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JEE Main Previous Year Questions (2026): Some Basic Concepts of Chemistry
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Page 1
JEE Main Previous Year Qs (2025):
Some Basic Concepts of Chemistry
Q1: 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final
concentration of the solution is
____ × 10
-2
M. (Nearest integer)
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 57
Solution:
To determine the final concentration of the NaOH solution, we use the formula for mixing
solutions :
?? ?? =
?? 1
× ?? 1
+ ?? 2
× ?? 2
?? 1
+ ?? 2
Where :
?? 1
= 2M and ?? 1
= 20 mL : Concentration and volume of the first solution.
?? 2
= 0.5M and ?? 2
= 400 mL : Concentration and volume of the second solution.
Substitute the values into the equation:
?? ?? =
2 × 20 + 0.5 × 400
420
Calculating each term:
2 × 20 = 40
0.5 × 400 = 200
Add these results:
?? ?? =
40 + 200
420
=
240
420
˜ 0.571M
Convert this to scientific notation as the problem specifies the answer should be in × 10
-2
form:
?? ?? = 57.1 × 10
-2
M
Rounded to the nearest integer, the final concentration is:
Q2: During " S " estimation, 160 mg of an organic compound gives 466 mg of barium
sulphate. The percentage of Sulphur in the given compound is ____ %.
(Given molar mass in ?? ?????? -?? of ???? : ?????? , ?? : ???? , ?? : ???? )
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 40
Solution:
Page 2
JEE Main Previous Year Qs (2025):
Some Basic Concepts of Chemistry
Q1: 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final
concentration of the solution is
____ × 10
-2
M. (Nearest integer)
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 57
Solution:
To determine the final concentration of the NaOH solution, we use the formula for mixing
solutions :
?? ?? =
?? 1
× ?? 1
+ ?? 2
× ?? 2
?? 1
+ ?? 2
Where :
?? 1
= 2M and ?? 1
= 20 mL : Concentration and volume of the first solution.
?? 2
= 0.5M and ?? 2
= 400 mL : Concentration and volume of the second solution.
Substitute the values into the equation:
?? ?? =
2 × 20 + 0.5 × 400
420
Calculating each term:
2 × 20 = 40
0.5 × 400 = 200
Add these results:
?? ?? =
40 + 200
420
=
240
420
˜ 0.571M
Convert this to scientific notation as the problem specifies the answer should be in × 10
-2
form:
?? ?? = 57.1 × 10
-2
M
Rounded to the nearest integer, the final concentration is:
Q2: During " S " estimation, 160 mg of an organic compound gives 466 mg of barium
sulphate. The percentage of Sulphur in the given compound is ____ %.
(Given molar mass in ?? ?????? -?? of ???? : ?????? , ?? : ???? , ?? : ???? )
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 40
Solution:
Millimoles of BaSO
4
=
466
233
= 2mmol
% S =
466
233
× 32
160
× 100 = 40%
Q3 : When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the
mass of aluminium oxide produced in grams is ____ .
(Nearest integer)
Given :
Molar mass of Al is 27.0 g mol
-1
Molar mass of ?? is 16.0 g mol
-1
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 153
Solution:
To determine the mass of aluminum oxide produced, we first consider the chemical reaction
between aluminum and oxygen:
4Al + 3O
2
? 2Al
2
O
3
Given:
Molar mass of Al = 27.0 g/mol
Molar mass of O = 16.0 g/mol
81.0 g of Al and 128.0 g of O
2
Calculating moles of reactants:
For aluminum (Al), using its molar mass:
81 g
27 g/mol
= 3 moles of Al
For oxygen ( O
2
), using its molar mass ( O
2
is composed of two oxygen atoms):
128 g
32 g/mol
= 4 moles of O
2
Identifying the limiting reagent:
From the stoichiometry of the reaction:
4 moles of Al react with 3 moles of O
2
3 moles of Al require
3
4
× 3 moles of O
2
= 2.25 moles of O
2
Since only 2.25 moles of O
2
are needed by 3 moles of Al , and we have 4 moles of O
2
, aluminum
is the limiting reagent.
