Q. What is the major product of the reaction of 2-butyne with H2 in the presence of a catalyst?
A.
1-butyne
B.
butane
C.
2-butyne
D.
3-butyne
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Solution
The major product of hydrogenation of 2-butyne is butane.
Correct Answer:
B
— butane
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Q. What is the major product of the reaction of propene with HBr?
A.
1-Bromopropane
B.
2-Bromopropane
C.
Propyl bromide
D.
Bromopropene
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Solution
The major product of the reaction of propene with HBr is 2-bromopropane due to Markovnikov's rule.
Correct Answer:
B
— 2-Bromopropane
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Q. What is the major product when 1-bromobutane reacts with sodium ethoxide in ethanol?
A.
1-ethoxybutane
B.
2-ethoxybutane
C.
butane
D.
butyl ethyl ether
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Solution
The reaction is an example of an SN2 reaction, leading to the formation of 1-ethoxybutane.
Correct Answer:
A
— 1-ethoxybutane
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Q. What is the major product when propene is treated with bromine?
A.
1-bromopropane
B.
2-bromopropane
C.
Bromopropane
D.
No reaction
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Solution
The major product of the reaction of propene with bromine is 2-bromopropane due to anti-addition.
Correct Answer:
B
— 2-bromopropane
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Q. What is the major product when propene reacts with HBr?
A.
1-Bromopropane
B.
2-Bromopropane
C.
Propyl bromide
D.
Bromopropene
Show solution
Solution
The major product of the reaction of propene with HBr is 2-Bromopropane due to Markovnikov's rule.
Correct Answer:
B
— 2-Bromopropane
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Q. What is the mass of 0.25 moles of glucose (C6H12O6)? (Molar mass = 180 g/mol)
A.
45 g
B.
90 g
C.
60 g
D.
75 g
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Solution
Mass = moles × molar mass = 0.25 moles × 180 g/mol = 45 g.
Correct Answer:
B
— 90 g
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Q. What is the mass of 0.5 moles of CaCO3?
A.
50 g
B.
25 g
C.
100 g
D.
75 g
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Solution
Molar mass of CaCO3 = 40 + 12 + 16*3 = 100 g/mol. Mass = moles x molar mass = 0.5 moles x 100 g/mol = 50 g.
Correct Answer:
B
— 25 g
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Q. What is the mass of 0.5 moles of NaCl?
A.
29 g
B.
58 g
C.
14 g
D.
0.5 g
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Solution
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol. Mass = moles x molar mass = 0.5 moles x 58.5 g/mol = 29.25 g.
Correct Answer:
A
— 29 g
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Q. What is the mass of 0.5 moles of sodium chloride (NaCl)?
A.
29 g
B.
35 g
C.
58 g
D.
70 g
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Solution
The molar mass of NaCl is 58 g/mol. Therefore, 0.5 moles of NaCl = 0.5 x 58 g = 29 g.
Correct Answer:
C
— 58 g
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Q. What is the mass of 0.75 moles of sulfuric acid (H2SO4)?
A.
75 g
B.
98 g
C.
147 g
D.
50 g
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Solution
Molar mass of H2SO4 = 2*1 + 32 + 4*16 = 98 g/mol. Mass = moles x molar mass = 0.75 moles x 98 g/mol = 73.5 g.
Correct Answer:
B
— 98 g
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Q. What is the mass of 1 mole of water (H2O)?
A.
18 g
B.
16 g
C.
20 g
D.
22 g
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Solution
The molar mass of water (H2O) is 18 g/mol (2*1 for H + 16 for O).
Correct Answer:
A
— 18 g
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Q. What is the mass of 2 moles of carbon dioxide (CO2)?
A.
44 g
B.
22 g
C.
88 g
D.
66 g
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Solution
The molar mass of CO2 is 12 g/mol (C) + 16 g/mol x 2 (O) = 44 g/mol. Therefore, 2 moles of CO2 weigh 2 x 44 g = 88 g.
Correct Answer:
A
— 44 g
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Q. What is the mass of 2 moles of sulfuric acid (H2SO4)?
A.
98 g
B.
196 g
C.
294 g
D.
392 g
Show solution
Solution
Molar mass of H2SO4 = 2*1 + 32 + 4*16 = 98 g/mol. Mass = 2 moles x 98 g/mol = 196 g.
Correct Answer:
B
— 196 g
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Q. What is the mass of 3 moles of carbon dioxide (CO2)?
A.
44 g
B.
66 g
C.
132 g
D.
22 g
Show solution
Solution
The molar mass of CO2 is 12 g/mol (C) + 16 g/mol x 2 (O) = 44 g/mol. Therefore, 3 moles of CO2 weigh 3 x 44 g = 132 g.
Correct Answer:
A
— 44 g
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Q. What is the mass of 3 moles of NaCl?
A.
87 g
B.
58.5 g
C.
174 g
D.
3 g
Show solution
Solution
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol. Mass = moles x molar mass = 3 moles x 58.5 g/mol = 174 g.
Correct Answer:
C
— 174 g
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Q. What is the mass of 3 moles of sulfuric acid (H2SO4)?
A.
98 g
B.
196 g
C.
294 g
D.
392 g
Show solution
Solution
Molar mass of H2SO4 = 2*1 + 32 + 4*16 = 98 g/mol. Mass = 3 moles x 98 g/mol = 294 g.
Correct Answer:
B
— 196 g
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Q. What is the mass of KCl produced when 10 g of K reacts with excess Cl2?
