Explanation: Blood maintains a stable internal Environment through buffer systems that resist changes in acidity or alkalinity. This stability is essential for enzyme function, oxygen Transport by hemoglobin, and normal metabolic reactions. The bicarbonate buffer system is the most important in blood, where carbonic Acid and bicarbonate ions work together to neutralize added Acids or Bases. The body also regulates pH using respiratory control of carbon dioxide and renal control of hydrogen ions and bicarbonate. Even slight deviations from the optimal physiological range can disrupt cellular processes and lead to serious medical conditions. Therefore, blood is tightly regulated within a narrow pH range to ensure proper biological functioning and homeostasis.
Option a – 7.4
The pH of the buffer solution is 5. If the pKa of a weak Acid is 4. The ratio of Acid to Salt in the solution is ( Simplified ICSEChemistryClass 9mcq )
(A) 1 : 1
(B) 1:10
(C) 10 : 1
(D) 1 : 2
Explanation: Buffer systems rely on a weak Acid and its conjugate Base to maintain pH stability. The relationship between pH, pKa, and the ratio of Acid to Salt is described through equilibrium behavior of dissociation. When the pH is higher than pKa, the conjugate Base form becomes more dominant, while a lower pH favors the undissociated Acid form. This balance allows the solution to resist changes when small amounts of Acid or Base are added. The system works through reversible ionization, where both forms continuously interconvert to maintain equilibrium. Such buffer behavior is important in biological systems, industrial processes, and analytical Chemistry where maintaining constant pH is critical for reactions and stability.
Option b – 1:10
The pH of the buffer solution prepared by adding 10 ml of 0.1 M CH3COOH and 20 ml of 0.1 M sodium acetate will be (pKa of CH3COOH = 4.74)
(A) 4.05
(B) 3.04
(C) 5.04
(D) 3.05
Explanation: A buffer system is formed when a weak Acid and its conjugate Base are present together in significant amounts. In such a system, the final pH depends on the relative amounts of Acid and Salt after mixing, not just their initial concentrations. When volumes and molarities are combined, moles of each component determine the equilibrium ratio. The weak Acid partially dissociates, while the Salt provides a common ion that suppresses further dissociation. This interaction stabilizes the hydrogen ion concentration, preventing drastic pH changes. The buffer works through reversible equilibrium between acetic acid and acetate ions, continuously adjusting to maintain balance. Such systems are widely used in biochemical reactions and laboratory preparations where maintaining a constant acidic Environment is essential for reaction control and stability.
Option c – 5.04
In a mixture of sodium acetate and acetic acid, some amount of water is added, then the pH of the solution ( Simplified ICSEChemistryClass 9mcq )
(A) Increases
(B) Decreases
(C) Unchanged
(D) Neutral
Explanation: Buffer solutions resist changes in pH even when diluted with water. This resistance occurs because both the weak acid and its conjugate Base are diluted equally, so their ratio remains almost unchanged. Since buffer pH depends primarily on this ratio rather than absolute concentration, adding water does not significantly disturb the equilibrium. The dissociation equilibrium simply shifts slightly but maintains overall balance. The system continues to neutralize small additions of acid or Base due to the presence of both components. However, extreme dilution can weaken buffering capacity, reducing effectiveness. In general, dilution affects buffer strength more than pH value. This concept is important in biological systems where fluids may undergo volume changes but must maintain stable pH for proper enzymatic and cellular functions.
Option c – Unchanged
A solution containing an equimolar concentration of CH3COOH and CH3COONa. If the pKa of acetic acid is 4.8. The pH of the solution will be
(A) 4.8
(B) 5.101
(C) 9.2
(D) 5
Explanation: When a weak acid and its Salt are present in equal amounts, the buffer system reaches a balanced equilibrium where neither form dominates. The pH in such systems is directly related to the acid dissociation constant and reflects the intrinsic strength of the acid. The equilibrium involves continuous interconversion between acid and conjugate Base, maintaining stability against small disturbances. This condition represents an ideal buffer state where resistance to pH change is maximum. Such equimolar buffer systems are widely used in chemical and biological applications because they provide predictable and stable environments for reactions. The pH remains steady unless a strong acid or Base is added in significant quantity, which would shift the equilibrium.
Option a – 4.8
Which one of the following mixtures will act as an acidic buffer solution?
(A) CH3COOH + CH3COONH4
(B) HCI + NH4Cl
(C) HCOOH + HCOOK
(D) HCI + NaCl
Explanation: Buffer solutions are formed when a weak acid is combined with its conjugate Base or a Salt containing that conjugate Base. This combination allows the system to neutralize small additions of acid or base through equilibrium shifts. Acidic buffers specifically operate in a region where hydrogen ion concentration is relatively higher but still controlled. The weak acid supplies protons, while the conjugate base regulates excess hydrogen ions, maintaining stability. Such systems are essential in biological fluids and chemical processes where pH must remain slightly acidic for optimal reaction conditions. The effectiveness depends on the presence of both components in comparable concentrations that can respond dynamically to changes in the Environment.
Option c – HCOOH + HCOOK
A solution is 2 M of NH4OH and 2 M NH4Cl for NH4OH pKb = 4.7. The pH of the solution is ( Simplified ICSEChemistryClass 9mcq )
(A) 4.7
(B) 9.3
(C) 7
(D) 9
Explanation: A buffer system involving a weak base and its Salt works through equilibrium between the base and its conjugate acid. The weak base partially accepts protons from water, forming hydroxide ions, while the Salt provides the conjugate acid that moderates this effect. The pKb value reflects the strength of the base and influences the overall basicity of the solution. When both components are present in equal concentration, the system achieves a stable equilibrium where hydroxide ion concentration is regulated. This balance prevents large changes in pH when small amounts of acid or base are introduced. Such buffer systems are important in maintaining biological and chemical stability in slightly basic environments.
Option b – 9.3
The pH of the mixture of acetic acid and potassium acetate will be
(A) 0
(B) 7
(C) Less than 7
(D) More than 7
Explanation: A mixture of a weak acid and its Salt forms a buffer system capable of maintaining pH stability. The acetate ion from the Salt suppresses further dissociation of acetic acid due to the common ion effect. This creates a dynamic equilibrium where both acid and base forms coexist in balance. The resulting pH depends on the relative proportions of the two components and the acid dissociation constant. Such systems resist changes in hydrogen ion concentration when small amounts of acid or base are added. They are widely used in laboratories and biological systems to maintain stable chemical environments necessary for controlled reactions and metabolic processes.
