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CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
CBSE 
Class XII Chemistry ? 
Board Paper – 2019 Solution 
 
   
 
SECTION A 
 
Ans.1.  1-Phenyl butan-2-one 
 
Ans.2. Chloroform in the presence of light and air gets oxidised to phosgene, a 
poisonous gas. It is carbonyl chloride and represented as COCl
2
. To prevent 
the formation of phosgene, chloroform is stored in dark-coloured bottles. 
This reaction is represented as   CHCl
3
 + 1/2 O
2 
? COCl
2
 + HCl 
 
Ans.3.  Cis-dichlorobis(ethylenediamine)cobalt(III)ion 
 
OR 
           Na
2
[NiCl
4
] 
 
Ans.4. A glycosidic bond is a type of covalent bond formed between two 
monosaccharides of carbohydrate or sugar, whereas a peptide bond is a type 
of covalent bond formed between two amino acids of proteins. 
  
 OR 
 
     A nucleotide is composed of a nitrogenous base, sugar and phosphate group, 
whereas a nucleoside is composed of only a nitrogenous base and a 
phosphate group.    
 
Ans.5. The increasing order of acidic character: 
                  Ethanol < Water < Phenol 
 
SECTION B 
 
Ans.6.   
(a) In the same volume of water, 0.1 mole of KCl will have greater boiling point 
elevation, as boiling point elevation is a colligative property which depends 
on the relative amounts of constituents and not on their identity. 
      Since more ions will be produced by KCl (ionic compound) than glucose 
(covalent compound) as there are more component parts in the KCl 
molecule, the KCl solution will have a higher boiling point. 
 
 
 
Page 2


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
CBSE 
Class XII Chemistry ? 
Board Paper – 2019 Solution 
 
   
 
SECTION A 
 
Ans.1.  1-Phenyl butan-2-one 
 
Ans.2. Chloroform in the presence of light and air gets oxidised to phosgene, a 
poisonous gas. It is carbonyl chloride and represented as COCl
2
. To prevent 
the formation of phosgene, chloroform is stored in dark-coloured bottles. 
This reaction is represented as   CHCl
3
 + 1/2 O
2 
? COCl
2
 + HCl 
 
Ans.3.  Cis-dichlorobis(ethylenediamine)cobalt(III)ion 
 
OR 
           Na
2
[NiCl
4
] 
 
Ans.4. A glycosidic bond is a type of covalent bond formed between two 
monosaccharides of carbohydrate or sugar, whereas a peptide bond is a type 
of covalent bond formed between two amino acids of proteins. 
  
 OR 
 
     A nucleotide is composed of a nitrogenous base, sugar and phosphate group, 
whereas a nucleoside is composed of only a nitrogenous base and a 
phosphate group.    
 
Ans.5. The increasing order of acidic character: 
                  Ethanol < Water < Phenol 
 
SECTION B 
 
Ans.6.   
(a) In the same volume of water, 0.1 mole of KCl will have greater boiling point 
elevation, as boiling point elevation is a colligative property which depends 
on the relative amounts of constituents and not on their identity. 
      Since more ions will be produced by KCl (ionic compound) than glucose 
(covalent compound) as there are more component parts in the KCl 
molecule, the KCl solution will have a higher boiling point. 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
(b) (i) Acetic acid ionises as 
            CH
3
COOH ? CH
3
COO
?
 + H
+
 
       
           CH
3
COOH=1- ? 
                        H
+
= ? 
        It produces (1+ ?) species from one CH
3
COOH. 
        Since ? < 1, the van’t hoff factor will lie between 1 and 2. So, i > 1. 
       
     (ii) Acetic acid dimerises in benzene solution. 
             2CH
3
COOH  ?  (CH
3
COOH)
2 
               
2 moles of acetic acid associate to form one mole of dimer, so that 
              i = 1 - (1 - 1/2) ? = 1 - ?/2 
           For association van’t hoff factor: 
                    i < 1  
 
Ans.7.   
CdCl
2
 on addition to AgCl causes cation vacancy. Impurity defect is produced 
when AgCl is doped with CdCl
2
. 
 
