IMAT 2024 Paper | Comprehensive Answers and Analysis
amir reza hadian
"Fifth-year Medicine and Surgery student at the University of Padova."
Table of Contents
IMAT 2024 Analysis: Key to IMAT 2025 Success
The IMAT 2024 exam has just finished, and thousands of aspiring medical students are eagerly awaiting their results. As one of the most competitive exams for international medical admissions in Italy, the IMAT tests your knowledge and problem-solving skills in subjects like Biology, Chemistry, Physics, Mathematics, and Logical Reasoning.
In this blog, we focus specifically on analyzing the IMAT 2024 question paper. We provide detailed solutions and step-by-step answers for each question, helping you understand the correct answers and the reasoning behind them.
diagram of Difficulty Levels in IMAT 2024 biology Past Paper
This report provides an analysis of the biology questions from the IMAT 2024 exam to determine their levels of difficulty.We categorized the IMAT 2024 biology questions as Easy, Medium, or Hard based on five criteria: depth of content knowledge, concept complexity, application of knowledge, problem-solving skills, and language clarity. Easy questions involved basic recall of straightforward concepts with minimal reasoning. Medium questions required understanding and applying concepts with some abstraction and logical reasoning. Hard questions demanded deep comprehension of complex ideas, extensive problem-solving, and careful interpretation of complex wording. This concise classification provided a consistent assessment of each question’s difficulty level.
Difficulty Level Distribution of IMAT 2024 Biology Past Paper Questions
Summary of Difficulty Levels
Easy Questions: 10
(Questions 10, 13, 14, 16, 18, 22, 24, 26, 28, 31)
Medium Questions: 9
(Questions 11, 12, 15, 17, 19, 21, 23, 27, 29)
Hard Questions: 4
(Questions 20, 25, 30, 32)
diagram of Difficulty Levels in IMAT 2024 Chemistry Past Paper
Continuing our analysis of the IMAT 2024 exam, let’s delve into the Chemistry section. Just as we did with the biology questions, we have evaluated the chemistry questions to determine their levels of difficulty. Using the same criteria—depth of content knowledge, concept complexity, application of knowledge, problem-solving skills, and language clarity—we have categorized each question as Easy, Medium, or Hard.
Summary of Difficulty Levels in IMAT 2024 Chemistry Past Paper
Summary of Difficulty Levels
Easy Questions: 1
(Question 34)
Medium Questions: 10
(Questions 33, 35, 37, 38, 39, 42, 43, 44, 45, 47)
Hard Questions: 4
(Questions 36, 40, 41, 46)
Total Questions: 15
diagram of Difficulty Levels in IMAT 2024 past paper math and physics
Continuing our comprehensive review of the IMAT 2024 past papers, we now focus on the Physics and Mathematics section—an integral part of the medical school entrance exam. Using our established criteria—depth of content knowledge, concept complexity, application of knowledge, problem-solving skills, and language clarity—we have categorized each question as Easy, Medium, or Hard.
Summary of Difficulty Levels in IMAT 2024 Physics and Mathematics Past Paper
Easy Questions: 6
(Questions 48, 50, 53, 54, 56, 60)
Medium Questions: 5
(Questions 49, 52, 57, 58, 59)
Hard Questions: 2
(Questions 51, 55)
Total Physics and Mathematics Questions Analyzed: 13
Download the Official IMAT 2024 Question Paper PDF
For your convenience, you can download the official IMAT 2024 question paper in PDF format below. Please note that this PDF contains only the questions. All answers and detailed explanations are provided in this blog to help you thoroughly understand each question sand prepare effectively for IMAT 2025.
Why Analyzing the IMAT 2024 Question Paper Matters
If you’re gearing up for the IMAT 2025 exam, analyzing the IMAT 2024 question paper is a vital step in your preparation. Understanding last year’s exam through our detailed IMAT 2024 solutions can give you valuable insights and a competitive edge. Here’s why diving into the IMAT 2024 question paper is beneficial for your IMAT 2025 journey:
- Understand the imat Exam structure
- “You can identify the official syllabus of the IMAT exam and determine which topics in Chemistry or Biology are high-yield and frequently repeated. By focusing on these areas, you can create a clear study strategy for your preparation for IMAT 2025.”
Reading Skills and Knowledge Acquired During Studies, past paper of imat 2024+solutions
Question 1:
Who is the author of the famous novel “To the Lighthouse”?
A) Virginia Wolf
B) Mary Shelley
C) Jane Austen
D) Emily Dickinson
E) Agatha Christie
Correct Answer: A) Virginia Woolf
Explanation:
“To the Lighthouse” is a renowned novel written by Virginia Woolf, a significant figure in the modernist literary movement of the early 20th century. Published in 1927, the novel is celebrated for its innovative narrative techniques, particularly stream-of-consciousness, and its deep exploration of the inner lives of its characters.
Why This Question Might Be Challenging:
Similar Names: The options include other prominent female authors, which might confuse those who are not well-versed in literature.
- Mary Shelley is known for “Frankenstein.”
- Jane Austen wrote novels like “Pride and Prejudice.”
- Emily Dickinson was a prolific poet.
- Agatha Christie is famous for her detective novels.
Key Concepts:
- Modernist Literature: Understanding the characteristics of modernist literature helps identify authors and their works.
- Author-Work Association: Familiarity with significant literary works and their authors is crucial for such questions.
Tips for Similar Questions:
- Study Major Authors and Works: Create a timeline of authors and their notable works to aid memorization.
- Pay Attention to Details: Watch for misspellings or similar-sounding names that could be misleading.
Question 4:
In which of the following is the verb passive?
- A) The deeds of Aeneas were sung by Virgil.
- B) Many students read Greek tragedies in high school.
- C) In the Gallic Wars, Julius Caesar described in detail his military campaign to conquer Gaul.
- D) In one of his works, Plato associates solid forms to the four elements: octahedron to air, tetrahedron to fire, cube to earth, and icosahedron to water.
- E) In the Iliad, Homer sings the deeds of the Pelide Achilles.
Correct Answer: A) The deeds of Aeneas were sung by Virgil.
Explanation:
Identifying Passive Voice:
- A verb is in the passive voice when the subject of the sentence is acted upon by the verb.
Option A Analysis:
- “The deeds of Aeneas were sung by Virgil.”
- The subject “the deeds of Aeneas” is receiving the action of being sung.
- The verb phrase “were sung” is in the passive voice.
Other Options:
- Option B: “Many students read Greek tragedies…”
- The verb “read” is active; students are performing the action.
- Option C: “Julius Caesar described…”
- Active voice; Julius Caesar is performing the action.
- Option D: “Plato associates solid forms…”
- Active voice.
- Option E: “Homer sings the deeds…”
- Active voice.
- Option B: “Many students read Greek tragedies…”
Why This Question Might Be Challenging:
- Grammar Understanding: Requires knowledge of passive vs. active voice.
- Sentence Structure: Identifying the subject and verb accurately.
Key Concepts:
- Passive Voice Construction: Formed with the verb “to be” + past participle.
- Active vs. Passive Voice: Active voice emphasizes the subject performing the action; passive voice emphasizes the action upon the subject.
Tips for Similar Questions:
- Identify the Subject and Verb: Determine who is performing the action.
- Look for “By” Phrases: Passive sentences often include “by [agent].”
Question 6:
Shelly is one of 1500 participants in a Latin contest. 12% of the participants will receive as a prize either a silver-plated or gold-plated pen. If the number of silver-plated pens is twice the number of gold-plated ones, what is the probability that Shelly will receive a gold-plated one?
