Chemistry Syllabus (JEE Main)

Inorganic Chemistry Organic Chemistry Physical Chemistry

Q. If 1 mole of a non-volatile solute is dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.512 °C kg/mol)

A. 0.512 °C
B. 1.024 °C
C. 2.048 °C
D. 0.256 °C

Solution

Boiling point elevation = i * Kb * m = 1 * 0.512 * 1 = 0.512 °C.

Correct Answer: B — 1.024 °C

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Q. If 1 mole of NaCl is dissolved in 1 kg of water, what is the expected van 't Hoff factor (i)?

A. 1
B. 2
C. 3
D. 4

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor (i) is 2.

Correct Answer: B — 2

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Q. If 1 mole of solute is dissolved in 1 liter of solution, what is the concentration in terms of molarity?

A. 1 M
B. 2 M
C. 0.5 M
D. 0.25 M

Solution

Molarity (M) = moles of solute / liters of solution = 1 mole / 1 L = 1 M.

Correct Answer: A — 1 M

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Q. If 10 g of CaCO3 decomposes completely, how many grams of CO2 are produced?

A. 22 g
B. 10 g
C. 44 g
D. 20 g

Solution

10 g of CaCO3 = 0.1 moles. CaCO3 → CaO + CO2, so 0.1 moles of CO2 = 0.1 * 44 g = 4.4 g.

Correct Answer: C — 44 g

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Q. If 10 g of Na reacts with excess Cl2, what is the mass of NaCl produced?

A. 58.5 g
B. 10 g
C. 20 g
D. 30 g

Solution

10 g of Na = 0.43 moles. Na + Cl2 → NaCl, so 0.43 moles of NaCl = 0.43 * 58.5 g = 25.2 g.

Correct Answer: A — 58.5 g

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Q. If 10 grams of calcium carbonate (CaCO3) decomposes, how many grams of calcium oxide (CaO) are produced?

A. 5 g
B. 10 g
C. 8 g
D. 7 g

Solution

The balanced equation is CaCO3 → CaO + CO2. The molar mass of CaCO3 is 100 g/mol and CaO is 56 g/mol. Thus, 10 g of CaCO3 produces (10 g / 100 g/mol) x 56 g/mol = 5.6 g of CaO.

Correct Answer: C — 8 g

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Q. If 10 grams of NaCl are dissolved in water, how many moles of NaCl are present? (Molar mass of NaCl = 58.5 g/mol)

A. 0.17 moles
B. 0.5 moles
C. 1.0 moles
D. 1.5 moles

Solution

Moles of NaCl = mass (g) / molar mass (g/mol) = 10 g / 58.5 g/mol = 0.171 moles.

Correct Answer: A — 0.17 moles

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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the mass/volume percent concentration?

A. 1%
B. 2%
C. 5%
D. 10%

Solution

Mass/volume percent = (mass of solute / volume of solution) x 100 = (10 g / 500 mL) x 100 = 2%.

Correct Answer: B — 2%

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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the molality of the solution? (Molar mass of NaCl = 58.5 g/mol)

A. 0.34 m
B. 0.17 m
C. 0.85 m
D. 0.50 m

Solution

Molality (m) = moles of solute / kg of solvent. Moles of NaCl = 10 g / 58.5 g/mol = 0.171 moles. Mass of water = 0.5 kg. Molality = 0.171 moles / 0.5 kg = 0.342 m.

Correct Answer: A — 0.34 m

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Q. If 10 grams of NaCl is dissolved in enough water to make 500 mL of solution, what is the molality of the solution? (Molar mass of NaCl = 58.5 g/mol)

A. 0.34 m
B. 0.17 m
C. 0.85 m
D. 0.50 m

Solution

Moles of NaCl = 10 g / 58.5 g/mol = 0.171 moles. Mass of solvent (water) = 0.5 kg. Molality (m) = moles of solute / kg of solvent = 0.171 moles / 0.5 kg = 0.34 m.

