Some Basic Concepts of Chemistry: JEE Main Previous Year Questions (2021-2024) | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Some Basic Concepts of Chemistry 
2024 
Q1: ???? ???? of gaseous hydrocarbon on combustion gives ???? ???? of ????
?? ( ?? ) and ???? ???? of water 
vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is     [ JEE Main 2024 (Online) 
1st February Evening Shift] 
Ans: 14 
To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same 
conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, 
since the gas volumes given are at the same conditions, we can directly relate them to their 
stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon. 
The general equation for complete combustion of a hydrocarbon with a formula C
?? H
?? can be 
represented as: 
C
?? H
?? + (?? +
?? 4
) O
2
? ?? CO
2
+
?? 2
H
2
O 
Given: 
10 mL of hydrocarbon (C
?? H
?? ) 
40 mL of CO
2
 
50 mL of H
2
O 
Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the 
coefficients: 
?? 1
=
40mLCO
2
10mLC
?? H
?? = 4
?? 2
=
50mLH
2
O
10mLC
?? H
?? = 5
 
From the first ratio, we see that: 
?? = 4 
This tells us that there are four carbon atoms in the hydrocarbon molecule. 
From the second ratio, by multiplying both sides by 2 , we get: 
?? = 5 × 2 = 10 
This means there are ten hydrogen atoms in the hydrocarbon molecule. 
So, the total number of carbon and hydrogen atoms in the hydrocarbon is: 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Some Basic Concepts of Chemistry 
2024 
Q1: ???? ???? of gaseous hydrocarbon on combustion gives ???? ???? of ????
?? ( ?? ) and ???? ???? of water 
vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is     [ JEE Main 2024 (Online) 
1st February Evening Shift] 
Ans: 14 
To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same 
conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, 
since the gas volumes given are at the same conditions, we can directly relate them to their 
stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon. 
The general equation for complete combustion of a hydrocarbon with a formula C
?? H
?? can be 
represented as: 
C
?? H
?? + (?? +
?? 4
) O
2
? ?? CO
2
+
?? 2
H
2
O 
Given: 
10 mL of hydrocarbon (C
?? H
?? ) 
40 mL of CO
2
 
50 mL of H
2
O 
Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the 
coefficients: 
?? 1
=
40mLCO
2
10mLC
?? H
?? = 4
?? 2
=
50mLH
2
O
10mLC
?? H
?? = 5
 