A.
74.5 g
B.
10 g
C.
20 g
D.
30 g
Show solution
Solution
10 g of K = 0.25 moles. K + Cl2 → KCl, so 0.25 moles of KCl = 0.25 * 74.5 g = 18.625 g.
Correct Answer:
A
— 74.5 g
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Q. What is the mass of water produced when 2 moles of hydrogen react with 1 mole of oxygen?
A.
18 g
B.
36 g
C.
54 g
D.
9 g
Show solution
Solution
2 moles of H2 produce 2 moles of H2O. Mass of H2O = 2 moles * 18 g/mole = 36 g.
Correct Answer:
A
— 18 g
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Q. What is the mass percent of a solution containing 20 g of NaCl in 180 g of water?
A.
10%
B.
20%
C.
25%
D.
15%
Show solution
Solution
Mass percent = (mass of solute / total mass) x 100 = (20 g / (20 g + 180 g)) x 100 = 10%.
Correct Answer:
A
— 10%
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Q. What is the mass percent of a solution containing 20 g of solute in 180 g of solution?
A.
10%
B.
20%
C.
25%
D.
15%
Show solution
Solution
Mass percent = (mass of solute / mass of solution) x 100 = (20 g / 200 g) x 100 = 10%.
Correct Answer:
B
— 20%
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Q. What is the mass percent of a solution containing 20 g of solute in 200 g of solution?
A.
10%
B.
20%
C.
5%
D.
15%
Show solution
Solution
Mass percent = (mass of solute / mass of solution) × 100 = (20 g / 200 g) × 100 = 10%.
Correct Answer:
B
— 20%
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Q. What is the maximum number of electrons in the n=2 shell?
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Solution
The maximum number of electrons in the n=2 shell is 2(2^2) = 8.
Correct Answer:
B
— 4
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Q. What is the maximum number of electrons that can be accommodated in a shell with quantum number n=4?
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Solution
The maximum number of electrons in a shell is given by 2n^2. For n=4, it is 2(4^2) = 32.
Correct Answer:
C
— 32
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Q. What is the maximum number of electrons that can be accommodated in a shell with n=3?
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Solution
The maximum number of electrons in a shell is given by 2n^2. For n=3, it is 2(3^2) = 18.
Correct Answer:
B
— 18
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Q. What is the maximum number of electrons that can be accommodated in a shell with principal quantum number n=5?
Show solution
Solution
The maximum number of electrons in a shell is given by 2n^2. For n=5, it is 2(5^2) = 50.
Correct Answer:
C
— 32
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Q. What is the maximum number of electrons that can be accommodated in a subshell with the quantum number l = 2?
Show solution
Solution
The maximum number of electrons in a subshell is given by the formula 2(2l + 1). For l = 2, it is 2(2*2 + 1) = 10.
Correct Answer:
C
— 10
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Q. What is the maximum number of electrons that can be accommodated in the n=2 shell?
Show solution
Solution
The maximum number of electrons in a shell is given by 2n^2. For n=2, this is 2(2^2) = 8.
Correct Answer:
B
— 4
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Q. What is the maximum number of electrons that can be accommodated in the n=5 shell?
Show solution
Solution
The maximum number of electrons in a shell is given by 2n^2. For n=5, it is 2(5^2) = 50.
Correct Answer:
C
— 32
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Q. What is the maximum number of electrons that can occupy a p subshell?
Show solution
Solution
A p subshell can hold a maximum of 6 electrons.
Correct Answer:
B
— 6
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Q. What is the maximum number of electrons that can occupy a single orbital?
Show solution
Solution
A single orbital can hold a maximum of 2 electrons, with opposite spins.
Correct Answer:
B
— 2
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Chemistry Syllabus (JEE Main) MCQ & Objective Questions
The Chemistry Syllabus for JEE Main is crucial for students aiming to excel in their exams. Understanding this syllabus not only helps in grasping fundamental concepts but also enhances performance in objective questions and MCQs. Regular practice with these types of questions is essential for scoring better and mastering important topics.
What You Will Practise Here
Basic Concepts of Chemistry
Atomic Structure and Chemical Bonding
States of Matter: Gases and Liquids
Thermodynamics and Thermochemistry
Equilibrium: Chemical and Ionic
Redox Reactions and Electrochemistry
Hydrocarbons and Environmental Chemistry
Exam Relevance
The Chemistry syllabus is a significant part of CBSE, State Boards, NEET, and JEE exams. Questions from this syllabus often appear in various formats, including multiple-choice questions, assertion-reason type questions, and numerical problems. Familiarity with the common question patterns can greatly enhance your exam preparation and confidence.
Common Mistakes Students Make
Misunderstanding the periodic trends and their implications.
Confusing different types of chemical bonds and their properties.
Neglecting to balance redox reactions properly.
Overlooking the significance of units in thermodynamic calculations.
Failing to apply concepts of equilibrium in problem-solving.
FAQs
Question: What are the key topics I should focus on in the Chemistry syllabus for JEE Main?Answer: Focus on atomic structure, chemical bonding, thermodynamics, and equilibrium as they are frequently tested.
Question: How can I improve my performance in Chemistry MCQs?Answer: Regular practice with past papers and understanding concepts deeply will help you tackle MCQs effectively.
Start your journey towards mastering the Chemistry Syllabus (JEE Main) by solving practice MCQs today. Test your understanding and build confidence for your exams!