Option c – Less than 7
Which one of the following is the purest form of Carbon?
(a) Charcoal
(b) Coke
(c) Fullerene
(d) Carbon black
Explanation: Carbon exists in several allotropic forms where atoms are arranged differently, leading to variations in purity, structure, and properties. Some forms contain impurities or amorphous structures, while others have highly ordered atomic arrangements. The purest form is characterized by a well-defined crystalline structure composed entirely of carbon atoms without significant contamination. Such structures exhibit unique physical and chemical properties due to uniform Bonding. Understanding allotropes helps in distinguishing between naturally occurring carbon forms and engineered nanostructures. These variations influence hardness, electrical conductivity, and thermal stability. The study of carbon allotropes is fundamental in materials science and nanotechnology due to their wide-ranging applications.
(a) By heating wood at a high temperature in the absence of air.
(b) By heating coal at a high temperature in the absence of air.
(c) By burning Hydrocarbons in a limited supply of air.
(d) By heating coal at a high temperature in the presence of air.
Explanation: Carbon black is produced through controlled combustion or thermal decomposition of hydrocarbon materials in limited oxygen conditions. This process prevents complete oxidation, resulting in finely divided carbon particles. The method involves careful regulation of air supply to ensure incomplete combustion. The resulting material has a high surface area and is used widely in industrial applications such as rubber reinforcement and pigment production. The formation process highlights the importance of oxygen availability in determining combustion products. Limited oxygen leads to carbon-rich residues instead of fully oxidized gases like carbon dioxide. This principle is important in industrial Chemistry and environmental control processes.
Option c – By burning Hydrocarbons in a limited supply of air
Incomplete combustion of a fuel gives poisonous ……… gas.
(a) carbon dioxide
(b) isocyanate
(c) carbon monoxide
(d) nitrogen
Explanation: When fuels undergo combustion with insufficient oxygen, the oxidation process is incomplete, leading to the formation of partially oxidized carbon compounds. These compounds can interfere with oxygen Transport in the human body by binding strongly with hemoglobin, reducing oxygen delivery to tissues. This makes them highly toxic even at low concentrations. Complete combustion, in contrast, produces carbon dioxide and water, which are less harmful. The danger of incomplete combustion is especially significant in poorly ventilated environments where fuel-burning devices operate. Understanding this process is crucial for safety in domestic and industrial settings, emphasizing the importance of proper ventilation and oxygen supply during fuel usage.
Option c – carbon monoxide
What is the bond order of the CO group?
(a) 1
(b) 2.5
(c) 3.5
(d) 3
Explanation: Bond order is a measure of the strength and stability of a chemical bond, derived from Molecular orbital theory. It reflects the difference between Bonding and antibonding electrons in a Molecule. A higher bond order indicates a stronger and shorter bond. In diatomic molecules like carbon monoxide, electron distribution leads to a bond that is stronger than a typical single bond due to multiple Bonding interactions. Molecular orbital considerations show significant overlap between atomic orbitals, increasing bond stability. This concept is important in predicting Molecular properties such as bond length, bond energy, and chemical reactivity. CO is known for its strong Bonding characteristics and stability compared to many other small molecules.
Option d – 3
Which of the following statements is not true? ( Simplified ICSE Chemistry Class 9mcq )
(a) Buckminsterfullerene is an allotropic of carbon.
(d) Each carbon Atom in Graphite is attached to the other three carbon atoms.
Explanation: Carbon exhibits allotropy, meaning it can exist in multiple structural forms such as diamond, graphite, and fullerene. Each allotrope has a different arrangement of carbon atoms, leading to distinct physical and chemical properties. Diamond is known for its strong three-dimensional Network of covalent bonds, making it extremely hard and electrically insulating. Graphite, on the other hand, has layered structures with delocalized electrons, allowing it to conduct Electricity. Fullerenes are Molecular forms of carbon with cage-like structures. Understanding these structural differences is important because properties like conductivity, hardness, and reactivity depend on Bonding and geometry rather than composition alone. Incorrect statements usually arise when properties of one allotrope are mistakenly attributed to another.
Option b – Diamond is a good conductor of Electricity
Graphene is
(a) An alloy of carbon
(b) Nanostructure of carbon
(c) Isotope of carbon
(d) None of the above
Explanation: Graphene is a single-layer, two-dimensional form of carbon where atoms are arranged in a hexagonal lattice. It is derived from graphite but consists of only one atomic layer, giving it exceptional mechanical strength, electrical conductivity, and thermal properties. The structure allows electrons to move freely across the sheet, making it highly conductive. Its unique properties arise from the sp² hybridization of carbon atoms and the delocalized π-electron system. Graphene is considered a nanostructure because it exists at the atomic scale and exhibits quantum effects. It has significant applications in electronics, materials science, and nanotechnology due to its flexibility, strength, and conductivity.
Option b – Nano structure of carbon
The order of electron affinity of B, C, N, and O is ( Simplified ICSE Chemistry Class 9mcq )
(A) O> B> C>N
(B) O> C> B>N
(C) B > N> C>O
(D) O> C>N> B
Explanation: Electron affinity refers to the energy change when an Atom gains an electron. It depends on atomic size, nuclear charge, and electron configuration stability. Across a period, electron affinity generally increases due to increasing nuclear attraction. However, exceptions occur due to stable half-filled and fully filled subshells, especially in nitrogen. Oxygen has a higher tendency to gain electrons compared to nitrogen because adding an electron to nitrogen disrupts its stable half-filled configuration. Boron and carbon show lower values due to their electron arrangements and effective nuclear charge differences. These trends explain Periodic variations in chemical reactivity and Bonding behavior among elements.
Explanation: Oxygen Molecule (O₂) shows unique magnetic behavior due to its electronic configuration. According to Molecular orbital theory, oxygen has two unpaired electrons in its antibonding orbitals. These unpaired electrons create a NET magnetic moment, allowing the Molecule to be attracted to a magnetic field. This property is characteristic of paramagnetic substances. Most molecules with all electrons paired are diamagnetic and are weakly repelled by magnetic fields. The paramagnetism of oxygen is an important experimental confirmation of Molecular orbital theory. It also explains why liquid oxygen can be influenced by magnetic fields, demonstrating quantum mechanical effects at a macroscopic level.