OR 
 
(a) Schottky defect 
(b) Metal excess defect 
 
Ans.8.  
(a)   8NH
3
(excess) + 3Cl
2
              6NH
4
Cl + N
2
 
(b)    XeF
6
 + 2H
2
O                  XeO
2
F
2
 + 4HF 
 
Ans.9.  
(a) Sulphurous acid  
         
(b) Sulphuric acid 
                       
 
 
 
 
 
Page 3


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
CBSE 
Class XII Chemistry ? 
Board Paper – 2019 Solution 
 
   
 
SECTION A 
 
Ans.1.  1-Phenyl butan-2-one 
 
Ans.2. Chloroform in the presence of light and air gets oxidised to phosgene, a 
poisonous gas. It is carbonyl chloride and represented as COCl
2
. To prevent 
the formation of phosgene, chloroform is stored in dark-coloured bottles. 
This reaction is represented as   CHCl
3
 + 1/2 O
2 
? COCl
2
 + HCl 
 
Ans.3.  Cis-dichlorobis(ethylenediamine)cobalt(III)ion 
 
OR 
           Na
2
[NiCl
4
] 
 
Ans.4. A glycosidic bond is a type of covalent bond formed between two 
monosaccharides of carbohydrate or sugar, whereas a peptide bond is a type 
of covalent bond formed between two amino acids of proteins. 
  
 OR 
 
     A nucleotide is composed of a nitrogenous base, sugar and phosphate group, 
whereas a nucleoside is composed of only a nitrogenous base and a 
phosphate group.    
 
Ans.5. The increasing order of acidic character: 
                  Ethanol < Water < Phenol 
 
SECTION B 
 
Ans.6.   
(a) In the same volume of water, 0.1 mole of KCl will have greater boiling point 
elevation, as boiling point elevation is a colligative property which depends 
on the relative amounts of constituents and not on their identity. 
      Since more ions will be produced by KCl (ionic compound) than glucose 
(covalent compound) as there are more component parts in the KCl 
molecule, the KCl solution will have a higher boiling point. 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
(b) (i) Acetic acid ionises as 
            CH
3
COOH ? CH
3
COO
?
 + H
+
 
       
           CH
3
COOH=1- ? 
                        H
+
= ? 
        It produces (1+ ?) species from one CH
3
COOH. 
        Since ? < 1, the van’t hoff factor will lie between 1 and 2. So, i > 1. 
       
     (ii) Acetic acid dimerises in benzene solution. 
             2CH
3
COOH  ?  (CH
3
COOH)
2 
               
2 moles of acetic acid associate to form one mole of dimer, so that 
              i = 1 - (1 - 1/2) ? = 1 - ?/2 
           For association van’t hoff factor: 
                    i < 1  
 
Ans.7.   
CdCl
2
 on addition to AgCl causes cation vacancy. Impurity defect is produced 
when AgCl is doped with CdCl
2
. 
 
OR 
 
(a) Schottky defect 
(b) Metal excess defect 
 
Ans.8.  
(a)   8NH
3
(excess) + 3Cl
2
              6NH
4
Cl + N
2
 
(b)    XeF
6
 + 2H
2
O                  XeO
2
F
2
 + 4HF 
 
Ans.9.  
(a) Sulphurous acid  
         
(b) Sulphuric acid 
                       
 
 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
Ans.10. Carbon monoxide and methane are used as fuels in a fuel cell. 
                 Advantages of fuel cells: 
(a) Fuel cells have a higher efficiency than diesel or gas engines. 
(b) The cells were used for electric power generation in the Apollo Space 
Programme, where water vapour produced in the cells was condensed and 
used for drinking purposes. 
 
Ans.11.  
(a)  
 
(b)  
 
 
Ans.12.  
 
(a) The method cannot be used for preparing aromatic primary amines because 
aryl halides do not undergo nucleophilic substitution with the anion formed by 
pthalimide. 
 