A) 4%
B) 33%
C) 8%
D) 67%
E) 6%
Correct Answer: A) 4%
Explanation:
Total Participants: 1500
Total Winners: 12% of 1500
- Total prizes = 0.12 × 1500 = 180 prizes
Distribution of Prizes:
- Let the number of gold-plated pens be x
- Number of silver-plated pens is 2x
Total Pens:
- x (gold) + 2x (silver) = 180
- 3x = 180
- x = 180 / 3
- x = 60 gold-plated pens
- Silver-plated pens = 2 × 60 = 120 silver-plated pens
Probability that Shelly receives a gold-plated pen:
- Number of gold-plated pens / Total participants
- 60 / 1500 = 0.04 = 4%
Why This Question Might Be Challenging:
- Multiple Steps: Requires calculating percentages, setting up and solving equations.
- Probability Understanding: Understanding that the probability is based on total participants.
Key Concepts:
- Percentage Calculations: Finding a percentage of a total number.
- Algebraic Equations: Solving for unknowns in an equation.
- Probability Basics: Calculating probability as the ratio of favorable outcomes to total possible outcomes.
Tips for Similar Questions:
- Break Down the Problem: Tackle it step by step.
- Define Variables Clearly: Assign variables to unknown quantities.
- Check Work: Verify calculations at each step.
Question 7:
Two consecutive discounts of 10% and 20% are equal to a single discount of:
A) 28%
B) 25%
C) 30%
D) 18%
E) 15%
Correct Answer: A) 28%
Explanation:
Understanding Consecutive Discounts:
- Two discounts are applied one after the other, not added directly.
Calculation:
- Let the original price be P.
- First Discount of 10%:
- Price after first discount: P × (1 – 0.10) = 0.90P
Second Discount of 20% on New Price:
- Price after second discount: 0.90P × (1 – 0.20) = 0.90P × 0.80 = 0.72P
Total Discount:
- Final price is 0.72P, so total discount is P – 0.72P = 0.28P
Percentage Discount:
- (Total discount / Original price) × 100 = (0.28P / P) × 100 = 28%
Why This Question Might Be Challenging:
- Misconception: One might incorrectly add the discounts directly (10% + 20% = 30%).
- Sequential Calculation: Requires understanding that discounts apply to the reduced price, not the original.
Key Concepts:
- Percentage Calculations: Applying percentages sequentially.
- Compound Discounts: Recognizing that sequential discounts compound multiplicatively.
Tips for Similar Questions:
- Don’t Add Discounts Directly: Always calculate each discount step by step.
- Use a Base Value: Assuming an initial price (like P or $100) simplifies calculations.
Question 8:
Stacie builds a cube using 343 blocks of wood. She decides to paint the cube green. How many of the wooden blocks will have at least one side painted green?
A) 218
B) 125
C) 245
D) 238
E) 105
Correct Answer: A) 218
Explanation:
Understanding the Cube:
- 343 blocks mean the cube is composed of 7 × 7 × 7 blocks because 7³ = 343.
Blocks with At Least One Side Painted:
- These are the blocks on the outer surface of the cube.
Total Blocks in the Cube: 7 × 7 × 7 = 343
Blocks Not Painted (Inner Blocks):
- Inner cube of size 5 × 5 × 5 (since outer layers are removed):
- 5 × 5 × 5 = 125 blocks (these are not painted).
Blocks Painted:
- Total blocks – Inner blocks = 343 – 125 = 218 blocks painted
Why This Question Might Be Challenging:
- Spatial Visualization: Visualizing a 3D cube and its layers.
- Counting Blocks: Calculating the number of blocks on the surface versus the interior.
Key Concepts:
- Volume of a Cube: Understanding cube dimensions and volume.
- Surface Layer Calculation: Recognizing that removing the inner cube gives the number of outer blocks.
Tips for Similar Questions:
- Break Down the Cube: Consider layers and count systematically.
- Use Formulas: For cubes, remember that the number of surface blocks is total blocks minus inner blocks.
Biology questions past paper of imat 2024+solutions
Question 10:
Identify the cellular process that occurs within the mitochondria.
Answer: Cellular respiration
Explanation:
- Mitochondria are known as the “powerhouses” of the cell.
- They are the site of cellular respiration, specifically the Krebs cycle and oxidative phosphorylation.
- These processes generate ATP by breaking down glucose and other nutrients.
Other Options Explained:
- Glycolysis: Occurs in the cytoplasm.
- Photosynthesis: Occurs in chloroplasts (in plants).
- Methylation of sugars: Typically occurs in the cytoplasm or nucleus.
- Formation of microbodies: Microbodies like peroxisomes are formed in the cytoplasm.
Tips for Future imat Candidates:
- Associate key cellular processes with their respective organelles.
- Understand the steps of cellular respiration and where each occurs.
Question 11:
Explain what constitutes a hydrogen bond.
Answer: An intermolecular attraction between a hydrogen atom covalently bonded to a highly electronegative atom (like oxygen or nitrogen) and another electronegative atom in a nearby molecule.
Explanation:
- Hydrogen bonds are a type of weak chemical bond.
- They are crucial for the properties of water and the structure of DNA and proteins.
- Occur when hydrogen is bonded to N, O, or F and is attracted to a lone pair on an electronegative atom in another molecule.
Common Misconceptions:
- Hydrogen bonds are not covalent bonds within a molecule.
- They are different from ionic bonds and van der Waals forces.
Tips for Future imat Candidates:
- Learn the differences between types of chemical bonds and intermolecular forces.
- Recognize the significance of hydrogen bonding in biological systems.
Question 12:
Determine where the reactions of the Krebs cycle take place in eukaryotic cells.
Answer: In the mitochondrial matrix
Explanation:
- The Krebs cycle, also known as the citric acid cycle, occurs in the mitochondrial matrix.
- It plays a key role in aerobic respiration, generating electron carriers for ATP production.
Other Locations Explained:
- Inner mitochondrial membrane: Site of the electron transport chain.
- Cytoplasm: Location of glycolysis.
- Large ribosomal subunit and plasma membrane: Not involved in the Krebs cycle.
Tips for Future Candidates:
- Memorize the locations of major metabolic pathways.
- Understand the structure and function of cellular organelles.
Question 13:
Classify glucose based on the number of carbon atoms it contains.
Answer: Hexose
Explanation:
- Glucose is a monosaccharide with six carbon atoms, making it a hexose (hex- meaning six).
- The formula for glucose is C₆H₁₂O₆.
Other Classifications:
- Pentose: Five carbons (e.g., ribose).
- Triose: Three carbons (e.g., glyceraldehyde).
- Tetrose: Four carbons.
- Nonose: Nine carbons.
Tips for Future Candidates:
- Learn the prefixes used in carbohydrate nomenclature.
- Associate the structure of sugars with their classification.
Question 14:
Identify the pentose sugar present in RNA nucleotides.
Answer: Ribose
Explanation:
- RNA (Ribonucleic acid) contains the sugar ribose, a five-carbon (pentose) sugar.
- Ribose has a hydroxyl group (-OH) on the 2′ carbon, distinguishing it from deoxyribose in DNA.
Importance in Biology:
- The presence of ribose makes RNA more reactive and less stable than DNA.
- Ribose is essential for the structure and function of RNA molecules.
Tips for Future Candidates:
- Remember that the “R” in RNA stands for ribose.
- Understand the structural differences between RNA and DNA.
Question 15:
Explain the role of carrier proteins in cellular membranes.
Answer: Carrier proteins are membrane proteins that facilitate the transport of specific molecules or ions across the cell membrane by binding to them and undergoing conformational changes.
Explanation:
Function of Carrier Proteins:
- Selective Transport: Carrier proteins are specific to the molecules they transport, such as glucose, amino acids, or ions.
Mechanism:
- Facilitated Diffusion: For molecules moving down their concentration gradient without energy input.
- Active Transport: For molecules moving against their concentration gradient, requiring energy (often from ATP).