From the first ratio, we see that: 
?? = 4 
This tells us that there are four carbon atoms in the hydrocarbon molecule. 
From the second ratio, by multiplying both sides by 2 , we get: 
?? = 5 × 2 = 10 
This means there are ten hydrogen atoms in the hydrocarbon molecule. 
So, the total number of carbon and hydrogen atoms in the hydrocarbon is: 
Total atoms = ?? + ?? = 4 + 10 = 14 
Therefore, the total number of carbon and hydrogen atoms in the hydrocarbon is 14 . 
Q2: Consider the following reaction : 
?? ???????? ?? + ?? (????
?? )
?? ????
?? ? ????
?? (????
?? )
?? + ?? ????
?? ???? 
If ???????????? of ???????? ?? is mixed with ???????????? of (????
?? )
?? ????
?? , then the amount of ????
?? (????
?? )
?? formed 
is mmol (nearest integer).  [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 24 
To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced 
equation shows that 3 moles of PbCl
2
 react with 2 moles of (NH
4
)
3
PO
4
 to produce 1 mole of 
Pb
3
(PO
4
)
2
. 
The reaction is: 
3PbCl
2
+ 2(NH
4
)
3
PO
4
? Pb
3
(PO
4
)
2
+ 6NH
4
Cl 
The molar ratio of PbCl
2
 to Pb
3
(PO
4
)
2
 is 3: 1, and the molar ratio of (NH
4
)
3
PO
4
 to Pb
3
(PO
4
)
2
 is 2: 1. 
We need to determine which reactant is the limiting reagent because it will dictate the amount of 
Pb
3
(PO
4
)
2
 produced. 
The stoichiometric calculations are as follows: 
For PbCl
2
 : 
Moles of Pb
3
(PO
4
)
2
 formed from PbCl
2
=
72mmol of  PbCl
2
3mmol of PbCl
2
/ mmol of Pb
3
(PO
4
)
2
= 24mmol For (NH
4
)
3
PO
4
 : 
Moles of Pb
3
(PO
4
)
2
 formed from (NH
4
)
3
PO
4
=
50 mmol of  (NH
4
)
3
PO
4
2mmol of  (NH
4
)
3
PO
4
/mmol of  Pb
3
(PO
4
)
2
= 25nmol 
Now we can identify the limiting reagent by comparing the two amounts of Pb
3
(PO
4
)
2
 that could be 
produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts 
the reaction. 
Since the PbCl
2
 can produce only 24mmol of Pb
3
(PO
4
)
2
 versus the 25mmol that could be produced by 
(NH
4
)
3
PO
4
, PbCl
2
 is the limiting reagent. 
Therefore, the amount of Pb
3
(PO
4
)
2
 formed is 24mmol (as a nearest integer). 
Q3: The molarity of ?? ?? orthophosphoric acid (?? ?? ????
?? ) having ???? % purity by weight (specific gravity 
?? . ???? ?? ????
-?? ) is M. 
(Molar mass of ?? ?? ????
?? = ???? ?? ?????? -?? )  [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 11 
Specific gravity (density) = 1.54 g/cc. 
Volume = 1 L = 1000ml 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Some Basic Concepts of Chemistry 
2024 
Q1: ???? ???? of gaseous hydrocarbon on combustion gives ???? ???? of ????
?? ( ?? ) and ???? ???? of water 
vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is     [ JEE Main 2024 (Online) 
1st February Evening Shift] 
Ans: 14 
To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same 
conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, 
since the gas volumes given are at the same conditions, we can directly relate them to their 
stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon. 
The general equation for complete combustion of a hydrocarbon with a formula C
?? H
?? can be 
represented as: 
C
?? H
?? + (?? +
?? 4
) O
2
? ?? CO
2
+
?? 2
H
2
O 
Given: 
10 mL of hydrocarbon (C
?? H
?? ) 
40 mL of CO
2
 
50 mL of H
2
O 
Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the 
coefficients: 
?? 1
=
40mLCO
2
10mLC
?? H
?? = 4
?? 2
=
50mLH
2
O
10mLC
?? H
?? = 5
 