Option a – Paramagnetism
A salt of sulfurous acid is called
(A) Sulphate
(B) Sulphurate
(C) Sulphite
(D) Sulphonate
Explanation: Sulfurous acid (H₂SO₃) is a weak acid that forms Salts through partial or complete replacement of hydrogen ions with metal ions. These Salts contain the sulfite ion (SO₃²⁻) or hydrogen sulfite ion depending on the degree of neutralization. Such Salts are commonly used in chemical processes and preservation reactions due to their reducing properties. The naming of Salts depends on the parent acid, and sulfurous acid specifically leads to sulfite compounds. Understanding acid–salt relationships helps in identifying compound behavior, solubility, and reactivity in aqueous solutions. These Salts often play roles in industrial and Environmental Chemistry.
Option c – Sulphite
Which of the following element does not show an oxidation state of +4? ( Simplified ICSE Chemistry Class 9 mcq )
(A) O
(B) S
(C) Se
(D) Te
Explanation: Elements in group 16 show varying oxidation states depending on their ability to expand octet and form covalent bonds. Heavier elements like sulfur, selenium, and tellurium can exhibit multiple oxidation states, including +4, due to accessible d-orbitals and Bonding flexibility. Oxygen, however, is highly electronegative and small in size, limiting its oxidation states mainly to −2 and occasional positive values in compounds with fluorine. It does not typically exhibit +4 oxidation state because it cannot expand its octet. These trends reflect Periodic variations in Atomic Structure and Bonding capability across the group.
Option a – O
The ozone layer is depleted by
(A) NO
(B) NO2
(C) N₂O5
(D) NO3
Explanation: Ozone layer depletion is caused by reactive nitrogen oxides and other pollutants that participate in catalytic destruction of ozone molecules in the stratosphere. These species break ozone into oxygen, reducing the protective layer that absorbs harmful ultraviolet radiation. Nitrogen oxides act as catalysts, meaning they are regenerated after reactions and can destroy many ozone molecules repeatedly. This process disrupts the natural balance between ozone formation and decomposition. The depletion leads to increased UV radiation reaching Earth’s surface, affecting Living Organisms and ecosystems. Understanding this mechanism is important for environmental protection and atmospheric chemistry.
Option a – NO
In pyro sulfurous acid, the oxidation state of sulfur is ( Simplified ICSE Chemistry Class 9 mcq )
(A) Only +2
(B) Only +4
(C) +2 and +6
(D) Only +6
Explanation: Pyrosulfurous acid is related to sulfur oxyacids where sulfur exists in intermediate oxidation states depending on Bonding with oxygen and hydrogen. Oxygen typically has an oxidation state of −2, and hydrogen is +1. By balancing the overall Molecular charge, the oxidation state of sulfur can be determined through algebraic calculation. Such compounds often show sulfur in intermediate oxidation states between its lower and higher oxidation limits. These oxidation states reflect the ability of sulfur to form multiple bonds and exhibit variable valency. This variability is a key feature of group 16 elements and explains their rich redox chemistry.
Option d – Only +6
Group 16 elements have a lower value of first ionization enthalpy as compared to group 15 elements because
(A) Half-filled p-orbitals in group 15 elements are more stable.
(B) Group 16 elements have a smaller size than Group 15 elements.
(C) Group 16 elements contain double bonds while Group 15 elements have triple bonds.
(D) Group 16 elements have more electrons in p-orbitals.
Explanation: Ionization enthalpy depends on atomic size, nuclear charge, and electron configuration stability. Group 15 elements have a stable half-filled p-orbital configuration, making their electrons more difficult to remove. Group 16 elements have one more electron, causing increased electron–electron repulsion in paired orbitals, which reduces the energy required to remove an electron. This repulsion outweighs the increase in nuclear charge across the period, resulting in lower ionization enthalpy compared to group 15. These Periodic trends help explain differences in chemical reactivity and Bonding behavior between neighboring groups in the Periodic Table.
Option a – Half-filled p-orbitals in group 15 elements are more stable
The electron affinity of sulfur is ( Simplified ICSE Chemistry Class 9 mcq )
(A) More than O and Se
(B) More than O but less than Se
(C) Less than O but more than Se
(D) Equal to O and Se
Explanation: Electron affinity trends depend on atomic size and nuclear attraction for incoming electrons. Sulfur, being larger than oxygen, experiences less electron–electron repulsion when gaining an electron, but also has lower nuclear attraction compared to oxygen. Selenium, being larger than sulfur, has even lower attraction for added electrons. These competing factors create a non-uniform trend in group 16 elements. Understanding electron affinity helps explain reactivity patterns, especially in redox reactions where elements tend to gain or lose electrons. The variations are influenced by Periodic structure and shielding effects within atomic orbitals.
Option a – More than O and Se
The minimum catenation ability is shown by
(A) Oxygen
(B) Sulphur
(C) Selenium
(D) Tellurium
Explanation: Catenation refers to the ability of an element to form bonds with itself, creating chains or rings. This property depends on bond strength between identical atoms and atomic size. Smaller atoms with strong bonds show higher catenation ability because their orbitals overlap effectively. As we move down group 16, atomic size increases and bond strength between atoms decreases due to poor orbital overlap. This weakens the ability to form long chains. Elements like oxygen show very limited catenation compared to sulfur, selenium, and tellurium. The trend reflects how Atomic Structure directly influences bonding patterns and structural diversity in elements of the same group.
Explanation: The oxygen family refers to group 16 elements in the Periodic Table, which share the same valence electron configuration. These elements show similar chemical behavior due to identical outer shell structure. However, their physical properties vary significantly down the group, shifting from non-metallic to more metallic character. Oxygen and sulfur are typical non-Metals, while heavier members show increasing metallic properties. This change occurs due to increasing atomic size and decreasing electronegativity. Despite these differences, they all form similar types of compounds such as oxides and hydrides. Their chemistry is strongly influenced by their ability to gain or share electrons in bonding.
