Physical Chemistry Study Resources for Competitive Exams

Physical Chemistry

Atomic Structure Basic Concepts of Chemistry Chemical Bonding Chemical Kinetics Electrochemistry Equilibrium Redox Reactions Solutions States of Matter Surface Chemistry Thermodynamics

Q. What is the molecular orbital configuration of the F2 molecule?

A. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
B. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴
C. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)¹
D. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³(π*2p)²

Solution

The correct configuration for F2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)².

Correct Answer: A — (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²

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Q. What is the molecular orbital configuration of the O2 molecule?

A. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
B. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)²
C. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)¹
D. (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)²

Solution

The correct configuration for O2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹.

Correct Answer: A — (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹

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Q. What is the molecular shape of a molecule with the formula AX3E?

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

Solution

AX3E indicates three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.

Correct Answer: C — Trigonal pyramidal

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Q. What is the molecular shape of BF3 according to VSEPR theory?

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

Solution

BF3 has three bonding pairs and no lone pairs, resulting in a trigonal planar shape.

Correct Answer: B — Trigonal planar

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Q. What is the molecular shape of NH3 according to VSEPR theory?

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

Solution

NH3 has three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.

Correct Answer: D — Trigonal pyramidal

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Q. What is the Nernst equation used for?

A. Calculating pH
B. Determining cell potential
C. Finding molarity
D. Measuring temperature

Solution

The Nernst equation is used to determine the cell potential under non-standard conditions.

Correct Answer: B — Determining cell potential

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Q. What is the normality of a solution containing 1 mole of H2SO4 in 1 liter of solution?

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

Solution

Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so N = 2 moles / 1 L = 2 N.

Correct Answer: B — 2 N

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Q. What is the normality of a solution containing 2 moles of H2SO4 in 1 liter of solution?

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

Solution

Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so 2 moles × 2 = 4 N.

Correct Answer: B — 4 N

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Q. What is the normality of a solution containing 3 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)

A. 3 N
B. 6 N
C. 1.5 N
D. 1 N

Solution

Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents per mole, so 3 moles = 6 equivalents. Normality = 6 equivalents / 2 L = 3 N.

Correct Answer: B — 6 N

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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution?

A. 4 N
B. 8 N
C. 2 N
D. 1 N

Solution

Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents, so 4 moles = 8 equivalents. N = 8 eq / 2 L = 4 N.

Correct Answer: B — 8 N

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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)

A. 4 N
B. 2 N
C. 8 N
D. 1 N

Solution

Normality (N) = number of equivalents / liters of solution. H2SO4 has 2 equivalents per mole, so 4 moles = 8 equivalents. Normality = 8 equivalents / 2 L = 4 N.

Correct Answer: C — 8 N

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Q. What is the normality of a solution that is 1 M in H2SO4?

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

Solution

Normality (N) = Molarity (M) x number of equivalents. H2SO4 has 2 acidic protons, so 1 M x 2 = 2 N.

Correct Answer: B — 2 N

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Q. What is the number of atoms in 2 moles of aluminum (Al)? (2044)

A. 6.022 x 10^23
B. 1.2044 x 10^24
C. 3.011 x 10^23
D. 12.044 x 10^24

Solution

Number of atoms = moles x Avogadro's number = 2 moles x 6.022 x 10^23 atoms/mole = 1.2044 x 10^24 atoms.

Correct Answer: B — 1.2044 x 10^24

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Q. What is the number of atoms in 2 moles of CaCO3?

A. 6.022 x 10^23
B. 1.2044 x 10^24
C. 3.011 x 10^23
D. 1.8066 x 10^24

Solution

Each CaCO3 has 5 atoms (1 Ca, 1 C, 3 O). Number of atoms = moles x atoms per molecule x Avogadro's number = 2 moles x 5 x 6.022 x 10^23 = 1.206 x 10^24 atoms.

Correct Answer: D — 1.8066 x 10^24

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Q. What is the number of atoms in 2 moles of Na2SO4?

A. 6.022 x 10^23
B. 1.2044 x 10^24
C. 1.2044 x 10^25
D. 3.011 x 10^23

Solution

Na2SO4 has 2 Na + 1 S + 4 O = 7 atoms. Number of atoms = moles x atoms per mole = 2 moles x 7 atoms = 14 atoms = 1.2044 x 10^24 atoms.

Correct Answer: B — 1.2044 x 10^24

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Q. What is the number of atoms in 2 moles of NaCl? (2044)

A. 6.022 x 10^23
B. 1.2044 x 10^24
C. 1.2044 x 10^25
D. 3.011 x 10^23

Solution

Each NaCl unit has 2 atoms (Na and Cl). Therefore, 2 moles of NaCl contain 2 x 6.022 x 10^23 x 2 = 1.2044 x 10^24 atoms.

Correct Answer: B — 1.2044 x 10^24

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Q. What is the number of moles in 10 grams of Na?

A. 0.43
B. 0.22
C. 0.5
D. 0.1

Solution

Molar mass of Na = 23 g/mol. Number of moles = mass / molar mass = 10 g / 23 g/mol = 0.43 moles.

Correct Answer: A — 0.43

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Q. What is the number of moles in 10 grams of NaOH?

A. 0.25
B. 0.5
C. 0.75
D. 1

Solution

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Number of moles = mass/molar mass = 10 g / 40 g/mol = 0.25 moles.

Correct Answer: B — 0.5

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Q. What is the number of moles in 100 grams of NaOH?