Calculating moles of Al
2
O
3
formed:
Page 3
JEE Main Previous Year Qs (2025):
Some Basic Concepts of Chemistry
Q1: 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final
concentration of the solution is
____ × 10
-2
M. (Nearest integer)
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 57
Solution:
To determine the final concentration of the NaOH solution, we use the formula for mixing
solutions :
?? ?? =
?? 1
× ?? 1
+ ?? 2
× ?? 2
?? 1
+ ?? 2
Where :
?? 1
= 2M and ?? 1
= 20 mL : Concentration and volume of the first solution.
?? 2
= 0.5M and ?? 2
= 400 mL : Concentration and volume of the second solution.
Substitute the values into the equation:
?? ?? =
2 × 20 + 0.5 × 400
420
Calculating each term:
2 × 20 = 40
0.5 × 400 = 200
Add these results:
?? ?? =
40 + 200
420
=
240
420
˜ 0.571M
Convert this to scientific notation as the problem specifies the answer should be in × 10
-2
form:
?? ?? = 57.1 × 10
-2
M
Rounded to the nearest integer, the final concentration is:
Q2: During " S " estimation, 160 mg of an organic compound gives 466 mg of barium
sulphate. The percentage of Sulphur in the given compound is ____ %.
(Given molar mass in ?? ?????? -?? of ???? : ?????? , ?? : ???? , ?? : ???? )
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 40
Solution:
Millimoles of BaSO
4
=
466
233
= 2mmol
% S =
466
233
× 32
160
× 100 = 40%
Q3 : When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the
mass of aluminium oxide produced in grams is ____ .
(Nearest integer)
Given :
Molar mass of Al is 27.0 g mol
-1
Molar mass of ?? is 16.0 g mol
-1
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 153
Solution:
To determine the mass of aluminum oxide produced, we first consider the chemical reaction
between aluminum and oxygen:
4Al + 3O
2
? 2Al
2
O
3
Given:
Molar mass of Al = 27.0 g/mol
Molar mass of O = 16.0 g/mol
81.0 g of Al and 128.0 g of O
2
Calculating moles of reactants:
For aluminum (Al), using its molar mass:
81 g
27 g/mol
= 3 moles of Al
For oxygen ( O
2
), using its molar mass ( O
2
is composed of two oxygen atoms):
128 g
32 g/mol
= 4 moles of O
2
Identifying the limiting reagent:
From the stoichiometry of the reaction:
4 moles of Al react with 3 moles of O
2
3 moles of Al require
3
4
× 3 moles of O
2
= 2.25 moles of O
2
Since only 2.25 moles of O
2
are needed by 3 moles of Al , and we have 4 moles of O
2
, aluminum
is the limiting reagent.
Calculating moles of Al
2
O
3
formed:
From the reaction, 4 moles of Al yield 2 moles of Al
2
O
3
. Thus:
3
4
× 2 =
3
2
moles of Al
2
O
3
Calculating mass of Al
2
O
3
:
The molar mass of Al
2
O
3
is:
(2 × 27) + (3 × 16) = 102 g/mol
So, the mass of
3
2
moles of Al
2
O
3
is:
3
2
× 102 = 153 grams
Thus, the mass of aluminum oxide produced is 153 grams.
Q4: 0.01 mole of an organic compound (?? ) containing ???? % hydrogen, on complete
combustion produced ?? . ?? ???? ?? ?? . Molar mass of (?? ) is ____ ?? ?????? -?? .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 100
Solution:
Organic compound ?
combustion
H
2
O
0.9gm
? mole of H
2
O =
0.9
18
= 0.05 mole
? mole of H in H
2
O = 0.05 × 2 = 0.1 mole
= mole of H in 0.01 mole
Organic compound
? wt of H atom in 0.01 mole compound = 0.1 × 1
= 0.1gm
? wt of H atom in one mole compound
=
0.1
0.01
= 10gm
? wt. % of ?? =
wt. of ?? in one mole compound
Molar mass of compound
× 1
10 =
10
?? × 100
? ?? = 100
Q5: Xg of benzoic acid on reaction with aq ?????????? ?? released ????