Option d – Polymorphic
When a slow dry stream of oxygen is passed through a silent electric discharge, the conversion of oxygen to ozone (10%) occurs. The product is known as P. Here P refers to ( Simplified ICSE Chemistry Class 9 mcq )
(A) Ozonized oxygen
(B) Diatomic oxygen
(C) Both ‘A’ and ‘B’
(D) Neither ‘A’ nor ‘B’
Explanation: Ozone formation occurs when oxygen molecules are subjected to high-energy conditions such as electric discharge, which splits O₂ molecules into atomic oxygen. These atoms then combine with oxygen molecules to form ozone. When only a small fraction of oxygen is converted, the resulting mixture contains both oxygen and ozone. This mixture has distinct chemical properties due to the presence of ozone, which is a more reactive allotrope of oxygen. Such mixtures are important in studying atmospheric chemistry and oxidation reactions. The process illustrates how energy input can alter Molecular structures and create new allotropes without changing elemental composition.
Option a – Ozonized oxygen
Pure ozone is pale blue …p…, dark blue ….Q…. and violet-black ….R…. Here P, Q, and R refer to
Explanation: Ozone exists in different physical states depending on temperature and pressure. In gaseous form, it appears pale blue due to its Molecular structure and Light absorption characteristics. When cooled, it becomes a darker blue liquid as Molecular interactions become stronger and density increases. In Solid form, ozone appears violet-black because of further condensation and structural ordering. These color changes are due to differences in how ozone molecules absorb and transmit Light in different phases. Understanding phase-dependent properties helps explain how temperature and pressure affect molecular behavior and physical appearance.
The physical properties of group 16 elements are : (I) Polonium is metal and radioactive. (II) All these elements exhibit allotropy. (III) With the increase in atomic number, the melting and boiling point increases down the group. The option with the correct SET of properties is ( Simplified ICSE Chemistry Class 9 MCQ )
(A) I and II
(B) II and III
(C) I and III
(D) I, II, and III
Explanation: Group 16 elements exhibit trends in physical and chemical properties as atomic number increases. Oxygen and sulfur show typical non-metallic behavior, while heavier elements like tellurium and polonium show metallic characteristics. Polonium is radioactive due to nuclear instability in heavy atoms. Allotropes exist for several members, especially sulfur, which forms different structural modifications. Melting and boiling points generally increase down the group due to stronger intermolecular forces from increasing atomic size and Mass. These trends illustrate how Periodic properties evolve systematically with changes in electronic structure and nuclear charge.
Option d – I, II, and III
Rhombic sulfur has a unit
(A) S4
(B) S6
(C) S2
(D) S8
Explanation: Sulfur exists in several allotropes, with rhombic sulfur being the most stable form at room temperature. It consists of cyclic molecules made of eight sulfur atoms arranged in a crown-shaped ring structure. These molecules pack into a crystalline lattice, giving rhombic sulfur its characteristic physical properties. The stability of this form is due to minimal strain in the S₈ ring structure. Other forms like monoclinic sulfur are stable only at higher temperatures. The existence of multiple allotropes demonstrates sulfur’s ability to form stable ring structures through catenation, which is a key feature of group 16 chemistry.
Option d – S8
Which of the following aqueous solution will have the highest freezing point? ( Simplified ICSE Chemistry Class 9 MCQ )
(A) 0.1 M urea
(B) 0.2 M cane sugar
(C) 6 g of glucose in 100 cm³
(D) 1.5 g urea in 500 cm³
Explanation: Freezing point depression depends on the number of dissolved particles in a solution. More solute particles result in a greater decrease in freezing point. Therefore, solutions with fewer particles or weaker dissociation show higher freezing points. Non-electrolytes like urea and glucose do not ionize in water, so they produce fewer particles compared to ionic compounds. Concentration also plays a key role, as higher molality leads to stronger freezing point depression. Comparing different solutions requires considering both the nature of solute and its amount. The one producing the least particle concentration will have the highest freezing point.
Option d – 1.5 g urea in 500 cm³
The molal depression constant for water is 1.86. The freezing point of 0.1 m NaCl is
(A) -1.86°C
(B) -3.72°C
(C) -0.186°C
(D) -0.372°C
Explanation: Freezing point depression is a colligative property that depends on the number of solute particles in solution. Ionic compounds like NaCl dissociate into ions, increasing the effective particle concentration. This dissociation enhances the lowering of freezing point compared to non-electrolytes. The extent of depression is directly proportional to molality and the van’t Hoff factor, which accounts for ion formation. Water’s cryoscopic constant determines the sensitivity of freezing point changes. These principles are widely used in determining molecular masses and studying solution behavior in physical chemistry.
Option d – -0.372°C
The freezing point is the temperature at which ( Simplified ICSE Chemistry Class 9 MCQ )
(A) the Vapour pressure of the liquid is less than its Solid.
(B) The vapor pressure of the liquid is higher than its Solid.
(C) The vapor pressure of the liquid is equal to its Solid.
(D) The vapor pressure of the liquid is equal to atmospheric pressure.
Explanation: Freezing point is defined by equilibrium between liquid and Solid phases of a substance. At this temperature, both phases coexist, and the rate of melting equals the rate of freezing. This equilibrium depends on vapor pressure equality between Solid and liquid phases. The condition ensures no NET change in phase occurs, even though molecular motion continues at the microscopic level. Understanding phase equilibrium is essential in Thermodynamics and physical chemistry because it explains transitions between states of Matter. The concept is widely applied in material science and Environmental Studies.
Option c – The vapor pressure of the liquid is equal to its solid
When a nonvolatile solute like urea is added to the pure solvent the freezing points of the solvent in the solution
(A) Increases
(B) Decreases
(C) Remains constant
(D) Becomes zero
Explanation: Adding a nonvolatile solute to a solvent lowers its freezing point due to disruption of crystal formation. Solute particles interfere with the arrangement of solvent molecules, making it harder for the liquid to solidify. This effect is independent of the chemical nature of the solute and depends only on the number of particles present. It is a colligative property, meaning it depends on particle concentration rather than identity. This principle is important in real-life applications such as antifreeze solutions used in vehicles, where lowering freezing point prevents freezing under cold conditions.