From the first ratio, we see that: 
?? = 4 
This tells us that there are four carbon atoms in the hydrocarbon molecule. 
From the second ratio, by multiplying both sides by 2 , we get: 
?? = 5 × 2 = 10 
This means there are ten hydrogen atoms in the hydrocarbon molecule. 
So, the total number of carbon and hydrogen atoms in the hydrocarbon is: 
Total atoms = ?? + ?? = 4 + 10 = 14 
Therefore, the total number of carbon and hydrogen atoms in the hydrocarbon is 14 . 
Q2: Consider the following reaction : 
?? ???????? ?? + ?? (????
?? )
?? ????
?? ? ????
?? (????
?? )
?? + ?? ????
?? ???? 
If ???????????? of ???????? ?? is mixed with ???????????? of (????
?? )
?? ????
?? , then the amount of ????
?? (????
?? )
?? formed 
is mmol (nearest integer).  [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 24 
To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced 
equation shows that 3 moles of PbCl
2
 react with 2 moles of (NH
4
)
3
PO
4
 to produce 1 mole of 
Pb
3
(PO
4
)
2
. 
The reaction is: 
3PbCl
2
+ 2(NH
4
)
3
PO
4
? Pb
3
(PO
4
)
2
+ 6NH
4
Cl 
The molar ratio of PbCl
2
 to Pb
3
(PO
4
)
2
 is 3: 1, and the molar ratio of (NH
4
)
3
PO
4
 to Pb
3
(PO
4
)
2
 is 2: 1. 
We need to determine which reactant is the limiting reagent because it will dictate the amount of 
Pb
3
(PO
4
)
2
 produced. 
The stoichiometric calculations are as follows: 
For PbCl
2
 : 
Moles of Pb
3
(PO
4
)
2
 formed from PbCl
2
=
72mmol of  PbCl
2
3mmol of PbCl
2
/ mmol of Pb
3
(PO
4
)
2
= 24mmol For (NH
4
)
3
PO
4
 : 
Moles of Pb
3
(PO
4
)
2
 formed from (NH
4
)
3
PO
4
=
50 mmol of  (NH
4
)
3
PO
4
2mmol of  (NH
4
)
3
PO
4
/mmol of  Pb
3
(PO
4
)
2
= 25nmol 
Now we can identify the limiting reagent by comparing the two amounts of Pb
3
(PO
4
)
2
 that could be 
produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts 
the reaction. 
Since the PbCl
2
 can produce only 24mmol of Pb
3
(PO
4
)
2
 versus the 25mmol that could be produced by 
(NH
4
)
3
PO
4
, PbCl
2
 is the limiting reagent. 
Therefore, the amount of Pb
3
(PO
4
)
2
 formed is 24mmol (as a nearest integer). 
Q3: The molarity of ?? ?? orthophosphoric acid (?? ?? ????
?? ) having ???? % purity by weight (specific gravity 
?? . ???? ?? ????
-?? ) is M. 
(Molar mass of ?? ?? ????
?? = ???? ?? ?????? -?? )  [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 11 
Specific gravity (density) = 1.54 g/cc. 
Volume = 1 L = 1000ml 
Mass of solution = 1.54 × 1000 
= 1540 g 
% purity of H
2
SO
4
 is 70% 
So weight of H
3
PO
4
= 0.7 × 1540 = 1078 g 
Mole of H
3
PO
4
=
1078
98
= 11 
Molarity =
11
1 L
= 11 
Q4: Number of moles of methane required to produce ???? ?? ????
?? ( ?? )
 after combustion is ?? × ????
-?? 
moles. The value of ?? is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 50 
CH
4( g)
+ 2O
2( g)
? CO
2( g)
+ 2H
2
O
(l)
n
CO
2
=
22
44
= 0.5 moles 
 
So moles of CH
4
 required = 0.5 moles i.e. 50 × 10
-2
 mole 
x = 50 
Q5: Molar mass of the salt from ???????? , ???????? ?? , ???? and ?????? ?? which does not evolve coloured vapours 
on heating with concentrated ?? ?? ????
?? is ?? ?????? -?? . 
(Molar mass in ?? ?????? -?? : ???? : ???? , ?? : ???? , ?? : ???? , ?? : ???? , ???? : ???? , ?? : ?????? , ?? : ???? , ???? : ???? )   [JEE Main 2024 
(Online) 31st January Morning Shift] 
Ans: 78 
?????? 2
 does not evolve any gas with concentrated H
2
SO
4
. 
NaBr ? evolve Br
2
 
NaNO
3
? evolve NO
2
 
KI ? evolve I
2
 
Q6: The mass of sodium acetate (????
?? ?????????? ) required to prepare ?????? ???? of ?? . ?????? aqueous 
solution is g. (Molar mass of ????
?? ?????????? is ???? . ???? ?? ?????? -?? )   [JEE Main 2024 (Online) 30th January 
Morning Shift] 
Ans: 7 
 Moles = Molarity × Volume in litres 
 = 0.35 × 0.25
 Mass = moles × molar mass 
 = 0.35 × 0.25 × 82.02 = 7.18 g
 