?? that occupied
???? . ???? volume at STP. X is ____ g.
JEE Main 2025 (Online) 24th January Morning Shift
Page 4
JEE Main Previous Year Qs (2025):
Some Basic Concepts of Chemistry
Q1: 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final
concentration of the solution is
____ × 10
-2
M. (Nearest integer)
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 57
Solution:
To determine the final concentration of the NaOH solution, we use the formula for mixing
solutions :
?? ?? =
?? 1
× ?? 1
+ ?? 2
× ?? 2
?? 1
+ ?? 2
Where :
?? 1
= 2M and ?? 1
= 20 mL : Concentration and volume of the first solution.
?? 2
= 0.5M and ?? 2
= 400 mL : Concentration and volume of the second solution.
Substitute the values into the equation:
?? ?? =
2 × 20 + 0.5 × 400
420
Calculating each term:
2 × 20 = 40
0.5 × 400 = 200
Add these results:
?? ?? =
40 + 200
420
=
240
420
˜ 0.571M
Convert this to scientific notation as the problem specifies the answer should be in × 10
-2
form:
?? ?? = 57.1 × 10
-2
M
Rounded to the nearest integer, the final concentration is:
Q2: During " S " estimation, 160 mg of an organic compound gives 466 mg of barium
sulphate. The percentage of Sulphur in the given compound is ____ %.
(Given molar mass in ?? ?????? -?? of ???? : ?????? , ?? : ???? , ?? : ???? )
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 40
Solution:
Millimoles of BaSO
4
=
466
233
= 2mmol
% S =
466
233
× 32
160
× 100 = 40%
Q3 : When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the
mass of aluminium oxide produced in grams is ____ .
(Nearest integer)
Given :
Molar mass of Al is 27.0 g mol
-1
Molar mass of ?? is 16.0 g mol
-1
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 153
Solution:
To determine the mass of aluminum oxide produced, we first consider the chemical reaction
between aluminum and oxygen:
4Al + 3O
2
? 2Al
2
O
3
Given:
Molar mass of Al = 27.0 g/mol
Molar mass of O = 16.0 g/mol
81.0 g of Al and 128.0 g of O
2
Calculating moles of reactants:
For aluminum (Al), using its molar mass:
81 g
27 g/mol
= 3 moles of Al
For oxygen ( O
2
), using its molar mass ( O
2
is composed of two oxygen atoms):
128 g
32 g/mol
= 4 moles of O
2
Identifying the limiting reagent:
From the stoichiometry of the reaction:
4 moles of Al react with 3 moles of O
2
3 moles of Al require
3
4
× 3 moles of O
2
= 2.25 moles of O
2
Since only 2.25 moles of O
2
are needed by 3 moles of Al , and we have 4 moles of O
2
, aluminum
is the limiting reagent.
Calculating moles of Al
2
O
3
formed:
From the reaction, 4 moles of Al yield 2 moles of Al
2
O
3
. Thus:
3
4
× 2 =
3
2
moles of Al
2
O
3
Calculating mass of Al
2
O
3
:
The molar mass of Al
2
O
3
is:
(2 × 27) + (3 × 16) = 102 g/mol
So, the mass of
3
2
moles of Al
2
O
3
is:
3
2
× 102 = 153 grams
Thus, the mass of aluminum oxide produced is 153 grams.
Q4: 0.01 mole of an organic compound (?? ) containing ???? % hydrogen, on complete
combustion produced ?? . ?? ???? ?? ?? . Molar mass of (?? ) is ____ ?? ?????? -?? .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 100
Solution:
Organic compound ?
combustion
H
2
O
0.9gm
? mole of H
2
O =
0.9
18
= 0.05 mole
? mole of H in H
2
O = 0.05 × 2 = 0.1 mole
= mole of H in 0.01 mole
Organic compound
? wt of H atom in 0.01 mole compound = 0.1 × 1
= 0.1gm
? wt of H atom in one mole compound
=
0.1
0.01
= 10gm
? wt. % of ?? =
wt. of ?? in one mole compound
Molar mass of compound
× 1
10 =
10
?? × 100
? ?? = 100
Q5: Xg of benzoic acid on reaction with aq ?????????? ?? released ????