(b)  
(i) Primary amine reacts with benzenesulphonyl chloride to form N-
ethylbenzenesulphonyl amide. The hydrogen bonded to nitrogen is strongly 
acidic due to the presence of a strong electron-withdrawing sulphonyl group 
and is hence soluble in alkali. 
(ii) With secondary amine, N,N-diethyl-benzenesulphonamide is formed. N, N-
diethylbenzene sulphonamide does not contain any H atom attached to the 
nitrogen atom, so it is not acidic and is therefore insoluble in alkali. 
 
 
 
 
 
 
 
Page 4


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
CBSE 
Class XII Chemistry ? 
Board Paper – 2019 Solution 
 
   
 
SECTION A 
 
Ans.1.  1-Phenyl butan-2-one 
 
Ans.2. Chloroform in the presence of light and air gets oxidised to phosgene, a 
poisonous gas. It is carbonyl chloride and represented as COCl
2
. To prevent 
the formation of phosgene, chloroform is stored in dark-coloured bottles. 
This reaction is represented as   CHCl
3
 + 1/2 O
2 
? COCl
2
 + HCl 
 
Ans.3.  Cis-dichlorobis(ethylenediamine)cobalt(III)ion 
 
OR 
           Na
2
[NiCl
4
] 
 
Ans.4. A glycosidic bond is a type of covalent bond formed between two 
monosaccharides of carbohydrate or sugar, whereas a peptide bond is a type 
of covalent bond formed between two amino acids of proteins. 
  
 OR 
 
     A nucleotide is composed of a nitrogenous base, sugar and phosphate group, 
whereas a nucleoside is composed of only a nitrogenous base and a 
phosphate group.    
 
Ans.5. The increasing order of acidic character: 
                  Ethanol < Water < Phenol 
 
SECTION B 
 
Ans.6.   
(a) In the same volume of water, 0.1 mole of KCl will have greater boiling point 
elevation, as boiling point elevation is a colligative property which depends 
on the relative amounts of constituents and not on their identity. 
      Since more ions will be produced by KCl (ionic compound) than glucose 
(covalent compound) as there are more component parts in the KCl 
molecule, the KCl solution will have a higher boiling point. 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
(b) (i) Acetic acid ionises as 
            CH
3
COOH ? CH
3
COO
?
 + H
+
 
       
           CH
3
COOH=1- ? 
                        H
+
= ? 
        It produces (1+ ?) species from one CH
3
COOH. 
        Since ? < 1, the van’t hoff factor will lie between 1 and 2. So, i > 1. 
       
     (ii) Acetic acid dimerises in benzene solution. 
             2CH
3
COOH  ?  (CH
3
COOH)
2 
               
2 moles of acetic acid associate to form one mole of dimer, so that 
              i = 1 - (1 - 1/2) ? = 1 - ?/2 
           For association van’t hoff factor: 
                    i < 1  
 
Ans.7.   
CdCl
2
 on addition to AgCl causes cation vacancy. Impurity defect is produced 
when AgCl is doped with CdCl
2
. 
 
OR 
 
(a) Schottky defect 
(b) Metal excess defect 
 
Ans.8.  
(a)   8NH
3
(excess) + 3Cl
2
              6NH
4
Cl + N
2
 
(b)    XeF
6
 + 2H
2
O                  XeO
2
F
2
 + 4HF 
 
Ans.9.  
(a) Sulphurous acid  
         
(b) Sulphuric acid 
                       
 
 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
Ans.10. Carbon monoxide and methane are used as fuels in a fuel cell. 
                 Advantages of fuel cells: 
(a) Fuel cells have a higher efficiency than diesel or gas engines. 
(b) The cells were used for electric power generation in the Apollo Space 
Programme, where water vapour produced in the cells was condensed and 
used for drinking purposes. 
 
Ans.11.  
(a)  
 
(b)  
 
 
Ans.12.  
 
(a) The method cannot be used for preparing aromatic primary amines because 
aryl halides do not undergo nucleophilic substitution with the anion formed by 
pthalimide. 
 
(b)  
(i) Primary amine reacts with benzenesulphonyl chloride to form N-
ethylbenzenesulphonyl amide. The hydrogen bonded to nitrogen is strongly 
acidic due to the presence of a strong electron-withdrawing sulphonyl group 
and is hence soluble in alkali. 
(ii) With secondary amine, N,N-diethyl-benzenesulphonamide is formed. N, N-
diethylbenzene sulphonamide does not contain any H atom attached to the 
nitrogen atom, so it is not acidic and is therefore insoluble in alkali. 
 