Ans. 7 
Q7: ?? . ???? ???? thick coating of silver is deposited on a plate of ?? . ???? ?? ?? area. The number of silver 
atoms deposited on plate are × ????
????
. (At mass ???? = ?????? , ?? = ?? . ?? ?? ????
-?? )  [JEE Main 2024 (Online) 
30th January Morning Shift] 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Some Basic Concepts of Chemistry 
2024 
Q1: ???? ???? of gaseous hydrocarbon on combustion gives ???? ???? of ????
?? ( ?? ) and ???? ???? of water 
vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is     [ JEE Main 2024 (Online) 
1st February Evening Shift] 
Ans: 14 
To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same 
conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, 
since the gas volumes given are at the same conditions, we can directly relate them to their 
stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon. 
The general equation for complete combustion of a hydrocarbon with a formula C
?? H
?? can be 
represented as: 
C
?? H
?? + (?? +
?? 4
) O
2
? ?? CO
2
+
?? 2
H
2
O 
Given: 
10 mL of hydrocarbon (C
?? H
?? ) 
40 mL of CO
2
 
50 mL of H
2
O 
Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the 
coefficients: 
?? 1
=
40mLCO
2
10mLC
?? H
?? = 4
?? 2
=
50mLH
2
O
10mLC
?? H
?? = 5
 
From the first ratio, we see that: 
?? = 4 
This tells us that there are four carbon atoms in the hydrocarbon molecule. 
From the second ratio, by multiplying both sides by 2 , we get: 
?? = 5 × 2 = 10 
This means there are ten hydrogen atoms in the hydrocarbon molecule. 
So, the total number of carbon and hydrogen atoms in the hydrocarbon is: 
Total atoms = ?? + ?? = 4 + 10 = 14 
Therefore, the total number of carbon and hydrogen atoms in the hydrocarbon is 14 . 
Q2: Consider the following reaction : 
?? ???????? ?? + ?? (????
?? )
?? ????
?? ? ????
?? (????
?? )
?? + ?? ????
?? ???? 
If ???????????? of ???????? ?? is mixed with ???????????? of (????
?? )
?? ????
?? , then the amount of ????
?? (????
?? )
?? formed 
is mmol (nearest integer).  [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 24 
To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced 
equation shows that 3 moles of PbCl
2
 react with 2 moles of (NH
4
)
3
PO
4
 to produce 1 mole of 
Pb
3
(PO
4
)
2
. 
The reaction is: 
3PbCl
2
+ 2(NH
4
)
3
PO
4
? Pb
3
(PO
4
)
2
+ 6NH
4
Cl 
The molar ratio of PbCl
2
 to Pb
3
(PO
4
)
2
 is 3: 1, and the molar ratio of (NH
4
)
3
PO
4
 to Pb
3
(PO
4
)
2
 is 2: 1. 
We need to determine which reactant is the limiting reagent because it will dictate the amount of 
Pb
3
(PO
4
)
2
 produced. 
The stoichiometric calculations are as follows: 
For PbCl
2
 : 
Moles of Pb
3
(PO
4
)
2
 formed from PbCl
2
=
72mmol of  PbCl
2
3mmol of PbCl
2
/ mmol of Pb
3
(PO
4
)
2
= 24mmol For (NH
4
)
3
PO
4
 : 
Moles of Pb
3
(PO
4
)
2
 formed from (NH
4
)
3
PO
4
=
50 mmol of  (NH
4
)
3
PO
4
2mmol of  (NH
4
)
3
PO
4
/mmol of  Pb
3
(PO
4
)
2
= 25nmol 
Now we can identify the limiting reagent by comparing the two amounts of Pb
3
(PO
4
)
2
 that could be 
produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts 
the reaction. 
Since the PbCl
2
 can produce only 24mmol of Pb
3
(PO
4
)
2
 versus the 25mmol that could be produced by 
(NH
4
)
3
PO
4
, PbCl
2
 is the limiting reagent. 
Therefore, the amount of Pb
3
(PO
4
)
2
 formed is 24mmol (as a nearest integer). 
Q3: The molarity of ?? ?? orthophosphoric acid (?? ?? ????
?? ) having ???? % purity by weight (specific gravity 
?? . ???? ?? ????
-?? ) is M. 
(Molar mass of ?? ?? ????
?? = ???? ?? ?????? -?? )  [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 11 
Specific gravity (density) = 1.54 g/cc. 
Volume = 1 L = 1000ml 
Mass of solution = 1.54 × 1000 
= 1540 g 
% purity of H
2
SO
4
 is 70% 
So weight of H
3
PO
4
= 0.7 × 1540 = 1078 g 
Mole of H
3
PO
4
=
1078
98
= 11 
Molarity =
11
1 L
= 11 
Q4: Number of moles of methane required to produce ???? ?? ????
?? ( ?? )
 after combustion is ?? × ????
-?? 
moles. The value of ?? is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 50 
CH
4( g)
+ 2O
2( g)
? CO
2( g)
+ 2H
2
O
(l)
n
CO
2
=
22
44
= 0.5 moles 
 