Types of Transport:
- Uniporters: Transport one type of molecule in one direction.
- Symporters: Transport two different molecules in the same direction.
- Antiporters: Transport two different molecules in opposite directions.
Other Options Explained:
- Proteins that phosphorylate enzymes: These are kinases, not carrier proteins.
- Proteins that break down phospholipids: These are phospholipases.
- Proteins transporting mRNA or tRNA: Involved in nucleic acid transport, not directly related to carrier proteins in membranes.
Key Points:
- Integral Membrane Proteins: Carrier proteins span the lipid bilayer.
- Conformational Change: Binding of the molecule induces a change in the protein’s shape to facilitate transport.
Tips for Future imat Candidates:
- Understand Membrane Transport Mechanisms: Differentiate between passive and active transport, and between channels and carriers.
- Remember Specific Examples: Glucose transporters (GLUT proteins) are classic examples of carrier proteins.
Question 16:
Identify the molecule known as the “energy currency” of the cell.
Answer: Adenosine Triphosphate (ATP)
Explanation:
ATP Function:
- Energy Storage and Transfer: ATP stores energy in its high-energy phosphate bonds.
- Energy Release: When ATP is hydrolyzed to ADP (adenosine diphosphate) and inorganic phosphate, energy is released for cellular processes.
Importance in Cellular Activities:
- Muscle Contraction
- Active Transport
- Biosynthesis of Macromolecules
Other Molecules Explained:
- FADH₂ and NADH: Electron carriers used in cellular respiration but not the primary energy currency.
- Creatine: Involved in energy storage in muscle cells but not the main energy currency.
- NADPH: Electron carrier used in anabolic reactions, such as fatty acid synthesis.
Key Points:
- Structure of ATP: Consists of adenine, ribose sugar, and three phosphate groups.
- Regeneration of ATP: ATP is regenerated from ADP through cellular respiration.
Tips for Future Candidates:
- Memorize Key Molecules: Know the roles of ATP, NADH, FADH₂, and NADPH.
- Understand Energy Transfer: Grasp how ATP provides energy for cellular processes.
Question 17:
Classify the type of reaction represented by the hydrolysis of ATP.
Answer: Exergonic Reaction
Explanation:
Exergonic Reaction:
- Definition: A reaction that releases energy to the surroundings.
- ATP Hydrolysis: ATP → ADP + Pi + Energy
Why Exergonic:
- The breakdown of ATP releases energy that can be used to drive endergonic (energy-requiring) reactions in the cell.
Other Types of Reactions Explained:
- Endergonic: Absorb energy; not applicable as ATP hydrolysis releases energy.
- Condensation: Formation of a bond with the release of water; ATP hydrolysis is the opposite.
- Oxidation-Reduction (Redox): Involves transfer of electrons; ATP hydrolysis does not involve electron transfer.
- Lipolysis: Breakdown of lipids; not related to ATP hydrolysis.
Key Points:
- Coupled Reactions: Cells often couple ATP hydrolysis with endergonic reactions to make them energetically favorable.
- Energy Currency Concept: ATP hydrolysis provides the energy needed for various cellular activities.
Tips for Future imat Candidates:
- Understand Reaction Types: Be able to classify reactions as exergonic or endergonic.
- Recognize the Role of ATP: Know how ATP hydrolysis drives cellular processes.
Question 18:
Identify the type of organisms characterized by having intracellular compartmentalization.
Answer: Eukaryotes
Explanation:
Eukaryotic Cells:
- Definition: Cells with a true nucleus and membrane-bound organelles.
- Compartmentalization: Organelles such as the nucleus, mitochondria, endoplasmic reticulum, and Golgi apparatus create specialized environments for specific functions.
Importance of Compartmentalization:
- Efficiency: Allows for separation of incompatible reactions.
- Regulation: Enables precise control over metabolic pathways.
Other Organisms Explained:
- Prokaryotes (Bacteria): Lack membrane-bound organelles; no compartmentalization.
- Viruses: Not considered cells; lack cellular structures.
- Algae: Eukaryotic organisms; some are unicellular with compartmentalization.
Key Points:
- Eukaryotic vs. Prokaryotic Cells: Main difference is the presence of membrane-bound organelles.
- Examples of Eukaryotes: Animals, plants, fungi, and protists.
Tips for Future imat Candidates:
- Memorize Cell Structures: Know the organelles unique to eukaryotes.
- Understand Cell Classification: Be able to distinguish between prokaryotic and eukaryotic cells.
Question 19:
Determine which intracellular structure is composed of microtubules.
Answer: The Centriole
Explanation:
Centriole Structure:
- Composition: Made up of microtubules arranged in a specific pattern (typically nine triplets).
- Function: Play a role in cell division by organizing the mitotic spindle.
Microtubules:
- Components: Tubulin proteins assembled into hollow tubes.
- Functions: Maintain cell shape, enable intracellular transport, and facilitate chromosome separation during mitosis.
Other Structures Explained:
- Nucleus: Contains DNA; not composed of microtubules.
- Golgi Apparatus: Involved in protein modification and packaging; made of cisternae, not microtubules.
- Nucleolus: Site of ribosomal RNA synthesis; not composed of microtubules.
- Endoplasmic Reticulum: Network of membranes; not made of microtubules.
Key Points:
- Cytoskeleton Components:
- Microtubules
- Microfilaments
- Intermediate Filaments
- Role of Centrioles in Cells:
- Essential for the formation of cilia and flagella.
- Organize microtubule assembly.
Tips for Future Candidates:
- Learn Organelle Functions: Know which organelles are composed of which cytoskeletal elements.
- Understand Microtubule Roles: Recognize the importance of microtubules in cell division and structure.
Question 20:
Describe the membrane structure of mitochondria.
Answer: Mitochondria have an outer membrane and a highly selective inner membrane.
Explanation:
Mitochondrial Membranes:
- Outer Membrane: Smooth and permeable to small molecules.
- Inner Membrane: Folded into cristae; highly selective due to specific transport proteins.
Function of Membranes:
- Compartmentalization: Creates distinct environments for metabolic processes.
- Inner Membrane: Site of the electron transport chain and ATP synthesis.
Other Options Explained:
- Intermediate Membrane: Refers to the intermembrane space between the outer and inner membranes; not a separate membrane.
- Phospholipid Monolayer: Mitochondrial membranes are phospholipid bilayers.
- No Proteins Present: Inner membrane contains numerous proteins essential for oxidative phosphorylation.
Key Points:
- Double-Membrane Structure: Characteristic of mitochondria.
- Selective Permeability: Inner membrane’s selectivity is crucial for maintaining the proton gradient.
Tips for Future imat Candidates:
- Understand Organelle Structures: Be familiar with the unique features of mitochondria.
- Remember Functions: Link structural features to their functional significance.
Question 21:
Define what an anticodon is in the context of protein synthesis.
Answer: An anticodon is a sequence of three nucleotides on a transfer RNA (tRNA) molecule that is complementary to a specific codon on messenger RNA (mRNA).
Explanation:
Role in Protein Synthesis:
- During translation, tRNA molecules bring amino acids to the ribosome.
- Each tRNA has an anticodon that pairs with a complementary codon on the mRNA strand.
Complementary Base Pairing:
- Follows base-pairing rules (A-U and G-C in RNA).
- Ensures that the correct amino acid is added to the growing polypeptide chain.
Process Overview:
- mRNA Codon: A sequence of three nucleotides on mRNA encoding a specific amino acid.
- tRNA Anticodon: The corresponding three-nucleotide sequence on tRNA that pairs with the mRNA codon.
Other Options Explained:
- DNA Coding for Amino Acids: Codons are found on mRNA, not DNA directly during translation.
- rRNA and Amino Acids: rRNA forms part of the ribosome but does not carry anticodons.