?? that occupied
???? . ???? volume at STP. X is ____ g.
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 61
Solution:
C
6
H
5
COOH+ NaHCO
3
? C
6
H
5
COONa
+
+H
2
O + CO
2
× ???? 11.2 L
mole of C
6
H
5
COOH= mole of CO
2
=
11.2
22.4
= 0.5
mass of C
6
H
5
COOH= x = 0.5 × 122 = 61gm
Q6: Consider the following reaction occurring in the blast furnace:
????
?? ?? ?? ( ?? )
+ ?? ????
(?? )
? ?? ????
(?? )
+ ?? ????
?? ( ?? )
'
??
'
???? of iron is produced when ?? . ???? ×
????
?? ???????? ?? ?? ?? and ?? . ?? × ????
?? ???????? are brought together in the furnace. The value of '
?? ' is . (nearest integer) {Given:molar mass of ????
?? ?? ?? = ?????? ?? ?????? -?? , molar mass of
???? = ???? ?? ?????? -?? , molar mass of ???? = ???? ?? ?????? -?? }
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 420
Solution:
moles of Fe
3
O
4
=
2.32×10
3
×10
3
232
= 10000 mol
moles of CO =
2.8×10
2
×10
3
28
= 10000 mol
Fe
3
O
4
+ 4CO? 3Fe + 4CO
2
10
4
mol 10
4
mol
CO is L.R.
mole of Fe =
3
4
× 10
4
mass of Fe =
3
4
×
10
4
×56
1000
kg = 420 kg
Q7: Quantitative analysis of an organic compound (X) shows following % composition.
C : ???? . ?? %
Cl : 64.46%
?? : ?? . ?? %
(Empirical formula mass of the compound (?? ) is ____ × ????
-??
(Given molar mass in ?? ?????? -?? of ?? : ???? , ?? : ?? , ?? : ???? , ???? : ???? . ?? )
JEE Main 2025 (Online) 28th January Morning Shift
Page 5
JEE Main Previous Year Qs (2025):
Some Basic Concepts of Chemistry
Q1: 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final
concentration of the solution is
____ × 10
-2
M. (Nearest integer)
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 57
Solution:
To determine the final concentration of the NaOH solution, we use the formula for mixing
solutions :
?? ?? =
?? 1
× ?? 1
+ ?? 2
× ?? 2
?? 1
+ ?? 2
Where :
?? 1
= 2M and ?? 1
= 20 mL : Concentration and volume of the first solution.
?? 2
= 0.5M and ?? 2
= 400 mL : Concentration and volume of the second solution.
Substitute the values into the equation:
?? ?? =
2 × 20 + 0.5 × 400
420
Calculating each term:
2 × 20 = 40
0.5 × 400 = 200
Add these results:
?? ?? =
40 + 200
420
=
240
420
˜ 0.571M
Convert this to scientific notation as the problem specifies the answer should be in × 10
-2
form:
?? ?? = 57.1 × 10
-2
M
Rounded to the nearest integer, the final concentration is:
Q2: During " S " estimation, 160 mg of an organic compound gives 466 mg of barium
sulphate. The percentage of Sulphur in the given compound is ____ %.
(Given molar mass in ?? ?????? -?? of ???? : ?????? , ?? : ???? , ?? : ???? )
JEE Main 2025 (Online) 23rd January Morning Shift
Ans: 40
Solution:
Millimoles of BaSO
4
=
466
233
= 2mmol
% S =
466
233
× 32
160
× 100 = 40%
Q3 : When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the
mass of aluminium oxide produced in grams is ____ .