 
 
 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
SECTION C 
 
Ans.13.     
(a) E°
(Ag
+
/Ag)
 = 0.80 V, so E°
(Ag/Ag
+
) = 
-0.80 V 
 
           E°
(Zn
+2
/Zn)
 = -0.76 V 
 
           E°
cell
= E°
cathode 
- E°
anode 
                 =-0.80-(-0.76) 
                  =-0.4 V 
 
           ??
r
G°=-nFE°
cell 
             ==   = -2×9660×(-0.4) 
                     = 7728 J mol
-1 
(b) Kohlrausch law helps us in the determination of limiting molar 
conductivities for any electrolyte. 
 
 
Ans.14.  
(a) Coagulation 
The process of changing colloidal particles in a sol into an insoluble 
precipitate by the addition of some suitable electrolyte. 
Example: Particles in milk can get coagulated to form yogurt. 
 
(b) Multimolecular colloid: 
A colloid in which colloidal particles consist of aggregates of atoms or 
small molecules with a diameter of less than 1 nm. 
Example: Gold sol/sulphur sol 
 
(c) Gel 
A liquid dispersed in a solid. A gel is a sol in which solid particles are 
meshed resulting in a rigid or semi-rigid mixture.  
Example: Fruit jelly 
 
OR 
 
(a)  Ferric hydroxide is coagulated easily because starch is a lyophilic sol and 
ferric hydroxide is lyophobic. Lyophobic sols can be easily coagulated, 
because unlike lyophilic sols, the stability of lyophobic sols is only because of 
the charges present on lyophobic sols. If the charges on the lyophobic sol are 
removed, then the particles which are present in them will come closer to 
each other, form aggregates and result in precipitation. Just by addition of 
charge, hydrophobic sols lose their charges and coagulate. 
Page 5


  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
CBSE 
Class XII Chemistry ? 
Board Paper – 2019 Solution 
 
   
 
SECTION A 
 
Ans.1.  1-Phenyl butan-2-one 
 
Ans.2. Chloroform in the presence of light and air gets oxidised to phosgene, a 
poisonous gas. It is carbonyl chloride and represented as COCl
2
. To prevent 
the formation of phosgene, chloroform is stored in dark-coloured bottles. 
This reaction is represented as   CHCl
3
 + 1/2 O
2 
? COCl
2
 + HCl 
 
Ans.3.  Cis-dichlorobis(ethylenediamine)cobalt(III)ion 
 
OR 
           Na
2
[NiCl
4
] 
 
Ans.4. A glycosidic bond is a type of covalent bond formed between two 
monosaccharides of carbohydrate or sugar, whereas a peptide bond is a type 
of covalent bond formed between two amino acids of proteins. 
  
 OR 
 
     A nucleotide is composed of a nitrogenous base, sugar and phosphate group, 
whereas a nucleoside is composed of only a nitrogenous base and a 
phosphate group.    
 
Ans.5. The increasing order of acidic character: 
                  Ethanol < Water < Phenol 
 
SECTION B 
 
Ans.6.   
(a) In the same volume of water, 0.1 mole of KCl will have greater boiling point 
elevation, as boiling point elevation is a colligative property which depends 
on the relative amounts of constituents and not on their identity. 
      Since more ions will be produced by KCl (ionic compound) than glucose 
(covalent compound) as there are more component parts in the KCl 
molecule, the KCl solution will have a higher boiling point. 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
(b) (i) Acetic acid ionises as 
            CH
3
COOH ? CH
3
COO
?
 + H
+
 
       
           CH
3
COOH=1- ? 
                        H
+
= ? 
        It produces (1+ ?) species from one CH
3
COOH. 
        Since ? < 1, the van’t hoff factor will lie between 1 and 2. So, i > 1. 
       