Option b – Decreases
Depression at the freezing point is directly proportional to the
(A) Lowering in vapor pressure
(B) Molality of solution
(C) Molarity of solution
(D) Both ‘a’ and ‘b’
Explanation: Freezing point depression depends on the number of solute particles present in a solution. This relationship is a colligative property, meaning it is independent of solute identity and depends only on particle concentration. As the number of dissolved particles increases, the disruption of solvent crystallization becomes greater, leading to a larger decrease in freezing point. This proportionality is used in determining molecular masses of solutes and understanding solution behavior. The concept is important in both theoretical chemistry and practical applications such as antifreeze and de-icing solutions.
Option d – Both ‘a’ and ‘b’
The cryoscopic constant depends on ( Simplified ICSE Chemistry Class 9 MCQ )
(A) the Nature of the Solute
(B) the Nature of the solvent
(C) the Volume of the solvent
(D) the amount of solvent
Explanation: The cryoscopic constant is a property of a solvent that relates the depression in freezing point to the molal concentration of a solute. It reflects how sensitive a solvent is to the presence of dissolved particles. This constant depends on intrinsic physical properties of the solvent, especially its enthalpy of fusion and freezing point. Solvents with stronger intermolecular forces or higher latent Heat of fusion show different responses to solute addition. It is independent of the nature of the solute, which is why it is useful in colligative property calculations. This concept helps in determining molecular masses and understanding how solvents behave under solute addition in physical chemistry systems.
Option b – the Nature of the solvent
Three liquids A, B, and C boils at 130°C, 160°C, and 190°C respectively. Which of them has higher vapor pressure?
(A) A
(B) C
(C) B
(D) All
Explanation: Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. Liquids with weaker intermolecular forces evaporate more easily and therefore have higher vapor pressure. Boiling point is inversely related to vapor pressure: a lower boiling point indicates higher volatility and higher vapor pressure at a given temperature. As temperature increases, vapor pressure increases until it equals atmospheric pressure at the boiling point. Comparing substances using boiling points helps determine relative intermolecular forces and volatility. This concept is widely used in distillation and separation techniques in chemistry.
Option a – A
The strongest Lewis acid among the following is ( Simplified ICSE Chemistry Class 9 MCQ )
(A) BF3
(B) BCl3
(C) BBr3
(D) Bl3
Explanation: Lewis Acids are species that accept electron pairs. Their strength depends on electron deficiency, size, and ability to accept electron density. In group 13 compounds like boron halides, electron acceptance is influenced by back bonding and electronegativity of substituents. As the halogen size increases, back bonding decreases, making the central Atom more electron deficient and a stronger Lewis acid. This trend helps compare acidity among similar compounds. Lewis acid strength is important in catalysis and Organic reactions where electron pair acceptance drives bond formation and reaction pathways.
Option d – BI3
According to Bronsted concept, the base is
(A) Proton doner
(B) Proton loser
(C) Proton accepter
(D) Electron accepter
Explanation: The Brønsted–Lowry theory defines Acids and Bases based on proton transfer. A base is a substance that accepts a proton (H⁺), while an acid donates a proton. This definition focuses on behavior during chemical reactions rather than structure alone. When a base accepts a proton, it forms its conjugate acid. This concept explains acid–base reactions in both aqueous and non-aqueous systems. It is widely applicable in chemistry and Biology, especially in understanding buffer systems, enzyme activity, and proton transfer mechanisms in reactions.
Option c – Proton accepter
Which of the following is not a Lewis acid? ( Simplified ICSE Chemistry Class 9 MCQ )
(A) KCI
(B) SnCl4
(C) BF3
(D) FeCl3
Explanation: Lewis Acids are electron pair acceptors, often characterized by incomplete octets or positive charge density. Metal ions and electron-deficient compounds typically act as Lewis Acids. However, species that are electron-rich or capable of donating electron pairs do not behave as Lewis Acids. Instead, they may act as Lewis Bases. Understanding this distinction is important in coordination chemistry, where interactions between electron donors and acceptors form complex compounds. The ability of a species to accept or donate electron pairs determines its role in Chemical Bonding and reactivity patterns.
Option a – KCl
The conjugate acid of a strong base is
(A) Strong acid
(B) Weak acid
(C) Strong base
(D) Weak base
Explanation: In acid–base chemistry, conjugate acid–base pairs differ by a single proton. When a strong base accepts a proton, it forms its conjugate acid. The strength relationship between a base and its conjugate acid is inverse: a strong base forms a very weak conjugate acid because it has little tendency to donate a proton back. This concept helps explain equilibrium behavior in acid–base reactions. It is widely used in buffer systems and pH calculations, where conjugate pairs maintain balance between proton donation and acceptance.
Option b – Weak acid
Water is a ( Simplified ICSE Chemistry Class 9 MCQ )
(A) protic solvent
(B) Amphoteric acid
(C) Proton accepter
(D) None of these
Explanation: Water is a unique substance in acid–base chemistry because it can behave both as an acid and a base depending on the reaction Environment. This dual behavior is due to its ability to donate a proton or accept a proton. Such substances are called amphoteric. Water’s polarity and hydrogen bonding capacity also contribute to its versatility in chemical reactions. It plays a central role in biological systems and chemical equilibria. Its amphoteric nature allows it to participate in self-ionization, forming hydronium and hydroxide ions, which is fundamental to the concept of pH.
Option c – Proton accepter
Among the following examples, Arrhenius acid is
(A) NH3
(B) SO2
(C) HNO3
(D) AICl3
Explanation: Arrhenius Acids are substances that increase hydrogen ion concentration in aqueous solution. According to this definition, Acids must release H⁺ ions when dissolved in water. This concept is limited to aqueous systems but is useful for understanding basic acid behavior. Substances that do not release hydrogen ions in water are not considered Arrhenius Acids. This theory laid the foundation for modern acid–base chemistry and helps classify substances based on their behavior in water rather than their molecular structure alone.
Option c – HNO3
Which of the following is not an Arrhenius base? ( Simplified ICSE Chemistry Class 9 MCQ )
(A) NaOH
(B) NH3
(C) Ba(OH)2
(D) KOH
Explanation: Arrhenius Bases are substances that increase hydroxide ion concentration when dissolved in water. Typically, metal hydroxides fall under this category because they dissociate to release OH⁻ ions. However, some compounds may act as Bases through proton acceptance without directly producing hydroxide ions, making them non-Arrhenius Bases. This distinction highlights the limitation of Arrhenius theory, which only applies to aqueous solutions. Later theories like Brønsted and Lewis broadened the definition of Bases to include more general electron or proton interactions.