(Nearest integer)
Given :
Molar mass of Al is 27.0 g mol
-1
Molar mass of ?? is 16.0 g mol
-1
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 153
Solution:
To determine the mass of aluminum oxide produced, we first consider the chemical reaction
between aluminum and oxygen:
4Al + 3O
2
? 2Al
2
O
3
Given:
Molar mass of Al = 27.0 g/mol
Molar mass of O = 16.0 g/mol
81.0 g of Al and 128.0 g of O
2
Calculating moles of reactants:
For aluminum (Al), using its molar mass:
81 g
27 g/mol
= 3 moles of Al
For oxygen ( O
2
), using its molar mass ( O
2
is composed of two oxygen atoms):
128 g
32 g/mol
= 4 moles of O
2
Identifying the limiting reagent:
From the stoichiometry of the reaction:
4 moles of Al react with 3 moles of O
2
3 moles of Al require
3
4
× 3 moles of O
2
= 2.25 moles of O
2
Since only 2.25 moles of O
2
are needed by 3 moles of Al , and we have 4 moles of O
2
, aluminum
is the limiting reagent.
Calculating moles of Al
2
O
3
formed:
From the reaction, 4 moles of Al yield 2 moles of Al
2
O
3
. Thus:
3
4
× 2 =
3
2
moles of Al
2
O
3
Calculating mass of Al
2
O
3
:
The molar mass of Al
2
O
3
is:
(2 × 27) + (3 × 16) = 102 g/mol
So, the mass of
3
2
moles of Al
2
O
3
is:
3
2
× 102 = 153 grams
Thus, the mass of aluminum oxide produced is 153 grams.
Q4: 0.01 mole of an organic compound (?? ) containing ???? % hydrogen, on complete
combustion produced ?? . ?? ???? ?? ?? . Molar mass of (?? ) is ____ ?? ?????? -?? .
JEE Main 2025 (Online) 23rd January Evening Shift
Ans: 100
Solution:
Organic compound ?
combustion
H
2
O
0.9gm
? mole of H
2
O =
0.9
18
= 0.05 mole
? mole of H in H
2
O = 0.05 × 2 = 0.1 mole
= mole of H in 0.01 mole
Organic compound
? wt of H atom in 0.01 mole compound = 0.1 × 1
= 0.1gm
? wt of H atom in one mole compound
=
0.1
0.01
= 10gm
? wt. % of ?? =
wt. of ?? in one mole compound
Molar mass of compound
× 1
10 =
10
?? × 100
? ?? = 100
Q5: Xg of benzoic acid on reaction with aq ?????????? ?? released ????
?? that occupied
???? . ???? volume at STP. X is ____ g.
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 61
Solution:
C
6
H
5
COOH+ NaHCO
3
? C
6
H
5
COONa
+
+H
2
O + CO
2
× ???? 11.2 L
mole of C
6
H
5
COOH= mole of CO
2
=
11.2
22.4
= 0.5
mass of C
6
H
5
COOH= x = 0.5 × 122 = 61gm
Q6: Consider the following reaction occurring in the blast furnace:
????
?? ?? ?? ( ?? )
+ ?? ????
(?? )
? ?? ????
(?? )
+ ?? ????
?? ( ?? )
'
??
'
???? of iron is produced when ?? . ???? ×
????
?? ???????? ?? ?? ?? and ?? . ?? × ????
?? ???????? are brought together in the furnace. The value of '
?? ' is . (nearest integer) {Given:molar mass of ????
?? ?? ?? = ?????? ?? ?????? -?? , molar mass of
???? = ???? ?? ?????? -?? , molar mass of ???? = ???? ?? ?????? -?? }
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 420
Solution:
moles of Fe
3
O
4
=
2.32×10
3
×10
3
232
= 10000 mol
moles of CO =
2.8×10
2
×10
3
28
= 10000 mol
Fe
3
O
4
+ 4CO? 3Fe + 4CO
2
10
4
mol 10
4
mol
CO is L.R.
mole of Fe =
3
4
× 10
4
mass of Fe =
3
4
×
10
4
×56
1000
kg = 420 kg
Q7: Quantitative analysis of an organic compound (X) shows following % composition.
C : ???? . ?? %
Cl : 64.46%
?? : ?? . ?? %
(Empirical formula mass of the compound (?? ) is ____ × ????
-??