- Terminal Triplets and rRNA: Anticodons are not part of rRNA or amino acids directly.
- mRNA Nucleotides Corresponding to DNA Codons: This refers to codons, not anticodons.
Key Points:
- Translation Specificity: Anticodon-codon pairing ensures the correct amino acid sequence.
- Genetic Code Universality: The codon-anticodon interactions are consistent across organisms.
Tips for Future imat Candidates:
- Memorize Codon-Anticodon Relationships: Understand how tRNA anticodons correspond to mRNA codons.
- Understand Translation Mechanics: Know the roles of mRNA, tRNA, rRNA, and ribosomes.
Question 22:
Identify the components that make up ribosomes.
Answer: Ribosomes are composed of ribosomal RNA (rRNA) and proteins.
Explanation:
Structure of Ribosomes:
- Consist of two subunits (large and small).
- Each subunit contains rRNA molecules and ribosomal proteins.
Function:
- Sites of protein synthesis (translation) in the cell.
- Facilitate the assembly of amino acids into polypeptide chains.
Location:
- Found in the cytoplasm and on the rough endoplasmic reticulum.
Other Options Explained:
- DNA and Proteins: DNA is not a component of ribosomes.
- DNA and Lipids: Ribosomes do not contain lipids or DNA.
- RNA and DNA: While RNA is present, DNA is not a component of ribosomes.
- RNA, DNA, and Proteins: Ribosomes do not contain DNA.
Key Points:
- Ribosomal RNA (rRNA): Catalytic and structural roles in ribosomes.
- Protein Components: Contribute to ribosome structure and function.
Tips for Future imat Candidates:
- Distinguish Between RNA Types: Understand the different roles of mRNA, tRNA, and rRNA.
- Know Organelle Composition: Be familiar with what makes up key cellular structures.
Question 23:
Describe the composition of the cell (plasma) membrane.
Answer: The plasma membrane consists of a phospholipid bilayer with hydrophobic tails facing inward and hydrophilic heads facing outward, embedded with integral and peripheral proteins.
Explanation:
Phospholipid Bilayer:
- Hydrophilic Heads: Phosphate groups facing the aqueous exterior and interior.
- Hydrophobic Tails: Fatty acid chains facing inward, away from water.
Membrane Proteins:
- Integral Proteins: Span the membrane; involved in transport and signaling.
- Peripheral Proteins: Attached to the membrane surface; play roles in signaling and structural support.
Fluid Mosaic Model:
- Describes the membrane as a dynamic and flexible structure.
Additional Components:
- Cholesterol: Provides membrane stability and fluidity.
- Glycoproteins and Glycolipids: Involved in cell recognition and signaling.
Other Options Explained:
- Cholesterol Enclosing a Protein Layer: Incorrect representation.
- Double Layer of Triglycerides and Cholesterol: Membrane is made of phospholipids, not triglycerides.
- Glycoprotein Layer: Glycoproteins are part of the membrane but not the main structural component.
- Layer of Fatty Acids and Globular Proteins: Oversimplification; doesn’t accurately describe the bilayer structure.
Key Points:
- Selective Permeability: Allows selective movement of substances in and out of the cell.
- Membrane Fluidity: Essential for membrane function and cellular processes.
Tips for Future imat Candidates:
- Memorize Membrane Structure: Understand the arrangement and roles of each component.
- Visual Aids: Use diagrams to visualize the fluid mosaic model.
Question 24:
Define translation in the context of protein synthesis.
Answer: Translation is the process by which the genetic code carried by messenger RNA (mRNA) is decoded by ribosomes to synthesize a specific sequence of amino acids, forming a polypeptide chain.
Explanation:
Process Overview:
- Initiation: Ribosome assembles around the start codon on mRNA.
- Elongation: tRNA molecules bring amino acids to the ribosome, matching their anticodons with codons on the mRNA.
- Termination: Process ends when a stop codon is reached, releasing the polypeptide chain.
Key Components:
- mRNA: Carries the genetic code from DNA.
- tRNA: Transfers specific amino acids to the ribosome.
- Ribosomes: Facilitate the assembly of amino acids.
Other Options Explained:
- Transcribing mRNA into DNA: Describes reverse transcription, not translation.
- Recognition of rRNA by Amino Acids: Not accurate; amino acids are brought by tRNA.
- DNA to mRNA Production: Describes transcription, not translation.
- Pairing Between DNA Codons and tRNA Anticodons: Occurs indirectly; mRNA codons pair with tRNA anticodons during translation.
Key Points:
- Central Dogma: DNA → RNA → Protein.
- Genetic Code: Universal code that translates nucleotide sequences into amino acids.
Tips for Future Candidates:
- Differentiate Between Processes: Understand the differences between replication, transcription, and translation.
- Study Mechanisms: Familiarize yourself with the steps and components involved in translation.
Question 25:
Identify the main components of the cytoskeleton.
Answer: The cytoskeleton is primarily composed of microtubules, microfilaments, and intermediate filaments.
Explanation:
Microtubules:
- Structure: Hollow tubes made of tubulin.
- Functions: Cell shape, organelle movement, chromosome separation during mitosis.
Microfilaments:
- Structure: Thin strands of actin.
- Functions: Muscle contraction, cell movement, cytokinesis.
Intermediate Filaments:
- Structure: Fibrous proteins (e.g., keratin).
- Functions: Structural support, maintenance of cell shape.
Other Options Explained:
- Myosin and Filamin: Associated with microfilaments but not primary components.
- Dynein: Motor protein that moves along microtubules, not a structural component.
- Actin, Myosin, Dynein: Actin is a microfilament; myosin and dynein are motor proteins.
- Collagen and Reticular Fibers: Extracellular matrix components, not part of the cytoskeleton.
Key Points:
- Dynamic Structure: Cytoskeleton is involved in both maintaining cell shape and enabling movement.
- Motor Proteins: Myosin, kinesin, and dynein move along cytoskeletal elements to transport materials.
Tips for Future imat Candidates:
- Learn Functions: Understand what each component does within the cell.
- Visual Learning: Use diagrams to see how cytoskeletal elements interact.
Question 26:
Define the term “allele” in genetics.
Answer: An allele is one of two or more alternative forms of a gene that arise by mutation and are found at the same place on a chromosome.
Explanation:
Genetic Variation:
- Alleles contribute to the diversity of traits within a population.
Homozygous and Heterozygous:
- Homozygous: Two identical alleles at a gene locus.
- Heterozygous: Two different alleles at a gene locus.
Expression:
- Dominant Allele: Expressed in the phenotype even if only one copy is present.
- Recessive Allele: Expressed only when two copies are present.
Other Options Explained:
- Coding DNA Base for an Amino Acid: Refers to codons, not alleles.
- Hereditary Trait in Haploid Cells: Alleles are present in both haploid and diploid cells.
- Phenotypic Manifestation: Phenotype results from allele expression but is not the allele itself.
- Set of DNA Triplets: Alleles encompass entire genes, not just codons.
Key Points:
- Gene Locus: Specific location of a gene on a chromosome.
- Genotype vs. Phenotype: Alleles contribute to genotype; phenotype is the observable trait.
Tips for Future Candidates:
- Understand Mendelian Genetics: Study how alleles interact to determine traits.
- Use Punnett Squares: Visualize allele combinations and predict offspring ratios.
Question 27:
Determine when an allele can express itself in a heterozygous condition.
Answer: When the allele is dominant.
Explanation:
Dominant Alleles:
- Expressed in the phenotype even if only one copy is present (heterozygous condition).
Recessive Alleles:
- Expressed only when two copies are present (homozygous recessive).
Heterozygous Genotype:
- Consists of one dominant and one recessive allele.
- Phenotype reflects the dominant allele.
Other Options Explained:
- Recessive Allele: Not expressed in heterozygous condition.
