? ? ??
??
??
? ? ?
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
Given
21
21
3
21
m
(H ) 349.6Scm mol
(CH COO ) 40.9Scm mol
Molar conducvtivity( ) 39.05Scm mol
??
??
?
? ? ?
? ? ?
??
Hence, limiting molar conductivity of acetic acid is
m3
21
acetic acid (H ) (CH COO )
40.9 349.8
390.7cm mol
??
?
? ? ? ? ? ? ? ?
??
?
Degree of dissociation of acetic acid,
m m
39.05
390.7
0.0999
0.1
? ? ? ??
?
?
?
Degree of dissociation of acetic acid is 0.1.
10.
(i) Wolff–Kishner reduction
(ii) Etard reaction
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
11.
12 22 11
-1
f
Solution:
Molarmassof sugar(C H O )=12x12+22x1+11x16
=342gmol
?T = 2 7 3 . 1 5 -2 6 9 . 1 5
=4K
10% solutionof sucrosemeans10gof canesugar ispresent
=100-10=90gof water
10
No.of molesof cane sugar=
342
=029mol
molalityof the solu
-1
ff
f
-1
6 12 6
-1
0.029x1000
tion=
90
=0.324molkg
Weknow,
?T = K x m
4
K=
0.324
=12.334Kkgmol
Molarmassof glucose(C H O )=6x12+12x1+6x16
=180gmol
10% solutionof glucose means10gof canesugar ispresent
=100-10
=90gof water
Number of moles
-1
ff
10
of glucose= mol=0.055mol
180
0.055x1000
Molalityof the solution= =0.616molkg
90
Usingthe relation,
?T = K x m
=12.33x0.616
=7.6
Therefore,the freezing pointof10%glucose solution=273.15-7.6
=265.5K
CBSE XII CHEMISTRY
Board Paper – 2017 Solution
12.
(i) Solution:
t = 15 min
= 900 s
I= 2A
Molar mass of Ag = 108 gmol
-1
1 F = 96500 C mol
-1
Q=It
Q=2 900
Q=1800 C
?
From the reaction,
Ag
+
+ e
-
? Ag
Mass of Ag deposited
QM
nF
108 1800
1 96500
2.01 g
?
?
?
?
?
Mass of Ag deposited is 2.01 g.
(ii) A fuel cell is an electrochemical cell which converts chemical energy of a
spontaneous redox reaction to electrical energy.
13.
(i) Coordination isomerism
(ii) In [Ni(H2O)6]
2+
, Ni
2+
ion has two unpaired electrons, and hence, it is
coloured.
In [Ni(CN)4]
2-
, there is no unpaired electron, and hence, it is colourless.
(iii) Pentaamminecarbonatocobalt(III) chloride
14.
(i)
Lyophilic sol Lyophobic sol
Relatively stable as strong forces
of interaction exist between
colloidal particles and liquid.
Less stable as weak forces of
interaction exist between colloidal
particles and liquid.
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FAQs on CBSE Chemistry Past year paper (Solutions) - 2017, Class 12 - Chemistry for Grade 12
| 1. What is the CBSE Chemistry Past Year Paper (Solutions) - 2017 for Class 12? |  |
Ans. The CBSE Chemistry Past Year Paper (Solutions) - 2017 for Class 12 is a set of solved question papers from the CBSE board examination held in 2017 for the subject of Chemistry. These solutions provide the answers to the questions asked in the exam, helping students understand the concepts and improve their preparation for future exams.
| 2. How can the CBSE Chemistry Past Year Paper (Solutions) - 2017 help Class 12 students in their exam preparation? | |
Ans. The CBSE Chemistry Past Year Paper (Solutions) - 2017 can be a valuable resource for Class 12 students in their exam preparation. By going through the solved papers, students can familiarize themselves with the exam pattern, understand the type of questions asked, and learn the appropriate way to answer them. It helps them gauge their understanding of the subject, identify their weak areas, and work on them to improve their performance in the actual exam.
| 3. Are the solutions provided in the CBSE Chemistry Past Year Paper (Solutions) - 2017 accurate? | |
Ans. Yes, the solutions provided in the CBSE Chemistry Past Year Paper (Solutions) - 2017 are accurate. These solutions are prepared by subject experts who have extensive knowledge and experience in the field of Chemistry. They carefully analyze the questions and provide step-by-step explanations with accurate answers. Students can rely on these solutions to clarify their doubts and understand the correct approach to solving the problems.
| 4. Can the CBSE Chemistry Past Year Paper (Solutions) - 2017 be used as a revision tool? | |
Ans. Yes, the CBSE Chemistry Past Year Paper (Solutions) - 2017 can be an effective revision tool for Class 12 students. By solving the previous year's question papers and referring to the solutions, students can revise the entire syllabus, reinforce their understanding of the concepts, and practice answering different types of questions. It helps them assess their level of preparation and identify areas that require further revision, thereby enhancing their overall performance in the exam.
| 5. How can students make the best use of the CBSE Chemistry Past Year Paper (Solutions) - 2017? | |
Ans. To make the best use of the CBSE Chemistry Past Year Paper (Solutions) - 2017, students can follow these steps:
1. Start by solving the question paper on your own without referring to the solutions.
2. Once you have attempted all the questions, compare your answers with the solutions provided.
3. Analyze the mistakes you made and understand the correct approach to solving each question.
4. Focus on the topics or types of questions where you struggled the most and revise them thoroughly.
5. Practice solving similar questions from other sources to strengthen your understanding and improve your speed and accuracy.
6. Repeat the process with other question papers to gain more practice and enhance your exam preparation.
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