Option b – NH3
An electrolyte
(A) Gives complex ions in solution.
(B) Dissolved in water to give ions.
(C) Ionized in the solid state.
(D) Produce ions only when Electricity passed through it.
Explanation: Electrolytes are substances that produce ions when dissolved in water or molten state, allowing them to conduct Electricity. The presence of free ions enables charge Transport through solution. Strong electrolytes dissociate completely, while weak electrolytes dissociate partially. This behavior is important in electrochemistry, biological systems, and industrial processes. Electrolytes are essential for maintaining nerve function, Fluid balance, and electrical conductivity in Living Organisms. Their study helps understand conductivity, ionization, and chemical equilibrium in solutions.
Option b – Dissolved in water to give ions
The degree of dissociation does not depend on ( Simplified ICSE Chemistry Class 9 MCQ )
(A) the Nature of the solvent
(B) the Nature of the Solute
(C) Temperature
(D) the Electric current
Explanation: Degree of dissociation refers to the extent to which a compound breaks into ions in solution. It depends on factors like nature of solute, solvent polarity, and temperature. Stronger solvents and higher temperatures generally increase dissociation. However, it does not depend on external electrical current because ionization is a chemical equilibrium process, not driven by Electricity in solution under normal conditions. Understanding this helps distinguish between chemical ionization and electrolysis, where Electricity is used to force reactions.
Explanation: Lewis acid–base theory is based on electron pair interactions rather than proton transfer. A Lewis acid is defined as a species that can accept a pair of electrons to form a coordinate bond. This happens because the Atom or ion involved has an incomplete octet or vacant orbitals available for bonding. The concept is broader than Arrhenius and Brønsted definitions because it applies even in non-aqueous systems and in reactions without hydrogen ions. Lewis Acids play an important role in coordination chemistry, catalysis, and Organic reaction mechanisms where electron pair acceptance drives bond formation and reactivity changes.
Option c – Electron pair acceptor
Lewis acids are ( Simplified ICSE Chemistry Class 9 MCQ )
(A) Electron-rich molecules or species
(B) Electron deficient species.
(C) Nucleophiles
(D) Both ‘A’ and ‘C’
Explanation: Lewis acids are chemical species that act as electron pair acceptors due to electron deficiency or positive charge density. Their behavior depends on their ability to accept lone pairs from other species to form coordinate covalent bonds. Many metal ions and electron-deficient molecules fall into this category. These substances often have empty orbitals that allow them to accommodate incoming electron pairs. Understanding Lewis acidity is crucial in explaining complex formation, catalytic activity, and reaction mechanisms in both Inorganic and Organic Chemistry. It provides a unified explanation for many reactions that are not explained by proton-based acid–base theories.
Explanation: Degree of ionization refers to how much a substance breaks into ions when dissolved in a solvent. This depends mainly on the nature of the solute, the polarity of the solvent, and the strength of intermolecular interactions. Strong polar solvents stabilize ions better, increasing ionization. Temperature also influences ionization by providing energy to overcome bonding forces. The size and structure of solute particles also affect how easily they separate into ions. These factors together determine the equilibrium between ionized and unionized forms in solution, which is important in understanding conductivity and reactivity in chemical systems.
Option b – Nature of solute
The volume of solution containing 34.2 g of sugar which give rise to the osmotic pressure of 1.75 atm at 0℃ is
(A) 437.5 ml
(B) 347.5 ml
(C) 750.4 ml
(D) 504.7 ml
Explanation: Osmotic pressure is a colligative property that depends on the number of solute particles in solution. It is described using the relationship between pressure, volume, temperature, and number of moles of solute. Non-electrolytes like sugar do not dissociate, so the number of particles equals the number of molecules present. At constant temperature, osmotic pressure increases with concentration of solute particles. This principle is widely used to determine molecular Mass of unknown compounds. It is also important in biological systems where osmotic balance regulates movement of water across cell membranes.
Option a – 437.5 ml
The vapor pressure of pure water at 25°C is 30 mm. The vapor pressure of 10% glucose solution at the same temperature is ( Simplified ICSE Chemistry Class 9 MCQ )
(A) 31.5 mm
(B) 30.6 mm
(C) 29.67 mm
(D) 28.56 mm
Explanation: Vapor pressure lowering is a colligative property that occurs when a nonvolatile solute is added to a solvent. The solute reduces the number of solvent molecules escaping into the vapor phase, thereby lowering vapor pressure. This effect depends on the proportion of solute particles rather than their identity. In solutions like glucose in water, glucose does not evaporate, so only water contributes to vapor pressure. As solute concentration increases, vapor pressure decreases proportionally. This concept is important in solution chemistry and helps explain boiling point elevation and freezing point depression as related phenomena.
Option c – 29.67 mm
10 g of an Organic compound per liter produces osmotic pressure of 1.18 atm at 0°C. The molar Mass of the compound is (R = 0.082 lit atm)
(A) 78
(B) 94.5
(C) 189.94
(D) 120
Explanation: Osmotic pressure provides a method for determining molar Mass of solutes using colligative property relationships. It depends on solute concentration, temperature, and gas constant. Since the compound is Organic and assumed to be non-electrolytic, it does not dissociate into ions, so particle count equals molecular count. By measuring osmotic pressure at a known temperature and concentration, molar Mass can be calculated indirectly. This technique is widely used in chemistry for compounds that are difficult to analyze by other methods. It connects thermodynamic properties with molecular-level information about solutes in solution.
Option c – 189.94
6 g of urea is dissolved in 90 g of water. The vapor pressure of the solution in mm Hg is
(A) 745 mm
(B) 758 mm
(C) 761 mm
(D) 760 mm
Explanation: Vapor pressure of a solution depends on the relative number of solvent molecules present compared to solute particles. When a nonvolatile solute like urea is dissolved, it reduces the escaping tendency of solvent molecules from the surface. This leads to a decrease in vapor pressure compared to pure solvent. The effect is governed by Raoult’s law, which states that vapor pressure is proportional to the mole fraction of solvent. As solute concentration increases, vapor pressure decreases further. This principle is important in understanding colligative properties and solution behavior in physical chemistry.