     (ii) Acetic acid dimerises in benzene solution. 
             2CH
3
COOH  ?  (CH
3
COOH)
2 
               
2 moles of acetic acid associate to form one mole of dimer, so that 
              i = 1 - (1 - 1/2) ? = 1 - ?/2 
           For association van’t hoff factor: 
                    i < 1  
 
Ans.7.   
CdCl
2
 on addition to AgCl causes cation vacancy. Impurity defect is produced 
when AgCl is doped with CdCl
2
. 
 
OR 
 
(a) Schottky defect 
(b) Metal excess defect 
 
Ans.8.  
(a)   8NH
3
(excess) + 3Cl
2
              6NH
4
Cl + N
2
 
(b)    XeF
6
 + 2H
2
O                  XeO
2
F
2
 + 4HF 
 
Ans.9.  
(a) Sulphurous acid  
         
(b) Sulphuric acid 
                       
 
 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
Ans.10. Carbon monoxide and methane are used as fuels in a fuel cell. 
                 Advantages of fuel cells: 
(a) Fuel cells have a higher efficiency than diesel or gas engines. 
(b) The cells were used for electric power generation in the Apollo Space 
Programme, where water vapour produced in the cells was condensed and 
used for drinking purposes. 
 
Ans.11.  
(a)  
 
(b)  
 
 
Ans.12.  
 
(a) The method cannot be used for preparing aromatic primary amines because 
aryl halides do not undergo nucleophilic substitution with the anion formed by 
pthalimide. 
 
(b)  
(i) Primary amine reacts with benzenesulphonyl chloride to form N-
ethylbenzenesulphonyl amide. The hydrogen bonded to nitrogen is strongly 
acidic due to the presence of a strong electron-withdrawing sulphonyl group 
and is hence soluble in alkali. 
(ii) With secondary amine, N,N-diethyl-benzenesulphonamide is formed. N, N-
diethylbenzene sulphonamide does not contain any H atom attached to the 
nitrogen atom, so it is not acidic and is therefore insoluble in alkali. 
 
 
 
 
 
 
 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
SECTION C 
 
Ans.13.     
(a) E°
(Ag
+
/Ag)
 = 0.80 V, so E°
(Ag/Ag
+
) = 
-0.80 V 
 
           E°
(Zn
+2
/Zn)
 = -0.76 V 
 
           E°
cell
= E°
cathode 
- E°
anode 
                 =-0.80-(-0.76) 
                  =-0.4 V 
 
           ??
r
G°=-nFE°
cell 
             ==   = -2×9660×(-0.4) 
                     = 7728 J mol
-1 
(b) Kohlrausch law helps us in the determination of limiting molar 
conductivities for any electrolyte. 
 
 
Ans.14.  
(a) Coagulation 
The process of changing colloidal particles in a sol into an insoluble 
precipitate by the addition of some suitable electrolyte. 
Example: Particles in milk can get coagulated to form yogurt. 
 
(b) Multimolecular colloid: 
A colloid in which colloidal particles consist of aggregates of atoms or 
small molecules with a diameter of less than 1 nm. 
Example: Gold sol/sulphur sol 
 
(c) Gel 
A liquid dispersed in a solid. A gel is a sol in which solid particles are 
meshed resulting in a rigid or semi-rigid mixture.  
Example: Fruit jelly 
 
OR 
 
(a)  Ferric hydroxide is coagulated easily because starch is a lyophilic sol and 
ferric hydroxide is lyophobic. Lyophobic sols can be easily coagulated, 
because unlike lyophilic sols, the stability of lyophobic sols is only because of 
the charges present on lyophobic sols. If the charges on the lyophobic sol are 
removed, then the particles which are present in them will come closer to 
each other, form aggregates and result in precipitation. Just by addition of 
charge, hydrophobic sols lose their charges and coagulate. 
  
 
CBSE XII  | CHEMISTRY 
Board Paper – 2019 Solution (SET-3) 
 
     
(b) When an emulsion is centrifuged, its constituents separate out. This process 
is called demulsification. 
(c) Promotors enhance the activity of catalysts, whereas catalytic poisons or 
inhibitors decrease the activity of catalysts. 
 