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

Solution

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Number of moles = mass/molar mass = 100 g / 40 g/mol = 2.5 moles.

Correct Answer: A — 1

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Q. What is the number of moles in 180 grams of glucose (C6H12O6)?

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

Solution

Molar mass of C6H12O6 = 6*12 + 12*1 + 6*16 = 180 g/mol. Number of moles = mass/molar mass = 180 g / 180 g/mol = 1 mole.

Correct Answer: A — 1

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Q. What is the number of moles in 44 grams of CO2?

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

Solution

Molar mass of CO2 = 12 + 16*2 = 44 g/mol. Number of moles = mass/molar mass = 44 g / 44 g/mol = 1 mole.

Correct Answer: A — 1

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Q. What is the number of moles in 5 liters of a 2 M NaCl solution?

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

Solution

Moles = Molarity x Volume = 2 moles/L x 5 L = 10 moles.

Correct Answer: A — 2.5

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Q. What is the order of the reaction if the rate constant has the unit L/mol·s?

A. Zero order
B. First order
C. Second order
D. Third order

Solution

If the rate constant has the unit L/mol·s, the reaction is second order.

Correct Answer: C — Second order

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Q. What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L atm/(K mol))

A. 4.92 atm
B. 2.46 atm
C. 1.23 atm
D. 0.61 atm

Solution

Osmotic pressure = iCRT = 2 * 0.2 * 0.0821 * 298 = 4.92 atm (i = 2 for NaCl)

Correct Answer: A — 4.92 atm

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Q. What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

A. 4.92 atm
B. 1.64 atm
C. 0.82 atm
D. 2.46 atm

Solution

Osmotic pressure = (n/V)RT = (0.2/1) * 0.0821 * 298 = 4.92 atm.

Correct Answer: A — 4.92 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 L of water at 25 °C? (R = 0.0821 L·atm/(K·mol))

A. 12.3 atm
B. 1.23 atm
C. 0.5 atm
D. 2.5 atm

Solution

Osmotic pressure (π) = nRT/V = (0.5 moles * 0.0821 * 298) / 1 = 12.3 atm.

Correct Answer: B — 1.23 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

A. 12.3 atm
B. 0.5 atm
C. 1.0 atm
D. 2.5 atm

Solution

Osmotic pressure (π) = nRT/V = (0.5 mol)(0.0821 L·atm/(K·mol))(298 K) = 12.3 atm.

Correct Answer: A — 12.3 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 liter of water at 25 °C?

A. 12.3 atm
B. 24.6 atm
C. 6.1 atm
D. 3.1 atm

Solution

Osmotic pressure (π) = iCRT = 1 * 0.5 * 0.0821 * 298 = 12.3 atm.

Correct Answer: A — 12.3 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 L of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))

A. 12.3 atm
B. 10.2 atm
C. 8.2 atm
D. 6.1 atm

Solution

Osmotic pressure (π) = nRT/V = (0.5 moles)(0.0821 L·atm/(K·mol))(298 K) = 12.3 atm.

Correct Answer: A — 12.3 atm

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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 2 liters of solution? (R = 0.0821 L·atm/(K·mol), T = 298 K)

A. 6.14 atm
B. 12.28 atm
C. 3.07 atm
D. 1.54 atm

Solution

Osmotic pressure (π) = (n/V)RT = (0.5/2) * 0.0821 * 298 = 6.14 atm.

Correct Answer: A — 6.14 atm

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Physical Chemistry MCQ & Objective Questions

Physical Chemistry is a crucial branch of chemistry that combines principles of physics and chemistry to explain how matter behaves. Mastering this subject is essential for students preparing for school exams and competitive tests. Practicing MCQs and objective questions in Physical Chemistry not only enhances conceptual clarity but also boosts your chances of scoring better in exams. Engaging with practice questions helps identify important topics and reinforces learning.

What You Will Practise Here

Thermodynamics: Laws, concepts, and applications
Kinetics: Rate laws, reaction mechanisms, and factors affecting reaction rates
Equilibrium: Chemical equilibrium, Le Chatelier's principle, and equilibrium constants
Quantum Chemistry: Basic principles and applications in atomic structure
Electrochemistry: Redox reactions, electrochemical cells, and Nernst equation
Solutions: Colligative properties and their calculations
Acids and Bases: pH calculations, buffer solutions, and titration curves

Exam Relevance

Physical Chemistry is a significant part of the syllabus for CBSE, State Boards, NEET, and JEE. Questions often focus on theoretical concepts, numerical problems, and application-based scenarios. Common patterns include direct questions on laws of thermodynamics, calculations involving reaction rates, and conceptual questions on equilibrium. Familiarity with these patterns can greatly enhance your exam preparation.

Common Mistakes Students Make

Misunderstanding the application of thermodynamic laws in different scenarios
Confusing reaction rates with equilibrium constants
Overlooking the significance of units in calculations
Neglecting to practice numerical problems, leading to poor time management during exams

FAQs

Question: What are the key topics to focus on in Physical Chemistry for exams?
Answer: Focus on thermodynamics, kinetics, equilibrium, and electrochemistry as these are frequently tested.

Question: How can I improve my performance in Physical Chemistry MCQs?
Answer: Regular practice of MCQs and understanding the underlying concepts will significantly improve your performance.

Now is the time to enhance your understanding of Physical Chemistry! Dive into our practice MCQs and test your knowledge to excel in your exams. Your success starts with practice!

Physical Chemistry

Physical Chemistry MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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