So moles of CH
4
 required = 0.5 moles i.e. 50 × 10
-2
 mole 
x = 50 
Q5: Molar mass of the salt from ???????? , ???????? ?? , ???? and ?????? ?? which does not evolve coloured vapours 
on heating with concentrated ?? ?? ????
?? is ?? ?????? -?? . 
(Molar mass in ?? ?????? -?? : ???? : ???? , ?? : ???? , ?? : ???? , ?? : ???? , ???? : ???? , ?? : ?????? , ?? : ???? , ???? : ???? )   [JEE Main 2024 
(Online) 31st January Morning Shift] 
Ans: 78 
?????? 2
 does not evolve any gas with concentrated H
2
SO
4
. 
NaBr ? evolve Br
2
 
NaNO
3
? evolve NO
2
 
KI ? evolve I
2
 
Q6: The mass of sodium acetate (????
?? ?????????? ) required to prepare ?????? ???? of ?? . ?????? aqueous 
solution is g. (Molar mass of ????
?? ?????????? is ???? . ???? ?? ?????? -?? )   [JEE Main 2024 (Online) 30th January 
Morning Shift] 
Ans: 7 
 Moles = Molarity × Volume in litres 
 = 0.35 × 0.25
 Mass = moles × molar mass 
 = 0.35 × 0.25 × 82.02 = 7.18 g
 
Ans. 7 
Q7: ?? . ???? ???? thick coating of silver is deposited on a plate of ?? . ???? ?? ?? area. The number of silver 
atoms deposited on plate are × ????
????
. (At mass ???? = ?????? , ?? = ?? . ?? ?? ????
-?? )  [JEE Main 2024 (Online) 
30th January Morning Shift] 
Ans: 11 
Volume of silver coating = 0.05 × 0.05 × 10000 
= 25 cm
3
 
Mass of silver deposited = 25 × 7.9 g 
Moles of silver atoms =
25×7.9
108
 
Number of silver atoms =
25×7.9
108
× 6.023 × 10
23
 
= 11.01 × 10
23
 
Ans. 11 
Q8: Molality of ?? . ???? ?? ?? ????
?? solution (density ?? . ???? ?? ????
-?? ) is × ????
-?? ??    [JEE Main 2024 (Online) 
29th January Evening Shift] 
Ans: 815 
m =
M × 1000
 d
sol 
× 1000 - M × Molar mass   solute 
815 × 10
-3
 m
 