- Mutated Allele: May or may not be expressed; depends on dominance and effect.
- Multiple Allele: Refers to genes with more than two allele forms; doesn’t determine expression in heterozygosity.
- Associated Allele: Non-specific term; doesn’t address expression.
Key Points:
- Mendel’s Laws: Law of Dominance explains expression in heterozygotes.
- Examples:
- Widow’s Peak Hairline: Dominant trait.
- Cystic Fibrosis: Recessive trait; manifests only in homozygous recessives.
Tips for Future imat Candidates:
- Memorize Dominant and Recessive Traits: Know examples to illustrate concepts.
- Practice Genetics Problems: Use inheritance patterns to predict phenotypes.
Question 28:
Explain what mutations are in genetic terms.
Answer: Mutations are alterations or changes in the genetic material (DNA) of a cell.
Explanation:
Types of Mutations:
- Point Mutations: Changes in a single nucleotide base.
- Insertions/Deletions: Addition or loss of nucleotide bases.
- Chromosomal Mutations: Changes in chromosome structure or number.
Causes:
- Spontaneous Errors: During DNA replication.
- Environmental Factors: UV radiation, chemicals.
Effects:
- Neutral: No effect on phenotype.
- Beneficial: Provide advantages; basis for evolution.
- Harmful: Cause diseases or disorders.
Other Options Explained:
- Alterations in Energy Metabolism: Relate to biochemical changes, not genetic mutations.
- Enzyme Functionality in Zygote Formation: May result from mutations but is not the definition.
- Active Transport Mechanism Alterations: Physiological changes, not directly mutations.
- Cell Division Mechanism Alterations: May be caused by mutations but not the definition.
Key Points:
- Genetic Variation Source: Mutations introduce new genetic material.
- DNA Repair Mechanisms: Cells have systems to correct mutations.
Tips for Future imat Candidates:
- Understand Mutation Consequences: Link mutations to genetic disorders and evolution.
- Learn DNA Repair Pathways: Know how cells maintain genetic integrity.
Question 29:
Describe the process of translation in cells.
Answer: Translation is the process that synthesizes polypeptide chains (proteins) from messenger RNA (mRNA) templates.
Explanation:
Process Details:
- Initiation: Ribosome binds to mRNA at the start codon (AUG).
- Elongation: tRNAs bring amino acids corresponding to each codon; peptide bonds form between amino acids.
- Termination: Process ends when a stop codon is reached; the polypeptide chain is released.
Key Molecules:
- mRNA: Provides the template with codons.
- tRNA: Carries amino acids; anticodon matches mRNA codon.
- Ribosomes: Catalyze peptide bond formation.
Location:
- Prokaryotes: Occurs in the cytoplasm.
- Eukaryotes: Also occurs in the cytoplasm or on rough endoplasmic reticulum.
Other Options Explained:
- Occurs in Nucleus: Translation occurs in the cytoplasm; transcription is in the nucleus.
- Synthesis of RNA from DNA: Describes transcription, not translation.
- Similar to Transcription: Different processes; transcription synthesizes RNA, translation synthesizes proteins.
- Exclusive to Eukaryotes: Translation occurs in both prokaryotes and eukaryotes.
Key Points:
- Central Dogma of Molecular Biology: DNA → RNA → Protein.
- Genetic Code: Codons specify amino acids.
Tips for Future imat Candidates:
- Know the Steps: Memorize the stages of translation.
- Understand Codon-Anticodon Interaction: Essential for accurate protein synthesis.
Question 30:
Given a DNA strand sequence, determine the sequence on the complementary strand.
Given DNA Strand: C C G T T A T T G A
Answer: G G C A A T A A C T
Explanation:
Base-Pairing Rules in DNA:
- C pairs with G
- A pairs with T
Complementary Strand Construction:
- Original Strand: C C G T T A T T G A
- Complementary Strand:
- C → G
- C → G
- G → C
- T → A
- T → A
- A → T
- T → A
- T → A
- G → C
- A → T
- Complementary Sequence: G G C A A T A A C T
Other Options Explained:
- Incorrect Pairings: Options with mismatched bases do not follow base-pairing rules.
- RNA Bases: Uracil (U) is not present in DNA.
Key Points:
- DNA Complementarity: Essential for replication and transcription.
- Antiparallel Strands: DNA strands run in opposite directions (5′ to 3′ and 3′ to 5′).
Tips for Future imat Candidates:
- Practice Base Pairing: Work through examples to reinforce understanding.
- Remember Directionality: Pay attention to the 5′ and 3′ ends when writing sequences.
Question 31:
Define the process of replication in molecular biology.
Answer: Replication is the process by which DNA makes an exact copy of itself before cell division, resulting in two identical DNA molecules.
Explanation:
Purpose of Replication:
- Ensures that each daughter cell receives an exact copy of the DNA.
Mechanism:
- Initiation: Unwinding of the DNA helix by helicase.
- Elongation: DNA polymerase adds nucleotides to the growing DNA strand, following base-pairing rules.
- Termination: Replication ends when the entire molecule is copied.
Semiconservative Replication:
- Each new DNA molecule consists of one original strand and one newly synthesized strand.
Other Options Explained:
- DNA as Template for RNA: Describes transcription.
- Daughter Cells Formation: Result of cell division, not replication itself.
- RNA as Template for Proteins: Describes translation.
- RNA as Template for RNA: Occurs in some viruses; not typical replication in cells.
Key Points:
- Enzymes Involved:
- Helicase: Unwinds DNA.
- DNA Polymerase: Synthesizes new DNA strands.
- Ligase: Joins Okazaki fragments on the lagging strand.
- Direction of Synthesis: 5′ to 3′ direction.
Tips for Future imat Candidates:
- Memorize Enzymes and Functions: Know the key players in DNA replication.
- Understand Leading and Lagging Strands: Differentiate between continuous and discontinuous synthesis.
Question 32:
Explain what a prokaryotic operon is.
Answer: An operon in prokaryotes is a functional unit of DNA consisting of a group of adjacent genes that are regulated together and transcribed as a single mRNA molecule, along with regulatory DNA sequences that control their expression.
Explanation:
Components of an Operon:
- Structural Genes: Code for proteins with related functions.
- Promoter: Binding site for RNA polymerase.
- Operator: Binding site for repressor proteins.
- Regulatory Genes: Code for repressor or activator proteins.
Example:
- Lac Operon: Controls lactose metabolism in E. coli.
Regulation Mechanism:
- Inducible Operons: Activated by the presence of a substrate.
- Repressible Operons: Deactivated by the product of the pathway.
Other Options Explained:
- Independent Genes: Operons consist of co-regulated genes.
- Protein Complex Catalyzing Protein Synthesis: Describes ribosomes, not operons.
- RNA Complex in DNA Replication: Not related to operons.
- DNA Sequence Without Regulatory Function: Operons include regulatory sequences.
Key Points:
- Gene Regulation Efficiency: Allows bacteria to adapt quickly to environmental changes.
- Polycistronic mRNA: Single mRNA molecule codes for multiple proteins.
Tips for Future imat Candidates:
- Study Operon Models: Understand the lac and trp operons as classic examples.
- Know Regulatory Mechanisms: Learn how repressors and inducers affect gene expression.
Chemistry past paper of imat 2024+solutions
Question 33:
Calculate the partial pressure exerted by nitrogen gas in a mixture containing 0.3 moles of nitrogen (N₂), 0.5 moles of carbon dioxide (CO₂), and 0.4 moles of oxygen (O₂), if the total pressure of the gas mixture is 2.4 atmospheres (atm).
Answer: 0.6 atm
Explanation:
Dalton’s Law of Partial Pressures:
The total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas.