Option a – 745 mm
When NaCl is added to the urea solution the freezing point of the resulting solution as compared to the urea solution is ( Simplified ICSE Chemistry Class 9 MCQ )
(A) Lowered
(B) Raised
(C) No change
(D) Initially lowered then raised
Explanation: Freezing point depression increases when more solute particles are present in a solution. Urea is a non-electrolyte and produces fewer particles compared to ionic compounds. Sodium chloride dissociates into two ions in solution, increasing the total number of dissolved particles. This greater particle concentration enhances disruption of solvent crystal formation. As a result, the freezing point decreases further compared to a solution containing only urea. This illustrates the colligative nature of freezing point depression, where the effect depends on the total number of solute particles rather than their chemical identity.
Option c – No change
Which of the following sets of elements are called aerogens?
(A) He, Ne, Ar
(B) F. Cl, Br
(C) O. S. Se
(D) N. P. As
Explanation: Aerogens refer to noble gases, which are group 18 elements in the Periodic Table. These elements are characterized by completely filled valence shells, making them highly stable and chemically inert under normal conditions. Due to their stable electronic configuration, they rarely form compounds. They exist as monoatomic gases and are colorless, odorless, and tasteless. Their inertness makes them useful in applications requiring non-reactive environments, such as lighting, welding, and cryogenics. The term aerogens emphasizes their gaseous nature and chemical inactivity.
Option a – He, Ne, Ar
The noble gas used for the treatment of cancer is
(A) Helium
(B) Argon
(C) Krypton
(D) Radon
Explanation: Certain noble gases have medical applications due to their radioactive or inert properties. In cancer treatment, radioactive isotopes can be used in radiotherapy to target and destroy malignant cells. These gases emit radiation that damages cancerous tissues while minimizing harm to surrounding healthy cells when properly controlled. Noble gases are chemically inert, which makes them safe for controlled medical use because they do not react with biological molecules. Their application in medicine demonstrates how physical properties like radioactivity can be utilized in therapeutic procedures.
Option d – Radon
The compound that attacks Pyrex glass is
(A) XeF2
(B) XeF4
(C) XeF6
(D) All
Explanation: Pyrex glass is made primarily of silica, which is resistant to many chemicals but can be attacked by highly reactive fluorine compounds. Certain xenon fluorides are strong fluorinating agents due to the high electronegativity of fluorine and the ability of xenon to form unstable bonds. These compounds can react with silicon dioxide, breaking its strong Network structure. This demonstrates the reactivity of noble gas compounds, which was once considered impossible. Such reactions are studied in advanced Inorganic Chemistry to understand unusual oxidation states and bonding behavior of noble gases.
Option c – XeF6
The element having the highest ionization potential is
(A) H
(B) N
(C) O
(D) He
Explanation: Ionization potential refers to the energy required to remove an electron from an isolated gaseous Atom. It increases across a period due to increasing nuclear charge and decreases down a group due to increasing atomic size and shielding. Noble gases have the highest ionization energies because their electron configurations are extremely stable with filled valence shells. Among them, the smallest Atom with the strongest nuclear attraction for electrons exhibits the highest value. This makes them very resistant to electron removal, contributing to their chemical inertness and stability.
Option d – He
XeF4 on reaction with H₂ gives
(A) Xenon
(B) Hydrogen fluoride
(C) Both ‘A’ and ‘B’
(D) None of these
Explanation: Xenon fluorides are reactive compounds that can undergo redox reactions with hydrogen. In such reactions, fluorine atoms are reduced to form hydrogen fluoride, while xenon is reduced back to its elemental form. These reactions demonstrate that noble gas compounds can participate in chemical transformations under suitable conditions. The process involves breaking Xe–F bonds and forming strong H–F bonds, which drives the reaction forward. This illustrates the unusual chemistry of xenon compared to other noble gases, which are typically inert due to their stable electronic configurations.
Option c – Both ‘A’ and ‘B’
The noble gas forms an interstitial compound
(A) Neon
(B) Argon
(C) Helium
(D) Xenon
Explanation: Interstitial compounds are formed when small atoms occupy spaces within the crystal lattice of a metal or solid structure. Among noble gases, heavier ones with larger atomic size and higher polarizability are more likely to form such weakly bound compounds under extreme conditions. These interactions are generally rare due to the inert nature of noble gases. However, under specific high-pressure or low-temperature conditions, weak associations can occur. This concept highlights the limits of chemical inertness and shows that even noble gases can participate in unusual bonding situations under special environments.
Option d – Xenon
A monoatomic gas reacts with fluorine to form a fluoride that dissolves in HF to give a conducting solution. The fluoride is
(A) XeF2
(B) XeF4
(C) XeF6
(D) OF2
Explanation: Noble gases can form compounds under extreme conditions, especially with highly electronegative elements like fluorine. Among these gases, heavier members have larger atomic size and lower ionization energy, making them more reactive. Fluorides of such elements are often stable enough to exist in solid form but can undergo further reactions in polar solvents like hydrogen fluoride. When dissolved, they may ionize or form complex species that allow electrical conductivity. This behavior is unusual because noble gases are generally inert due to filled valence shells. Their reactivity increases down the group, making heavier noble gases capable of forming stable compounds with strong oxidizing agents like fluorine.
Option c – XeF6
What volume of hydrogen gas at 273 K and I atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic Mass = 10.8) from the reduction of boron trichloride by hydrogen?
(A) 67.2 L
(B) 44.8 L
(C) 22.4 L
(D) 89.6 L
Explanation: This type of problem is based on stoichiometry and gas laws, combining chemical equations with molar volume concepts. The reaction involves reduction of a boron compound using hydrogen gas, where balanced chemical equations determine mole ratios between reactants and products. Once the amount of boron formed is known, it is converted into moles using atomic Mass. From the reaction ratio, the corresponding moles of hydrogen required are calculated. Finally, using standard conditions where one mole of gas occupies a fixed volume, the required hydrogen volume is determined. Such calculations are important in understanding gas-phase reactions and quantitative relationships in chemistry.
