Chemistry Syllabus (JEE Main) - Exam Prep Guide

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Chemistry Syllabus (JEE Main)

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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!

Inorganic Chemistry Organic Chemistry Physical Chemistry

Q. A ladder leans against a wall and is in equilibrium. What forces are acting on the ladder?

A. Weight, normal force from the ground, and friction
B. Only weight and normal force
C. Only weight and friction
D. Only normal force and friction

Solution

The ladder experiences weight, a normal force from the ground, and friction at the base.

Correct Answer: A — Weight, normal force from the ground, and friction

Q. A particle is in equilibrium under the action of three forces. If two forces are known, how can the third force be determined?

A. By vector addition of the first two forces
B. By subtracting the first two forces
C. By multiplying the first two forces
D. By taking the average of the first two forces

Solution

The third force can be determined by vector addition of the first two forces to ensure the net force is zero.

Correct Answer: A — By vector addition of the first two forces

Q. A solution contains 20% (w/w) of glucose. If the total mass of the solution is 200 g, what is the mass of glucose in the solution?

A. 20 g
B. 40 g
C. 60 g
D. 80 g

Solution

Mass of glucose = (20/100) x 200 g = 40 g.

Correct Answer: B — 40 g

Q. A solution has a concentration of 0.1 M NaCl. How many grams of NaCl are present in 1 liter of this solution? (Molar mass of NaCl = 58.5 g/mol)

A. 5.85 g
B. 58.5 g
C. 0.1 g
D. 0.585 g

Solution

Mass = moles x molar mass = 0.1 moles x 58.5 g/mol = 5.85 g.

Correct Answer: B — 58.5 g

Q. A solution has a concentration of 0.2 M. How many moles of solute are present in 1.5 L of this solution?

A. 0.3 moles
B. 0.5 moles
C. 0.2 moles
D. 0.15 moles

Solution

Moles of solute = Molarity × Volume = 0.2 M × 1.5 L = 0.3 moles.

Correct Answer: B — 0.5 moles

Q. A solution has a density of 1.2 g/mL and contains 10% (w/v) NaCl. What is the mass of NaCl in 1 liter of this solution?

A. 100 g
B. 120 g
C. 80 g
D. 60 g

Solution

Mass of NaCl = (10/100) x 1000 mL = 100 g.

Correct Answer: B — 120 g

Q. A solution has a density of 1.2 g/mL and contains 10% (w/v) NaOH. What is the mass of NaOH in 1 L of this solution?

A. 100 g
B. 120 g
C. 80 g
D. 60 g

Solution

Mass of NaOH = (10/100) x 1000 mL = 100 g.

Correct Answer: B — 120 g

Q. A solution has a density of 1.2 g/mL and contains 30 g of solute. What is the molarity if the molar mass of the solute is 60 g/mol?

A. 0.5 M
B. 1 M
C. 2 M
D. 1.5 M

Solution

Volume of solution = mass / density = 30 g / 1.2 g/mL = 25 mL = 0.025 L. Moles of solute = 30 g / 60 g/mol = 0.5 moles. Molarity = 0.5 moles / 0.025 L = 20 M.

Correct Answer: B — 1 M

Q. A solution is prepared by dissolving 50 g of glucose (C6H12O6) in 250 g of water. What is the mass percent of glucose in the solution? (Molar mass of glucose = 180 g/mol)

A. 20%
B. 15%
C. 25%
D. 10%

Solution

Mass percent = (mass of solute / (mass of solute + mass of solvent)) × 100 = (50 g / (50 g + 250 g)) × 100 = 20%.

Correct Answer: A — 20%

Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. What is the concentration in terms of molarity? (Molar mass of NaCl = 58.5 g/mol)

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

Solution

Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.

Correct Answer: A — 1 M

Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)

A. 1 M
B. 2 M
C. 0.5 M
D. 0.1 M

Solution

Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = moles of solute / liters of solution = 1 mole / 1 L = 1 M.

Correct Answer: A — 1 M

Q. A solution is prepared by dissolving 58.5 g of NaCl in enough water to make 1 L of solution. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)

A. 1 M
B. 2 M
C. 0.5 M
D. 0.1 M

Solution

Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.

Correct Answer: A — 1 M

Q. A solution of 0.1 molal urea in water has a freezing point depression of how much? (K_f for water = 1.86 °C kg/mol)

A. 0.186 °C
B. 0.372 °C
C. 0.186 K
D. 0.372 K

Solution

Freezing point depression = K_f * m = 1.86 * 0.1 = 0.186 °C

Correct Answer: A — 0.186 °C

Q. According to Boyle's law, if the volume of a gas is doubled at constant temperature, what happens to the pressure?

A. It doubles
B. It halves
C. It remains constant
D. It quadruples

Solution

Boyle's law states that pressure is inversely proportional to volume at constant temperature, so if the volume is doubled, the pressure halves.