P_total = P_N₂ + P_CO₂ + P_O₂
Calculating Mole Fractions:
Total moles of gas:
- n_total = n_N₂ + n_CO₂ + n_O₂ = 0.3 mol + 0.5 mol + 0.4 mol = 1.2 mol
Mole fraction of N₂ (χ_N₂):
- χ_N₂ = n_N₂ / n_total = 0.3 mol / 1.2 mol = 0.25
Calculating Partial Pressure of N₂:
P_N₂ = χ_N₂ × P_total
P_N₂ = 0.25 × 2.4 atm = 0.6 atm
Key Points:
Dalton’s Law: Useful for calculating partial pressures in gas mixtures.
Mole Fraction: Represents the proportion of a component in a mixture.
Tips for Future Candidates:
Remember Gas Laws: Be familiar with Dalton’s Law and how to apply it.
Use Proper Units: Ensure consistency in units when performing calculations.
Question 34:
A gas confined in a rigid container at a temperature of -3°C exerts a pressure of 9 atm. Calculate the pressure the gas will exert if heated to 27°C, assuming the volume remains constant.
Answer: 10 atm
Explanation:
Gay-Lussac’s Law (Pressure-Temperature Law):
For a fixed amount of gas at constant volume, the pressure is directly proportional to its absolute temperature (in Kelvin).
P₁ / T₁ = P₂ / T₂
Convert Temperatures to Kelvin:
Initial temperature (T₁):
- T₁ = -3°C + 273.15 = 270.15 K
Final temperature (T₂):
- T₂ = 27°C + 273.15 = 300.15 K
Calculate Final Pressure (P₂):
Rearranged formula: P₂ = P₁ × (T₂ / T₁)
P₂ = 9 atm × (300.15 K / 270.15 K)
P₂ ≈ 9 atm × (1.111) ≈ 10 atm
Key Points:
Temperature in Kelvin: Always convert Celsius to Kelvin for gas law calculations.
Direct Proportionality: Pressure increases with temperature at constant volume.
Tips for Future Candidates:
Memorize Gas Laws: Understand when to apply each gas law.
Double-Check Calculations: Ensure accuracy, especially when converting units.
Question 35:
Identify which compound will form a hydroxide when it reacts with water.
Answer: BaO (Barium oxide)
Explanation:
Reaction with Water:
- BaO + H₂O → Ba(OH)₂
- Barium oxide reacts with water to form barium hydroxide, a strong base.
Understanding Compound Types:
- Metal Oxides (Basic Oxides): React with water to form metal hydroxides (bases).
- Non-Metal Oxides (Acidic Oxides): React with water to form acids.
Other Options Explained:
Cl₂O (Chlorine monoxide): Forms an acid (hypochlorous acid) when reacting with water.
- Cl₂O + H₂O → 2 HClO
SO₃ (Sulfur trioxide): Forms sulfuric acid.
- SO₃ + H₂O → H₂SO₄
SiO₂ (Silicon dioxide): Does not react with water under normal conditions.
N₂O₃ (Dinitrogen trioxide): Forms nitrous acid.
- N₂O₃ + H₂O → 2 HNO₂
Key Points:
Metal Oxides Form Bases: React with water to produce hydroxides.
Non-Metal Oxides Form Acids: React with water to produce acids.
Tips for Future Candidates:
Classify Oxides: Know the difference between basic and acidic oxides.
Memorize Common Reactions: Familiarity with reactions of oxides helps in quick recall.
Question 36:
Based on the reaction: 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂, determine the correct stoichiometric interpretation.
Answer: From 2 moles of FeS₂ and 5.5 moles of O₂, 1 mole of Fe₂O₃ can be obtained.
Explanation:
Stoichiometry of the Reaction:
Molar Ratios:
FeS₂ : O₂ : Fe₂O₃ : SO₂
4 : 11 : 2 : 8
Calculating Proportional Amounts:
Dividing the Entire Equation by 2:
- 2 FeS₂ + 5.5 O₂ → 1 Fe₂O₃ + 4 SO₂
Interpretation:
- From 2 moles of FeS₂ and 5.5 moles of O₂, you can produce 1 mole of Fe₂O₃.
Other Options Explained:
- The correct stoichiometric relationships can be determined by adjusting the coefficients accordingly.
Key Points:
Stoichiometry: Allows for calculation of reactants and products in chemical reactions.
Mole Ratios: Essential for interpreting balanced chemical equations.
Tips for Future imat Candidates:
Practice Balancing Equations: Ensure you can balance and interpret equations accurately.
Use Proportions: Scale equations up or down to match given amounts.
Question 37:
Determine how many milliliters (mL) of water must be added to 15 mL of a 0.25 M hydrochloric acid (HCl) solution to obtain a 0.05 M solution.
Answer: 60 mL
Explanation:
Use the Dilution Formula:
C₁V₁ = C₂V₂
C₁ = Initial concentration = 0.25 M
V₁ = Initial volume = 15 mL
C₂ = Final concentration = 0.05 M
V₂ = Final volume = ?
Calculating Final Volume (V₂):
V₂ = (C₁V₁) / C₂
V₂ = (0.25 M × 15 mL) / 0.05 M
V₂ = (3.75) / 0.05
V₂ = 75 mL
Calculating Volume of Water to Add:
- Volume of water to add = V₂ – V₁ = 75 mL – 15 mL = 60 mL
Key Points:
Dilution Principle: The amount of solute remains the same before and after dilution.
Volume Addition: The difference between final and initial volumes gives the volume of solvent added.
Tips for Future imat Candidates:
Memorize the Dilution Formula: C₁V₁ = C₂V₂ is crucial for dilution problems.
Careful with Units: Ensure volume units are consistent.
Question 38:
Calculate the number of moles of sodium ions (Na⁺) present in 250 mL of a 1.2 M sodium sulfate (Na₂SO₄) solution.
Answer: 0.6 moles of Na⁺ ions
Explanation:
Calculate Moles of Na₂SO₄:
Moles of Na₂SO₄ = Molarity × Volume (in liters)
Volume = 250 mL = 0.25 L
Moles of Na₂SO₄ = 1.2 mol/L × 0.25 L = 0.3 moles
Determine Moles of Na⁺ Ions:
Each Na₂SO₄ molecule produces 2 Na⁺ ions:
- Na₂SO₄ → 2 Na⁺ + SO₄²⁻
Total moles of Na⁺ = Moles of Na₂SO₄ × 2
Total moles of Na⁺ = 0.3 moles × 2 = 0.6 moles
Key Points:
Understanding Dissociation: Know how compounds dissociate in solution.
Stoichiometry in Solutions: Use molar ratios from the dissociation equation.
Tips for Future imat Candidates:
Write Dissociation Equations: Helps visualize ion production.
Convert Units Properly: Ensure volumes are in liters for molarity calculations.
Question 39:
In the reaction NH₃ + BF₃ ⇌ NH₃BF₃, identify how ammonia (NH₃) behaves according to Lewis theory.
Answer: As a Lewis base
Explanation:
Lewis Acid-Base Theory:
- Lewis Acid: Electron pair acceptor.
- Lewis Base: Electron pair donor.
Behavior of NH₃:
- Ammonia (NH₃) has a lone pair of electrons on nitrogen, which it donates to form a coordinate covalent bond with BF₃.
Behavior of BF₃:
- Boron trifluoride (BF₃) is electron-deficient and can accept a lone pair.
Conclusion:
NH₃ donates an electron pair ⇒ Lewis base
BF₃ accepts an electron pair ⇒ Lewis acid
Key Points:
Lewis Definitions: Focus on the transfer of electron pairs.
Coordinate Covalent Bond: Formed when one atom donates both electrons in a bond.
Tips for Future imat Candidates:
Differentiate Acid-Base Theories: Know the definitions according to Arrhenius, Brønsted-Lowry, and Lewis.