Ans.15.  
(a) a = 3 × 10
-8 
cm   
        d = 6.89 g cm
-3
 
         Z = 2 (In the BCC structure, two atoms are present.) 
         We know that d = ZM/a
3
 × N
A 
                                                
M = d × a
3
 × N
A
/z 
                               M = 6.89 × (3 × 10
-8
)
3
 × 6.022 × 10
23
/2 
                                  = 56 g 
(b)  
(i) When Germanium is doped with Indium, a P-type semiconductor is 
formed. 
(ii) The N-type semiconductor is obtained when silicon is doped with 
phosphorus. 
 
Ans.16. ? ? = iCRT 
         T = 27 + 273 = 300 K 
          i = 95/100 = 0.95 
          i = 1 - ? + n ? 
 
      For Na
2
SO
4, 
i = 1 - 0.95 + 3 × 0.95 
                       i = 1 - 0.95 + 2.85 
                       i = 1 + 1.9 
                       i = 2.9 
              ? ? = 2.9 × 0.1 × 0.0821 × 300 
                  = 7.1427 Pascal 
 
Ans.17.  
Following reactions take place at the upper part of the furnace in the range of 
temperature of 500–900 K. Here, ores are reduced to Fe by CO. 
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
573 673K
2 3 3 4 2
773 873K
3 4 2
773 873K
2 3 2
3Fe O s CO g 2Fe O s CO g
Fe O s 4CO g 3Fe s 4CO g
Fe O s CO g 2FeO s CO g
?
?
?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
 
Role of limestone: 
Limestone is decomposed to CaO and CO
2
. This CaO combines with 
silica to form fusible calcium silicate slag. 
 
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
1123K
32
1123K
23
Calcium Silicate (Slag)
CaCO s CaO s CO g ; H 179.9kJ
CaO s SiO s CaSiO s
 
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FAQs on CBSE Chemistry Past year paper (Solutions) - 2019, Class 12 - Chemistry for Grade 12

1. How can I access the CBSE Chemistry Past year paper (Solutions) for the 2019 NEET exam?
Ans. You can access the CBSE Chemistry Past year paper (Solutions) for the 2019 NEET exam by searching for them online on educational websites, official CBSE platforms, or by referring to NEET preparation books that provide solved papers. These resources will provide you with the necessary solutions to help you understand and analyze the questions asked in the exam.
2. How can solving past year papers help me prepare for the NEET exam?
Ans. Solving past year papers can be extremely beneficial in preparing for the NEET exam. By practicing these papers, you can become familiar with the exam pattern, understand the type of questions asked, and assess your current level of preparation. It allows you to identify your strengths and weaknesses, allowing you to focus on areas that require improvement. Additionally, solving past year papers boosts your confidence and helps you manage time effectively during the actual exam.
3. Are the solutions provided in the CBSE Chemistry Past year paper (Solutions) reliable?
Ans. Yes, the solutions provided in the CBSE Chemistry Past year paper (Solutions) are generally reliable. However, it is essential to cross-verify the solutions with other credible sources or consult subject experts to ensure accuracy. Mistakes can occur in any answer key, so it is always recommended to double-check and clarify any doubts you may have regarding the solutions.
4. Can solving the CBSE Chemistry Past year paper (Solutions) alone guarantee success in the NEET exam?
Ans. While solving the CBSE Chemistry Past year paper (Solutions) is an essential part of NEET exam preparation, it cannot guarantee success on its own. Along with solving past year papers, you must also focus on understanding the concepts, practicing different types of questions, and regularly revising the syllabus. It is crucial to have a comprehensive study plan, seek guidance from teachers or mentors, and consistently work hard to increase your chances of success in the NEET exam.
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Ans. To maximize the benefits of solving the CBSE Chemistry Past year paper (Solutions) for the NEET exam, you should solve them in a timed manner to simulate the actual exam conditions. After solving each paper, thoroughly analyze your performance, identify the areas where you made mistakes or faced difficulties, and work on improving them. Seek help from teachers or subject experts to clarify any doubts you may have. Regularly practice solving papers to enhance your speed, accuracy, and problem-solving skills.
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