(Given molar mass in ?? ?????? -?? of ?? : ???? , ?? : ?? , ?? : ???? , ???? : ???? . ?? )
JEE Main 2025 (Online) 28th January Morning Shift
Ans: 1655
Solution:
Identify the given weight percentages from a 100 g sample:
Carbon (C): 14.5 g
Chlorine (Cl): 64.46 g
Hydrogen (H): 1.8 g
Since the sample must sum to 100 g , the remaining mass is from oxygen (O) :
O: 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24 g.
Next, convert these masses to moles using the given atomic masses:
Moles of C:
14.5 g
12 g/ mol
˜ 1.2083 mol .
Moles of Cl :
64.46 g
35.5 g/mol
˜ 1.8171 mol .
Moles of H :
1.8 g
1 g/mol
= 1.8 mol .
Moles of O :
19.24 g
16 g/mol
˜ 1.2025 mol .
Now, find the simplest whole-number ratio by dividing each by the smallest number of moles
(approximately 1.2025 mol ):
Ratio for C:
1.2083 /1.2025 ˜ 1.00
Ratio for O:
1.2025 /1.2025 = 1.00
Ratio for H:
1.8/1.2025 ˜ 1.50
Ratio for Cl:
1.8171 /1.2025 ˜ 1.51 ˜ 1.50
Since the ratios for H and Cl are about 1.5, multiplying all ratios by 2 will give whole numbers:
C: 1 × 2 = 2
O: 1 × 2 = 2
H: 1.5 × 2 = 3
Cl : 1.5 × 2 = 3
Thus, the empirical formula of the compound is:
C
2
H
3
O
2
Cl
3
.
Calculate the empirical formula mass by summing the contributions:
C: 2 × 12 = 24 g/mo l
H: 3 × 1 = 3 g/mol
O: 2 × 16 = 32 g/mol
Cl : 3 × 35.5 = 106.5 g/mol
Total mass:
24 + 3 + 32 + 106.5 = 165.5 g/mol .
The problem asks for the empirical formula mass in the form " × 10
-1
". We can express
Read MoreFAQs on JEE Main Previous Year Questions (2026): Some Basic Concepts of Chemistry
| 1. What are the basic concepts of chemistry that students should understand for JEE preparation? |  |
Ans. Basic concepts of chemistry include fundamental principles such as atomic structure, mole concept, stoichiometry, chemical bonding, states of matter, and thermodynamics. Understanding these concepts is crucial as they form the foundation for more advanced topics in chemistry. Students should focus on grasping the definitions, formulas, and applications of these concepts to solve problems effectively.
| 2. How does the mole concept play a role in solving stoichiometry problems? | |
Ans. The mole concept is essential in stoichiometry as it allows chemists to relate the mass of substances to the number of particles involved in a chemical reaction. By using Avogadro's number (6.022 × 10²³), students can convert between moles and particles, facilitating calculations of reactants and products. Mastery of this concept enables the determination of quantities needed for reactions and yields produced.
| 3. What is the importance of balancing chemical equations in basic chemistry? | |
Ans. Balancing chemical equations is crucial because it reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. A balanced equation ensures that the number of atoms for each element is the same on both sides of the equation, allowing accurate predictions of the amounts of reactants required and products formed. This skill is vital for stoichiometric calculations and understanding reaction dynamics.
| 4. Can you explain the differences between ionic and covalent bonding? | |
Ans. Ionic bonding occurs when electrons are transferred from one atom to another, resulting in the formation of charged ions (cations and anions). This type of bond typically forms between metals and non-metals. In contrast, covalent bonding involves the sharing of electron pairs between atoms, often occurring between non-metal atoms. Understanding these differences helps students predict the properties of compounds and their behavior in reactions.
| 5. How do concepts of thermodynamics apply to chemical reactions in JEE? | |
Ans. Thermodynamics is fundamental in understanding energy changes during chemical reactions. Key concepts include enthalpy (ΔH), entropy (ΔS), and Gibbs free energy (ΔG). Students must learn how to calculate these values to determine the spontaneity of reactions and the energy required for reactions to occur. Mastery of thermodynamics helps in predicting reaction feasibility and is often tested in JEE exams.
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