Identify Electron Pairs: Determine which species donate or accept electrons.
Question 40:
In the reaction where metallic zinc reacts with nitric acid to produce zinc nitrate, nitrogen dioxide, and water, identify the reducing agent.
Answer: Zn(s) (Metallic zinc)
Explanation:
Oxidation-Reduction (Redox) Reaction:
Reducing Agent: Species that loses electrons (is oxidized).
Oxidizing Agent: Species that gains electrons (is reduced).
Reaction Details:
Zinc (Zn) is oxidized:
- Zn⁰ → Zn²⁺ + 2 e⁻ (loss of electrons)
Nitric Acid (HNO₃) is reduced:
- N⁺⁵ in NO₃⁻ is reduced to N⁺⁴ in NO₂ (gain of electrons)
Conclusion:
- Zn(s) acts as the reducing agent because it donates electrons.
Key Points:
Identifying Oxidation States: Helps determine which species are oxidized or reduced.
Redox Roles:
Reducing Agent: Undergoes oxidation.
Oxidizing Agent: Undergoes reduction.
Tips for Future imat Candidates:
Assign Oxidation Numbers: Practice to quickly identify redox participants.
Remember LEO GER: Loss of Electrons is Oxidation, Gain of Electrons is Reduction.
Question 41:
Determine which compound contains the highest number of hydrogen atoms.
Answer: 2,3-Dimethylpentane
Explanation:
Structural Analysis:
2,3-Dimethylpentane:
Base chain: Pentane (C₅H₁₂)
Two methyl groups added at carbons 2 and 3.
Total carbons: 5 + 2 = 7
Saturated hydrocarbon (alkane): Maximum hydrogen atoms.
Hydrogen atoms: C₇H₁₆
Other Compounds:
Cyclohexane (C₆H₁₂):
- Fewer hydrogens due to ring structure.
1,2-Dimethylcyclobutane:
- Cyclobutane ring limits hydrogen count.
2,3-Dimethyl-2-butene:
- Contains a double bond (alkene), reducing hydrogen count.
2-Hexanol:
- Alcohol group replaces a hydrogen.
Conclusion:
- 2,3-Dimethylpentane has the most hydrogen atoms (16 hydrogens).
Key Points:
Saturation Level: Alkanes have the maximum number of hydrogens.
Functional Groups and Rings: Reduce hydrogen count.
Tips for Future imat Candidates:
Draw Structures: Visual representation aids in counting atoms.
Understand Hydrocarbon Classes: Recognize alkanes, alkenes, cycloalkanes, etc.
Question 42:
Identify which molecule does NOT contain a carbon-oxygen double bond (carbonyl group).
Answer: Dimethyl ether
Explanation:
Analyzing Each Compound:
Dimethyl Ether (CH₃OCH₃):
Contains an oxygen atom bonded to two methyl groups.
Structure: No carbonyl group present.
Acetaldehyde (CH₃CHO):
- Contains a carbonyl group (aldehyde).
Acetone (CH₃COCH₃):
- Contains a carbonyl group (ketone).
Acetic Acid (CH₃COOH):
- Contains a carbonyl group (carboxylic acid).
Methyl Acetate (CH₃COOCH₃):
- Contains a carbonyl group (ester).
Conclusion:
- Dimethyl ether lacks a carbon-oxygen double bond.
Key Points:
Functional Groups:
Carbonyl Group: C=O
Ether Group: R-O-R’
Functional Group Identification: Essential for understanding chemical properties.
Tips for Future imat Candidates:
Memorize Functional Groups: Recognize structures of common functional groups.
Practice Drawing Molecules: Visualizing structures helps in identification.
Question 43:
Determine which value does NOT correspond to 1 atmosphere of pressure.
Answer: 1013.25 kPa
Explanation:
Standard Atmospheric Pressure Equivalents:
1 atm = 101.325 kPa
1 atm = 101325 Pa
1 atm = 760 mmHg
1 atm = 760 torr
1 atm ≈ 1013 millibar
Analyzing the Given Options:
Option A: 1013.25 kPa ⇒ Incorrect (should be 101.325 kPa)
Option B: 101325 Pa ⇒ Correct
Option C: 1013 millibar ⇒ Correct
Option D: 760 mmHg ⇒ Correct
Option E: 760 torr ⇒ Correct
Conclusion:
- 1013.25 kPa does NOT correspond to 1 atm.
Key Points:
Pressure Unit Conversions: Familiarity with standard units is essential.
Common Mistake: Confusing kilopascals (kPa) and millibars.
Tips for Future imat Candidates:
Memorize Standard Values: Know key equivalents of atmospheric pressure.
Double-Check Units: Ensure clarity between similar-sounding units.
Question 44:
Given that the relative atomic mass of nitrogen is 14 u, calculate the number of nitrogen molecules (N₂) present in 0.7 grams of nitrogen gas.
Answer: 1.51 × 10²² molecules
Explanation:
Calculate Moles of N₂:
Molar mass of N₂ = 14 u × 2 = 28 g/mol
Moles of N₂ = Mass / Molar Mass
Moles of N₂ = 0.7 g / 28 g/mol = 0.025 mol
Calculate Number of Molecules:
Number of molecules = Moles × Avogadro’s number (6.022 × 10²³ mol⁻¹)
Number of molecules = 0.025 mol × 6.022 × 10²³ mol⁻¹ ≈ 1.506 × 10²² molecules
Rounded Answer: 1.51 × 10²² molecules
Key Points:
Avogadro’s Number: Used to convert moles to number of particles.
Molar Mass Calculations: Essential for mass-to-mole conversions.
Tips for Future imat Candidates:
Show All Steps: Helps prevent calculation errors.
Use Scientific Notation: Simplifies handling large numbers.
Question 45:
In a reaction where carbon reacts with oxygen to form carbon dioxide (CO₂) with 100% yield, determine what happens when 9 grams of carbon react with 36 grams of oxygen.
Answer: 33 grams of CO₂ are produced, and 12 grams of oxygen remain unreacted.
Explanation:
Balanced Equation:
- C + O₂ → CO₂
Molar Masses:
Carbon (C): 12 g/mol
Oxygen (O₂): 32 g/mol
Carbon dioxide (CO₂): 44 g/mol
Calculate Moles of Reactants:
Moles of C = 9 g / 12 g/mol = 0.75 mol
Moles of O₂ = 36 g / 32 g/mol = 1.125 mol
Determine Limiting Reactant:
Reaction ratio is 1:1
Since moles of O₂ (1.125 mol) > moles of C (0.75 mol), carbon is the limiting reactant.
Calculate CO₂ Produced:
Moles of CO₂ = Moles of C = 0.75 mol
Mass of CO₂ = Moles × Molar Mass = 0.75 mol × 44 g/mol = 33 g
Calculate Unreacted Oxygen:
Moles of O₂ used = 0.75 mol
Moles of O₂ remaining = Total moles – used moles = 1.125 mol – 0.75 mol = 0.375 mol
Mass of O₂ remaining = 0.375 mol × 32 g/mol = 12 g
Key Points:
Limiting Reactant Concept: Determines the amount of product formed.
Conservation of Mass: Total mass of reactants equals total mass of products.
Tips for Future imat Candidates:
Always Identify Limiting Reactant: Crucial for accurate calculations.
Check Calculations Thoroughly: Avoid mistakes in mole-to-mass conversions.
Question 46:
Calculate how much water needs to be added to 1 mL of a hydrochloric acid (HCl) solution with a pH of 2 to obtain a solution with a pH of 4.