Correct Answer: B — It halves

Q. According to Boyle's law, if the volume of a gas is doubled, what happens to its pressure?

A. It doubles
B. It halves
C. It remains constant
D. It quadruples

Solution

Boyle's law states that pressure is inversely proportional to volume at constant temperature, so if volume is doubled, pressure is halved.

Correct Answer: B — It halves

Q. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to what?

A. Molar mass
B. Temperature
C. Pressure
D. Volume

Solution

Graham's law states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas.

Correct Answer: A — Molar mass

Q. According to the ideal gas law, which of the following is the correct equation?

A. PV = nRT
B. PV = nR
C. P = nRT/V
D. V = nP/RT

Solution

The ideal gas law is represented by the equation PV = nRT.

Correct Answer: A — PV = nRT

Q. According to the kinetic molecular theory, which of the following statements is true?

A. Gas particles have significant volume.
B. Gas particles are in constant motion.
C. Gas particles attract each other.
D. Gas particles do not collide.

Solution

According to the kinetic molecular theory, gas particles are in constant motion.

Correct Answer: B — Gas particles are in constant motion.

Q. According to VSEPR theory, what is the shape of the molecule with the formula AX2E2?

A. Linear
B. Bent
C. Trigonal planar
D. Tetrahedral

Solution

AX2E2 indicates two bonding pairs and two lone pairs, resulting in a bent shape.

Correct Answer: B — Bent

Q. At constant temperature and pressure, if ΔH is positive and ΔS is negative, what can be said about ΔG?

A. ΔG is positive
B. ΔG is negative
C. ΔG is zero
D. ΔG can be either positive or negative

Solution

If ΔH is positive and ΔS is negative, ΔG will be positive.

Correct Answer: A — ΔG is positive

Q. At constant temperature and pressure, if ΔH is positive and ΔS is negative, what is the sign of ΔG?

A. Always negative
B. Always positive
C. Depends on temperature
D. Zero

Solution

If ΔH is positive and ΔS is negative, ΔG will always be positive.

Correct Answer: B — Always positive

Q. At constant temperature, the volume of a gas is inversely proportional to its pressure. This is known as which law?

A. Boyle's Law
B. Charles's Law
C. Avogadro's Law
D. Ideal Gas Law

Solution

Boyle's Law states that for a given mass of gas at constant temperature, the volume is inversely proportional to the pressure.

Correct Answer: A — Boyle's Law

Q. At what temperature does a reaction become spontaneous if ΔH = 50 kJ and ΔS = 0.1 kJ/K?

A. 500 K
B. 250 K
C. 1000 K
D. 200 K

Solution

Set ΔG = 0: 0 = ΔH - TΔS; T = ΔH/ΔS = 50 kJ / 0.1 kJ/K = 500 K.

Correct Answer: A — 500 K

Q. At what temperature does the Gibbs Free Energy change from negative to positive?

A. At absolute zero
B. At the melting point
C. At the boiling point
D. At the transition temperature

Solution

The Gibbs Free Energy changes from negative to positive at the transition temperature, where the system shifts from one phase to another.

Correct Answer: D — At the transition temperature

Q. At what temperature does the volume of a gas become zero according to Charles's Law?

A. 0 K
B. -273.15 °C
C. 273.15 K
D. None of the above

Solution

According to Charles's Law, the volume of a gas approaches zero at absolute zero, which is -273.15 °C.

Correct Answer: B — -273.15 °C

Q. Calculate the molality of a solution if the boiling point elevation is 1.024 °C. (K_b for water = 0.512 °C kg/mol)

A. 1 mol/kg
B. 2 mol/kg
C. 0.5 mol/kg
D. 0.25 mol/kg

Solution

Molality = ΔT_b / (i * K_b) = 1.024 / (2 * 0.512) = 1 mol/kg

Correct Answer: B — 2 mol/kg

Q. Calculate the pH of a 0.1 M acetic acid solution (Ka = 1.8 x 10^-5).

A. 2.87
B. 3.87
C. 4.87
D. 5.87

Solution

Using the formula for weak acids, pH = 0.5(pKa - log[C]), where pKa = -log(1.8 x 10^-5) ≈ 4.74. Thus, pH = 0.5(4.74 - log(0.1)) = 3.87.

Correct Answer: B — 3.87

Q. Calculate the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate.

A. 4.76
B. 5.76
C. 6.76
D. 7.76

Solution

Using Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]); pKa of acetic acid = 4.76, so pH = 4.76 + log(1) = 4.76

Correct Answer: B — 5.76

Q. Calculate the pH of a solution that is 0.1 M in acetic acid (Ka = 1.8 x 10^-5).

A. 2.87
B. 3.87
C. 4.87
D. 5.87

Solution

Using the formula for weak acids, pH = 0.5(pKa - logC) = 0.5(4.74 - log(0.1)) = 3.87.

Correct Answer: B — 3.87

Q. Determine the hybridization of the central atom in BF3.

A. sp
B. sp2
C. sp3
D. dsp3

Solution

Boron in BF3 is sp2 hybridized, forming three equivalent sp2 hybrid orbitals.

Correct Answer: B — sp2

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