Correct Answer: A — 0.34 m

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Q. If 10 grams of NaOH are dissolved in water, how many moles of NaOH are present? (Molar mass of NaOH = 40 g/mol)

A. 0.25 moles
B. 0.5 moles
C. 1 mole
D. 2.5 moles

Solution

To find the number of moles, use the formula: moles = mass/molar mass. Thus, 10 g / 40 g/mol = 0.25 moles.

Correct Answer: B — 0.5 moles

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Q. If 10 grams of NaOH is dissolved in 500 mL of solution, what is the molality of the solution? (Molar mass of NaOH = 40 g/mol)

A. 0.5 m
B. 1 m
C. 2 m
D. 0.25 m

Solution

Moles of NaOH = 10 g / 40 g/mol = 0.25 moles. Mass of solvent (water) = 0.5 kg. Molality (m) = moles of solute / kg of solvent = 0.25 moles / 0.5 kg = 0.5 m.

Correct Answer: B — 1 m

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Q. If 10 grams of NaOH is dissolved in enough water to make 500 mL of solution, what is the molarity of the solution? (Molar mass of NaOH = 40 g/mol)

A. 0.5 M
B. 1 M
C. 2 M
D. 0.25 M

Solution

Moles of NaOH = 10 g / 40 g/mol = 0.25 moles. Molarity = 0.25 moles / 0.5 L = 0.5 M.

Correct Answer: B — 1 M

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Q. If 100 g of glucose (C6H12O6) is dissolved in 1 L of solution, what is the molarity of the solution? (Molar mass of glucose = 180 g/mol)

A. 0.56 M
B. 1.0 M
C. 0.33 M
D. 0.75 M

Solution

Moles of glucose = 100 g / 180 g/mol = 0.556 moles. Molarity = moles of solute / liters of solution = 0.556 moles / 1 L = 0.56 M.

Correct Answer: A — 0.56 M

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Q. If 100 J of heat is added to a system at a constant temperature of 300 K, what is the change in entropy?

A. 0.33 J/K
B. 0.25 J/K
C. 0.5 J/K
D. 0.75 J/K

Solution

The change in entropy ΔS = Q/T = 100 J / 300 K = 0.33 J/K.

Correct Answer: A — 0.33 J/K

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Q. If 15 grams of HCl is dissolved in 500 mL of solution, what is the molarity of the solution? (Molar mass of HCl = 36.5 g/mol)

A. 0.82 M
B. 1.0 M
C. 0.5 M
D. 1.5 M

Solution

Moles of HCl = 15 g / 36.5 g/mol = 0.41096 moles. Molarity = 0.41096 moles / 0.5 L = 0.82192 M.

Correct Answer: A — 0.82 M

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Q. If 15 grams of sugar is dissolved in 250 mL of solution, what is the molarity of the solution? (Molar mass of sugar = 180 g/mol)

A. 0.33 M
B. 0.5 M
C. 0.25 M
D. 0.75 M

Solution

Moles of sugar = 15 g / 180 g/mol = 0.0833 moles. Molarity = 0.0833 moles / 0.25 L = 0.333 M.

Correct Answer: A — 0.33 M

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Q. If 15 grams of sugar is dissolved in 250 mL of water, what is the molarity of the solution? (Molar mass of sugar = 180 g/mol)

A. 0.33 M
B. 0.5 M
C. 0.25 M
D. 0.75 M

Solution

Moles of sugar = 15 g / 180 g/mol = 0.0833 moles. Molarity = 0.0833 moles / 0.25 L = 0.333 M.

Correct Answer: A — 0.33 M

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Q. If 2 moles of a gas occupy 44.8 L at STP, what is the molar volume of the gas?

A. 22.4 L
B. 44.8 L
C. 11.2 L
D. 33.6 L

Solution

At STP, 1 mole of gas occupies 22.4 L. Therefore, the molar volume of the gas is 22.4 L.