Q9: A solution of ?? ?? ?? ?? ?? is ???? . ?? %?? ?? ????
?? by mass and has a density of ?? . ???? ?? /???? . The molarity of 
the ?? ?? ????
?? solution is ?? (nearest integer) 
[Given molar mass of ?? ?? ????
?? = ???? ?? ?????? -?? ]    [JEE Main 2024 (Online) 29th January Morning Shift] 
Ans: 4 
Explanation : Currently no explanation available 
Q10: ?? . ?? ?? of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. 
The mass of acetanilide produced if the reaction is ?????? % completed is × ????
-?? ?? . 
(Given molar mass in ?? ?????? -?? )   [JEE Main 2024 (Online) 27th January Evening Shift] 
?? : ???? , ?? : ???? ,
?? : ???? , ?? : ?? )
 
Ans:  135 
Explanation: Currently no explanation available 
Q11: Volume of ???????????? (formula weight ???? ?? ?????? -?? ) which can be prepared from ???? ?? of NaOH is 
× ????
-?? ????
?? .  [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: 7 
Explanation: Currently no explanation available 
Q12: Mass of methane required to produce ???? ?? of ????
?? after complete combustion is g. (Given Molar 
mass in g mol-???? = ???? . ?? , ?? = ?? . ?? , ?? = ???? . ?? )     [JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 8 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Some Basic Concepts of Chemistry 
2024 
Q1: ???? ???? of gaseous hydrocarbon on combustion gives ???? ???? of ????
?? ( ?? ) and ???? ???? of water 
vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is     [ JEE Main 2024 (Online) 
1st February Evening Shift] 
Ans: 14 
To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same 
conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, 
since the gas volumes given are at the same conditions, we can directly relate them to their 
stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon. 
The general equation for complete combustion of a hydrocarbon with a formula C
?? H
?? can be 
represented as: 
C
?? H
?? + (?? +
?? 4
) O
2
? ?? CO
2
+
?? 2
H
2
O 
Given: 
10 mL of hydrocarbon (C
?? H
?? ) 
40 mL of CO
2
 
50 mL of H
2
O 
Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the 
coefficients: 
?? 1
=
40mLCO
2
10mLC
?? H
?? = 4
?? 2
=
50mLH
2
O
10mLC
?? H
?? = 5
 