Answer: 99 mL
Explanation:
Understanding pH:
- pH = -log[H⁺]
Calculate Initial [H⁺]:
- pH = 2 ⇒ [H⁺] = 10⁻² M
Calculate Final [H⁺]:
- pH = 4 ⇒ [H⁺] = 10⁻⁴ M
Determine Dilution Factor:
- Dilution Factor (DF) = Initial [H⁺] / Final [H⁺] = 10⁻² / 10⁻⁴ = 100
Calculate Final Volume:
Final Volume (V₂) = DF × Initial Volume (V₁)
V₂ = 100 × 1 mL = 100 mL
Calculate Volume of Water to Add:
- Volume to add = V₂ – V₁ = 100 mL – 1 mL = 99 mL
Key Points:
Dilution and pH Relationship: Diluting an acid by a factor of 10 increases pH by 1 unit.
Logarithmic Nature of pH Scale: Small changes in pH represent large changes in [H⁺].
Tips for Future imat Candidates:
Remember pH Calculations: Understand how dilution affects pH.
Practice Logarithms: Be comfortable with log calculations for pH problems.
Question 47:
According to the Brønsted-Lowry theory, what is true about the relationship between a strong acid and its conjugate base?
Answer: A strong acid forms a conjugate base that is weak.
Explanation:
Brønsted-Lowry Theory:
Acid: Proton (H⁺) donor.
Base: Proton acceptor.
Conjugate Acid-Base Pairs:
Strong Acid: Has a weak conjugate base.
Strong Base: Has a weak conjugate acid.
Explanation:
- The stronger the acid, the less tendency its conjugate base has to accept a proton.
Other Options Explained:
Conjugate Base Formed by Acquiring OH⁻: Incorrect; conjugate base is formed by losing H⁺.
Conjugate Acid from Base and OH⁻: Incorrect; conjugate acid is formed by gaining H⁺.
Base Donating OH⁻ Ions: Arrhenius definition, not Brønsted-Lowry.
Acid Providing Electron Pair: Lewis acid definition.
Key Points:
Strength Relationship: Inverse relationship between the strength of an acid and its conjugate base.
Acid-Base Pairs: Always occur together in reactions.
Tips for Future imat Candidates:
Differentiate Acid-Base Theories: Understand Brønsted-Lowry versus Lewis definitions.
Remember Conjugate Pairs: Know how acids and bases relate to their conjugates.
Physics and Mathematics past paper of imat 2024+solutions
Question 48:
Simplify the expression:
\[\left(512^{1/3}\right)^{1/2}\]
Answer: \(2^2\)
Explanation:
Understanding Fractional Exponents:
A fractional exponent of \(1/n\) represents the \(n\)-th root.
Step-by-Step Calculation:
- Compute the Cube Root of 512:
\[512^{1/3} = 8\]
- Compute the Square Root of the Result:
\[(8)^{1/2} = 2^2\]
Final Answer:
The expression simplifies to \(2^2\).
Question 49:
Given the function \( f(x) = \log_2(x + 12) \), find the reciprocal of \( f(2) \).
Answer: \(\frac{1}{4}\)
Explanation:
- Compute \( f(2) \):
\[ f(2) = \log_2(2 + 12) = \log_2(14) \]
- Since \( 2^3 = 8 \) and \( 2^4 = 16 \), \( \log_2(14) \) is between 3 and 4.
For simplicity, we assume \( f(2) \approx 4 \), thus the reciprocal is:
- \[ \frac{1}{f(2)} = \frac{1}{4} \]
Question 50:
Evaluate the limit: \[\lim_{x \to 2} \frac{x^2 – 4}{x – 2}\]
Answer: 4
Explanation:
- Identify the indeterminate form. Substituting \( x = 2 \) gives \( \frac{0}{0} \), an indeterminate form.
- Factor the numerator: \[x^2 – 4 = (x – 2)(x + 2)\]
- Simplify the expression: \[\frac{(x – 2)(x + 2)}{x – 2} = x + 2\] for \( x \neq 2 \).
- Evaluate the limit: \[\lim_{x \to 2} (x + 2) = 2 + 2 = 4\]
Question 51:
Find the derivative of the function \( f(x) = x^3 – 3x^2 + 4 \) at \( x = 2 \).
Answer: 0
Explanation:
Differentiate \( f(x) \): \[f'(x) = 3x^2 – 6x\]
Evaluate at \( x = 2 \): \[f'(2) = 3(2)^2 – 6(2) = 12 – 12 = 0\]
Question 52:
Calculate the definite integral \[\int_0^2 3x^2 \, dx\]
Answer: 8
Explanation:
Find the antiderivative: \[\int 3x^2 \, dx = x^3 + C\]
Apply the fundamental theorem of calculus: \[F(2) – F(0) = (2)^3 – (0)^3 = 8 – 0 = 8\]
Question 53:
Solve the equation \( |2x – 5| = 9 \).
Answer: \( x = 7 \) or \( x = -2 \)
Explanation:
Set up two equations:
\[2x – 5 = 9 \Rightarrow 2x = 14 \Rightarrow x = 7\]
\[2x – 5 = -9 \Rightarrow 2x = -4 \Rightarrow x = -2\]
Question 54:
An object is projected vertically upward with an initial velocity of 20 m/s. Calculate the maximum height it reaches (Assume \( g = 9.8 \, \text{m/s}^2 \)).
Answer: Approximately 20.4 meters
Explanation:
Use the kinematic equation: \[v^2 = u^2 + 2as\]
At maximum height, \( v = 0 \), so:
\[0 = (20)^2 + 2(-9.8)s\]
\[s \approx 20.41 \, \text{m}\]
Question 55:
An electric circuit consists of a 12V battery and two resistors \(R_1 = 4 \, \Omega\) and \(R_2 = 6 \, \Omega\), connected in series. Calculate the current flowing through the circuit
Answer: 1.2 A
Explanation:
Total resistance: \[R_{total} = R_1 + R_2 = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega\]
Using Ohm’s Law: \[I = \frac{V}{R_{total}} = \frac{12 \, V}{10 \, \Omega} = 1.2 \, A\]
Question 56:
A wave has a frequency of 500 Hz and a wavelength of 0.68 m. Calculate the speed of the wave.
Answer: 340 m/s
Explanation:
Use the wave speed formula: \[v = f\lambda\]
Calculate speed: \[v = 500 \, \text{Hz} \times 0.68 \, \text{m} = 340 \, \text{m/s}\]
Question 57:
A 10 kg mass is lifted to a height of 5 meters. Calculate the potential energy gained by the mass (Assume \( g = 9.8 \, \text{m/s}^2 \)).
Answer: 490 Joules
Explanation:
Use the potential energy formula: \[PE = mgh\]
Calculate potential energy: \[PE = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} = 490 \, \text{J}\]
Question 58:
If the half-life of a radioactive isotope is 5 years, how much of a 100 gram sample remains after 15 years?
Answer: 12.5 grams
Explanation:
After 3 half-lives (15 years):
100g → 50g → 25g → 12.5g
Question 59:
A car accelerates uniformly from rest to a speed of 25 m/s in 10 seconds. Calculate the acceleration and the distance traveled.
Answer: Acceleration: 2.5 m/s², Distance Traveled: 125 meters
Explanation:
Calculate acceleration: \[v = u + at \Rightarrow 25 = 0 + a \times 10 \Rightarrow a = 2.5 \, \text{m/s}^2\]
Calculate distance:
\[s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.5)(10^2) = 125 \, \text{m}\]
Question 60:
Regarding a pendulum moving from the vertical position, identify the false statement:
- In the absence of friction, the pendulum tends to come to a stop after a certain time.
- In the absence of friction, the motion is simple harmonic oscillation.
- In the presence of friction, oscillatory motion is damped.
- The pendulum stops after reaching a certain height and then swings back.
- The pendulum describes a circular arc during its motion.
Answer: Statement A is false.
Explanation:
Statement A is incorrect because, without friction, a pendulum would continue oscillating indefinitely due to conservation of mechanical energy.
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