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JEE Main MCQ & Objective Questions

The JEE Main exam is a crucial step for students aspiring to enter prestigious engineering colleges in India. It tests not only knowledge but also the ability to apply concepts effectively. Practicing MCQs and objective questions is essential for scoring better, as it helps in familiarizing students with the exam pattern and enhances their problem-solving skills. Engaging with practice questions allows students to identify important questions and strengthen their exam preparation.

What You Will Practise Here

Fundamental concepts of Physics, Chemistry, and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theories relevant to JEE Main
Diagrams and graphical representations for better understanding
Numerical problems and their step-by-step solutions
Previous years' JEE Main questions for real exam experience
Time management strategies while solving MCQs

Exam Relevance

The topics covered in JEE Main are not only significant for the JEE exam but also appear in various CBSE and State Board examinations. Many concepts are shared with the NEET syllabus, making them relevant across multiple competitive exams. Common question patterns include conceptual applications, numerical problems, and theoretical questions that assess a student's understanding of core subjects.

Common Mistakes Students Make

Misinterpreting the question stem, leading to incorrect answers
Neglecting units in numerical problems, which can change the outcome
Overlooking negative marking and not managing time effectively
Relying too heavily on rote memorization instead of understanding concepts
Failing to review and analyze mistakes from practice tests

FAQs

Question: How can I improve my speed in solving JEE Main MCQ questions?
Answer: Regular practice with timed quizzes and focusing on shortcuts can significantly enhance your speed.

Question: Are the JEE Main objective questions similar to previous years' papers?
Answer: Yes, many questions are based on previous years' patterns, so practicing them can be beneficial.

Question: What is the best way to approach JEE Main practice questions?
Answer: Start with understanding the concepts, then attempt practice questions, and finally review your answers to learn from mistakes.

Now is the time to take charge of your preparation! Dive into solving JEE Main MCQs and practice questions to test your understanding and boost your confidence for the exam.

Chemistry Syllabus (JEE Main) Mathematics Syllabus (JEE Main) Physics Syllabus (JEE Main)

Q. If 1 liter of a 2 M solution is diluted to 3 liters, what is the new molarity?

A. 0.67 M
B. 1 M
C. 1.5 M
D. 2 M

Solution

Using the dilution formula M1V1 = M2V2, (2 M)(1 L) = M2(3 L) => M2 = 2/3 = 0.67 M.

Correct Answer: A — 0.67 M

Q. If 1 mol of NaCl is dissolved in 1 kg of water, how many particles are present in solution?

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

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl produces 2 mol of particles in solution.

Correct Answer: B — 2

Q. If 1 mol 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 = 2.

Correct Answer: B — 2

Q. If 1 mol of NaCl is dissolved in water, how many particles are present in solution?

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

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl results in 2 mol of particles.

Correct Answer: B — 2

Q. If 1 mole of a gas occupies 22.4 L at STP, how many liters will 0.5 moles occupy?

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

Solution

Volume = moles x volume per mole = 0.5 moles x 22.4 L/mole = 11.2 L.

Correct Answer: A — 11.2 L

Q. If 1 mole of a gas occupies 22.4 L at STP, how much volume will 0.5 moles occupy?

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

Solution

Volume = moles x volume per mole = 0.5 moles x 22.4 L/mole = 11.2 L.

Correct Answer: A — 11.2 L

Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the freezing point depression?

A. 0 °C
B. 1.86 °C
C. 3.72 °C
D. 5.58 °C

Solution

The freezing point depression is calculated using the formula ΔTf = i * Kf * m. For a non-electrolyte, i = 1, Kf for water = 1.86 °C kg/mol, and m = 1 mol/kg gives ΔTf = 1.86 °C.

Correct Answer: B — 1.86 °C

Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the expected change in freezing point?

A. 0.0 °C
B. -1.86 °C
C. -3.72 °C
D. -5.58 °C

Solution

The freezing point depression for 1 mole of a non-electrolyte solute in 1 kg of water is -1.86 °C.

Correct Answer: B — -1.86 °C

Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the expected freezing point depression?

A. -1.86 °C
B. -3.72 °C
C. -0.52 °C
D. -2.00 °C

Solution

The freezing point depression is calculated using the formula ΔTf = Kf * m, where Kf for water is 1.86 °C kg/mol.

Correct Answer: A — -1.86 °C

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

Q. If 1 mole of an ideal gas occupies 22.4 L at STP, what is the pressure exerted by the gas?

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

Solution

At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 L at a pressure of 1 atm.

Correct Answer: A — 1 atm

Q. If 1 mole of an ideal gas occupies 22.4 L at STP, what is the volume occupied by 2 moles at the same conditions?

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 56.8 L

Solution

According to Avogadro's Law, 2 moles of gas will occupy double the volume, which is 44.8 L at STP.

Correct Answer: C — 44.8 L

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

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

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

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

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

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

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%

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

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

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

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

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

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

Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature of the mixture?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

Using the principle of conservation of energy, the final temperature will be 50°C.

Correct Answer: A — 50°C

Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

The final temperature will be 50°C due to equal masses and specific heat capacities.

Correct Answer: A — 50°C

Q. If 100 g of water at 80°C is mixed with 200 g of water at 20°C, what will be the final temperature?

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy, the final temperature can be calculated to be 40°C.

Correct Answer: B — 40°C

Q. If 100 J of heat is added to a system and 40 J of work is done by the system, what is the change in internal energy?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

According to the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

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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