Major Competitive Exams

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Major Competitive Exams

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Major Competitive Exams MCQ & Objective Questions

Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.

What You Will Practise Here

Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams

Exam Relevance

The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.

Common Mistakes Students Make

Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam

FAQs

Question: How can I improve my performance in Major Competitive Exams?
Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.

Question: What types of questions should I focus on for these exams?
Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.

Question: Are there specific strategies for tackling objective questions?
Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.

Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!

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Q. If 0.6 is expressed as a fraction, what is it?

A. 3/5
B. 2/3
C. 1/2
D. 4/5

Solution

0.6 = 6/10 = 3/5 after simplification.

Correct Answer: A — 3/5

Q. If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)

A. 3360 J
B. 4000 J
C. 4200 J
D. 4800 J

Solution

Q = m*c*ΔT = 1000 g * 4.2 J/g°C * (100°C - 20°C) = 4200 J.

Correct Answer: C — 4200 J

Q. If 1 kg of water is heated from 25°C to 75°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C) (2021)

A. 21000 J
B. 42000 J
C. 84000 J
D. 105000 J

Solution

Q = mcΔT = (1000 g)(4.2 J/g°C)(50°C) = 210000 J.

Correct Answer: B — 42000 J

Q. If 1 kg of water is mixed with 1 kg of sugar, what is the expected change in boiling point? (2023)

A. No change
B. Increase
C. Decrease
D. Depends on the sugar type

Solution

The addition of sugar (a non-volatile solute) will raise the boiling point of water due to boiling point elevation.

Correct Answer: B — Increase

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

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

Solution

Using the dilution formula M1V1 = M2V2, we have 2 M * 1 L = M2 * 3 L, thus M2 = 2/3 = 0.67 M.

Correct Answer: A — 0.67 M

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

A. 1.5 M
B. 3 M
C. 6 M
D. 0.5 M

Solution

Using the dilution formula M1V1 = M2V2, we have 3 M × 1 L = M2 × 2 L. Thus, M2 = 1.5 M.

Correct Answer: A — 1.5 M

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 gas occupies 22.4 liters at STP, how many liters will 0.5 moles occupy? (2023)

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

Solution

At STP, 1 mole of gas occupies 22.4 L. Therefore, 0.5 moles will occupy 0.5 x 22.4 L = 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 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-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-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 a non-volatile solute is dissolved in 1 kg of water, what is the expected freezing point depression? (2019)

A. 1.86 °C
B. 3.72 °C
C. 0.93 °C
D. 2.0 °C

Solution

Freezing point depression (ΔTf) = i * Kf * m = 1 * 1.86 °C kg/mol * 1 mol/kg = 1.86 °C.

Correct Answer: A — 1.86 °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 1/5 of a number is 20, what is the number?

A. 100
B. 80
C. 60
D. 40

Solution

Let the number be x. Then, 1/5 * x = 20, so x = 20 * 5 = 100.

Correct Answer: A — 100

Q. If 10 g of a substance reacts completely with 20 g of another substance, what is the total mass of the products formed? (2020)

A. 30 g
B. 20 g
C. 10 g
D. 40 g

Solution

According to the law of conservation of mass, the total mass of the products is equal to the total mass of the reactants, which is 10 g + 20 g = 30 g.

Correct Answer: A — 30 g

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 glucose (C6H12O6) is dissolved in 200 g of water, what is the mass percentage of the solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of sugar in the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: B — 5.00%

Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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