From the first ratio, we see that: 
?? = 4 
This tells us that there are four carbon atoms in the hydrocarbon molecule. 
From the second ratio, by multiplying both sides by 2 , we get: 
?? = 5 × 2 = 10 
This means there are ten hydrogen atoms in the hydrocarbon molecule. 
So, the total number of carbon and hydrogen atoms in the hydrocarbon is: 
Total atoms = ?? + ?? = 4 + 10 = 14 
Therefore, the total number of carbon and hydrogen atoms in the hydrocarbon is 14 . 
Q2: Consider the following reaction : 
?? ???????? ?? + ?? (????
?? )
?? ????
?? ? ????
?? (????
?? )
?? + ?? ????
?? ???? 
If ???????????? of ???????? ?? is mixed with ???????????? of (????
?? )
?? ????
?? , then the amount of ????
?? (????
?? )
?? formed 
is mmol (nearest integer).  [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: 24 
To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced 
equation shows that 3 moles of PbCl
2
 react with 2 moles of (NH
4
)
3
PO
4
 to produce 1 mole of 
Pb
3
(PO
4
)
2
. 
The reaction is: 
3PbCl
2
+ 2(NH
4
)
3
PO
4
? Pb
3
(PO
4
)
2
+ 6NH
4
Cl 
The molar ratio of PbCl
2
 to Pb
3
(PO
4
)
2
 is 3: 1, and the molar ratio of (NH
4
)
3
PO
4
 to Pb
3
(PO
4
)
2
 is 2: 1. 
We need to determine which reactant is the limiting reagent because it will dictate the amount of 
Pb
3
(PO
4
)
2
 produced. 
The stoichiometric calculations are as follows: 
For PbCl
2
 : 
Moles of Pb
3
(PO
4
)
2
 formed from PbCl
2
=
72mmol of  PbCl
2
3mmol of PbCl
2
/ mmol of Pb
3
(PO
4
)
2
= 24mmol For (NH
4
)
3
PO
4
 : 
Moles of Pb
3
(PO
4
)
2
 formed from (NH
4
)
3
PO
4
=
50 mmol of  (NH
4
)
3
PO
4
2mmol of  (NH
4
)
3
PO
4
/mmol of  Pb
3
(PO
4
)
2
= 25nmol 
Now we can identify the limiting reagent by comparing the two amounts of Pb
3
(PO
4
)
2
 that could be 
produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts 
the reaction. 
Since the PbCl
2
 can produce only 24mmol of Pb
3
(PO
4
)
2
 versus the 25mmol that could be produced by 
(NH
4
)
3
PO
4
, PbCl
2
 is the limiting reagent. 
Therefore, the amount of Pb
3
(PO
4
)
2
 formed is 24mmol (as a nearest integer). 
Q3: The molarity of ?? ?? orthophosphoric acid (?? ?? ????
?? ) having ???? % purity by weight (specific gravity 
?? . ???? ?? ????
-?? ) is M. 
(Molar mass of ?? ?? ????
?? = ???? ?? ?????? -?? )  [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 11 
Specific gravity (density) = 1.54 g/cc. 
Volume = 1 L = 1000ml 
Mass of solution = 1.54 × 1000 
= 1540 g 
% purity of H
2
SO
4
 is 70% 
So weight of H
3
PO
4
= 0.7 × 1540 = 1078 g 
Mole of H
3
PO
4
=
1078
98
= 11 
Molarity =
11
1 L
= 11 
Q4: Number of moles of methane required to produce ???? ?? ????
?? ( ?? )
 after combustion is ?? × ????
-?? 
moles. The value of ?? is   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 50 
CH
4( g)
+ 2O
2( g)
? CO
2( g)
+ 2H
2
O
(l)
n
CO
2
=
22
44
= 0.5 moles 
 
So moles of CH
4
 required = 0.5 moles i.e. 50 × 10
-2
 mole 
x = 50 
Q5: Molar mass of the salt from ???????? , ???????? ?? , ???? and ?????? ?? which does not evolve coloured vapours 
on heating with concentrated ?? ?? ????
?? is ?? ?????? -?? . 
(Molar mass in ?? ?????? -?? : ???? : ???? , ?? : ???? , ?? : ???? , ?? : ???? , ???? : ???? , ?? : ?????? , ?? : ???? , ???? : ???? )   [JEE Main 2024 
(Online) 31st January Morning Shift] 
Ans: 78 
?????? 2
 does not evolve any gas with concentrated H
2
SO
4
. 
NaBr ? evolve Br
2
 
NaNO
3
? evolve NO
2
 
KI ? evolve I
2
 
Q6: The mass of sodium acetate (????
?? ?????????? ) required to prepare ?????? ???? of ?? . ?????? aqueous 
solution is g. (Molar mass of ????
?? ?????????? is ???? . ???? ?? ?????? -?? )   [JEE Main 2024 (Online) 30th January 
Morning Shift] 
Ans: 7 
 Moles = Molarity × Volume in litres 
 = 0.35 × 0.25
 Mass = moles × molar mass 
 = 0.35 × 0.25 × 82.02 = 7.18 g
 
Ans. 7 
Q7: ?? . ???? ???? thick coating of silver is deposited on a plate of ?? . ???? ?? ?? area. The number of silver 
atoms deposited on plate are × ????
????
. (At mass ???? = ?????? , ?? = ?? . ?? ?? ????
-?? )  [JEE Main 2024 (Online) 
30th January Morning Shift] 
Ans: 11 
Volume of silver coating = 0.05 × 0.05 × 10000 
= 25 cm
3
 