Correct Answer: A — 22.4 L

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Q. If 2 moles of glucose are dissolved in 1 liter of water, what is the molarity of the solution?

A. 1 M
B. 2 M
C. 0.5 M
D. 3 M

Solution

Molarity = moles of solute / liters of solution = 2 moles / 1 L = 2 M.

Correct Answer: B — 2 M

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Q. If 2 moles of KCl are dissolved in 1 liter of solution, what is the molarity of the solution?

A. 1 M
B. 2 M
C. 0.5 M
D. 3 M

Solution

Molarity (M) = moles of solute / liters of solution = 2 moles / 1 L = 2 M.

Correct Answer: B — 2 M

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Q. If 2 moles of NaCl are dissolved in 1 kg of water, what is the van 't Hoff factor (i)?

A. 1
B. 2
C. 3
D. 4

Solution

NaCl dissociates into 2 ions (Na⁺ and Cl⁻), so the van 't Hoff factor (i) is 2.

Correct Answer: C — 3

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Q. If 20 grams of glucose (C6H12O6) is dissolved in 200 mL of solution, what is the mass percent concentration? (Molar mass of glucose = 180 g/mol)

A. 10%
B. 5%
C. 20%
D. 15%

Solution

Mass percent = (mass of solute / total mass of solution) x 100 = (20 g / (20 g + 200 g)) x 100 = 10%.

Correct Answer: A — 10%

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Q. If 3 moles of a gas occupy 60 liters at STP, what is the molar volume of the gas?

A. 20 L
B. 30 L
C. 40 L
D. 60 L

Solution

Molar volume = total volume / number of moles = 60 L / 3 moles = 20 L/mole.

Correct Answer: B — 30 L

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Q. If 3 moles of a gas occupy 60 liters, what is the volume occupied by 1 mole of the gas at the same conditions?

A. 20 L
B. 30 L
C. 15 L
D. 10 L

Solution

Volume per mole = total volume / number of moles = 60 L / 3 moles = 20 L/mole.

Correct Answer: A — 20 L

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Q. If 4 moles of aluminum react with 3 moles of oxygen, how many moles of aluminum oxide are formed?

A. 2 moles
B. 3 moles
C. 4 moles
D. 6 moles

Solution

The balanced equation is 4Al + 3O2 → 2Al2O3, which shows that 4 moles of Al produce 2 moles of Al2O3.

Correct Answer: A — 2 moles

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Q. If 4 moles of solute are dissolved in 2 kg of solvent, what is the molality of the solution?

A. 2 m
B. 4 m
C. 1 m
D. 0.5 m

Solution

Molality (m) = moles of solute / kg of solvent = 4 moles / 2 kg = 2 m.

Correct Answer: A — 2 m

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Q. If 5 g of CaCO3 decomposes completely, how many grams of CaO are produced?

A. 2 g
B. 3 g
C. 4 g
D. 5 g

Solution

CaCO3 → CaO + CO2. Molar mass of CaCO3 = 100 g, CaO = 56 g. 5 g of CaCO3 produces (5 g * 56 g) / 100 g = 2.8 g of CaO.

Correct Answer: C — 4 g

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Q. If 5 moles of KCl are dissolved in 1 kg of water, what is the molality of the solution?

A. 5 m
B. 2.5 m
C. 10 m
D. 1 m

Solution

Molality (m) = moles of solute / kg of solvent = 5 moles / 1 kg = 5 m.

Correct Answer: A — 5 m

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Q. If 5 moles of KCl are dissolved in 3 kg of water, what is the molality of the solution?

A. 1.67 m
B. 2.5 m
C. 1.25 m
D. 0.5 m

Solution

Molality (m) = moles of solute / kg of solvent = 5 moles / 3 kg = 1.67 m.

Correct Answer: A — 1.67 m

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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!

Chemistry Syllabus (JEE Main)

Chemistry Syllabus (JEE Main) MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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