Mass of silver deposited = 25 × 7.9 g 
Moles of silver atoms =
25×7.9
108
 
Number of silver atoms =
25×7.9
108
× 6.023 × 10
23
 
= 11.01 × 10
23
 
Ans. 11 
Q8: Molality of ?? . ???? ?? ?? ????
?? solution (density ?? . ???? ?? ????
-?? ) is × ????
-?? ??    [JEE Main 2024 (Online) 
29th January Evening Shift] 
Ans: 815 
m =
M × 1000
 d
sol 
× 1000 - M × Molar mass   solute 
815 × 10
-3
 m
 
Q9: A solution of ?? ?? ?? ?? ?? is ???? . ?? %?? ?? ????
?? by mass and has a density of ?? . ???? ?? /???? . The molarity of 
the ?? ?? ????
?? solution is ?? (nearest integer) 
[Given molar mass of ?? ?? ????
?? = ???? ?? ?????? -?? ]    [JEE Main 2024 (Online) 29th January Morning Shift] 
Ans: 4 
Explanation : Currently no explanation available 
Q10: ?? . ?? ?? of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. 
The mass of acetanilide produced if the reaction is ?????? % completed is × ????
-?? ?? . 
(Given molar mass in ?? ?????? -?? )   [JEE Main 2024 (Online) 27th January Evening Shift] 
?? : ???? , ?? : ???? ,
?? : ???? , ?? : ?? )
 
Ans:  135 
Explanation: Currently no explanation available 
Q11: Volume of ???????????? (formula weight ???? ?? ?????? -?? ) which can be prepared from ???? ?? of NaOH is 
× ????
-?? ????
?? .  [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: 7 
Explanation: Currently no explanation available 
Q12: Mass of methane required to produce ???? ?? of ????
?? after complete combustion is g. (Given Molar 
mass in g mol-???? = ???? . ?? , ?? = ?? . ?? , ?? = ???? . ?? )     [JEE Main 2024 (Online) 27th January Morning Shift] 
Ans: 8 
To solve this problem, we can use stoichiometry. First, we need to write down the balanced chemical 
equation for the complete combustion of methane (CH
4
) . 
The balanced equation for combustion of methane is: 
CH
4
+ 2O
2
? CO
2
+ 2H
2
O 
This equation tells us that one mole of methane reacts with two moles of oxygen to produce one mole 
of carbon dioxide and two moles of water. 
Next, we should find the molar mass of methane (CH
4
) using the given molar masses: 
Molar mass of CH
4
= molar mass of C + 4 * molar mass of H 
Molar mass of CH
4
= 12.0 g/mol (for C ) +4 * 1.0 g/mol (for H ) 
Molar mass of CH
4
= 12.0 g/mol+ 4.0 g/mol 
Molar mass of CH
4
= 16.0 g/mol 
Now, to determine the mass of methane required to produce 22 g of CO
2
, we should find out how many 
moles of CO
2
 there are in 22 g and use the mole ratio from the balanced equation to find the moles of 
methane required. 
Moles of CO
2
= mass of CO
2
/ molar mass of CO
2
 
Molar mass of CO
2
= molar mass of C + 2 * molar mass of 0 
Molar mass of CO
2
= 12.0 g/mol+ 2 * 16.0 g/mol 
Molar mass of CO
2
= 12.0 g/mol+ 32.0 g/mol 
Molar mass of CO
2
= 44.0 g/mol 
Moles of CO
2
=
22 g
44.0 g/mol
 
Moles of CO
2
= 0.5moles 
We will use the stoichiometric ratio from the balanced chemical equation to find the moles of CH
4
 
required to produce 0.5 moles of CO
2
 : 
1 mole of CH
4
 : 1 mole of CO
2
 
This means that we also require 0.5 moles of CH
4
. 
Finally, we need to convert moles of methane to grams to find the mass: 
Mass of CH
4
= moles of CH
4
* molar mass of CH
4
 
Mass of CH
4
= 0.5 moles * 16.0 g/mol 
Mass of CH
4
= 8.0 g