Option a – 67.2 L
Carbon reacts with oxygen to form two compounds: carbon monoxide and carbon dioxide. The ratio of masses of oxygen that combine with the same Mass of carbon in carbon monoxide and carbon dioxide is ………
(A) 1 : 2
(B) 2 : 1
(C) 1 : 3
(D) 3 : 4
Explanation: This question is based on the law of multiple proportions, which states that when two elements form more than one compound, the masses of one element that combine with a fixed Mass of the other are in simple whole number ratios. Carbon forms two oxides with oxygen, where different amounts of oxygen combine with the same Mass of carbon. Carbon monoxide contains less oxygen per carbon Atom compared to carbon dioxide, which has a higher oxygen content. By comparing the oxygen masses in both compounds for the same carbon mass, a simple ratio emerges, illustrating fundamental chemical combination laws.
Option a – 1 : 2
84 g of a gas occupies the same volume as 96 g of oxygen under similar conditions of temperature and pressure. The molecular weight of the gas is ………
(A) 36.5 g
(B) 28 g
(C) 14 g
(D) 18.2 g
Explanation: This problem uses Avogadro’s law, which states that equal volumes of gases under identical conditions contain equal numbers of molecules. Therefore, the ratio of masses of gases occupying the same volume is proportional to their molar masses. By comparing a known gas like oxygen with the unknown gas under identical conditions, the molecular mass can be determined through proportional reasoning. Oxygen’s molar mass serves as a reference point for calculation. This approach is commonly used in gas stoichiometry to determine molecular weights experimentally.
Option b – 28 g
The volume of CO₂ obtained on the complete decomposition of 19.73 g of BaCO3 at STP is (Atomic mass: Ba = 137.3, C = 12,0=16)
(A) 2.24 L
(B) 1.12 L
(C) 0.84 L
(D) 0.56 L
Explanation: Thermal decomposition of carbonates produces carbon dioxide and metal oxides in a fixed stoichiometric ratio. The amount of CO₂ formed depends on the number of moles of the carbonate decomposed. First, the mass of the carbonate is converted into moles using its molar mass. Since one mole of carbonate produces one mole of CO₂, the moles of gas formed are directly determined. At standard temperature and pressure, one mole of any gas occupies a fixed volume, allowing conversion into gas volume. This principle links solid decomposition reactions with gas behavior under standard conditions.
Option a – 2.24 L
The molecular mass and empirical formula of a compound are 180 g and CH₂O respectively. What will be the molecular formula of the compound?
(A) C9H1809
(B) CH₂O
(C) C6H12O6
(D) C₂H4O₂
Explanation: The molecular formula is a multiple of the empirical formula, determined by comparing molar mass with empirical formula mass. First, the empirical formula mass is calculated by adding atomic masses of elements in CH₂O. Then, the ratio of molecular mass to empirical mass gives a whole number multiplier. This multiplier is used to scale the empirical formula into the molecular formula. This method is widely used in chemistry to determine actual molecular composition from experimental data. It connects percentage composition analysis with molecular structure determination.
Option c – C6H12O6
The number of grams of oxygen in 0.10 mole of Na₂CO3.10H₂O is ………..
(A) 13 g
(B) 20.8 g
(C) 18 g
(D) 208 g
Explanation: Hydrated Salts contain both the main compound and water of crystallization. To determine the mass of a specific element like oxygen, all oxygen atoms present in both parts must be counted. The total number of oxygen atoms is calculated from the chemical formula, including those in carbonate and water molecules. This number is then converted into moles and multiplied by atomic mass to obtain mass in grams. Such calculations are important in stoichiometry and help in understanding composition of hydrated compounds in chemistry.
Option b – 20.8 g
The maximum volume of 0.25 M HCl which can be made by using one liter of 0.15 M HCl and one liter of 0.40 M HCl without adding water is ……….. Assume that volumes are additive.
(A) 2 L
(B) 2.2 L
(C) 1.667 L
(D) 0.55 L
Explanation: This problem is based on solution mixing and concentration principles. When two solutions of different molarities are combined, the total moles of solute are added together, while total volume is the sum of individual volumes. The final concentration depends on the ratio of total moles to total volume. Since no water is added, only dilution and mixing affect the final molarity. Such calculations are important in laboratory preparation of solutions where precise concentration control is required for reactions and experiments.
Option c – 1.667 L
What is the mass of carbon dioxide which contains the same number of molecules as are present in 40 g of oxygen?
(A) 40 g
(B) 55 g
(C) 32 g
(D) 44 g
Explanation: Equal number of molecules means equal number of moles based on Avogadro’s principle. First, the number of moles in the given oxygen sample is determined using its molar mass. Since carbon dioxide must contain the same number of molecules, it must have the same number of moles. This mole value is then converted into mass using molar mass of carbon dioxide. This method highlights the relationship between mass, moles, and molecular count in gases and is widely used in chemical calculations involving stoichiometry.
Option b – 55 g
Amount of NO(g) formed in grams by oxidation of 1 mole of NH3(g) by 1 mole of O2(g) is ………….
(A) 30.0 g
(B) 54.0 g
(C) 60.0 g
(D) 24.0 g
Explanation: This is a stoichiometric reaction problem involving oxidation of ammonia to nitric oxide. The balanced chemical equation determines the mole ratio between reactants and products. Since the number of moles of reactants is given, the limiting reactant concept may be applied to determine how much product can form. Once the moles of nitric oxide are found, they are converted into mass using molar mass. Such reactions are important in industrial chemistry, especially in nitrogen cycle processes and fertilizer production.
Option d – 24.0 g
A mixture of gases contains H₂ and O. gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the H₂ and O₂ gases in the mixture?
(A) 16 : 1
(B) 2 : 1
(C) 1 : 4
(D) 4 : 1
Explanation: This problem involves converting mass ratios into mole ratios using molar masses. Hydrogen and oxygen gases have different molecular weights, so equal masses correspond to different numbers of moles. To find the molar ratio, each mass is divided by its molar mass to determine moles, and then compared. This conversion is important because chemical reactions depend on mole ratios rather than mass ratios. Such calculations are widely used in gas mixture problems and reaction stoichiometry where composition must be expressed in terms of particle count rather than weight.
Option d – 4 : 1
A mixture of Nal and NaCl on reaction with H₂SO4 gave Na₂SO4 equal to the weight of the original mixture taken. The mass percentage of Nal in the original mixture is ……… (Atomic weight of Na= 23, 1= 127, S = 32, Cl = 35.5, O=16)
(A) 15.38
(B) 28.38
(C) 62.38
(D) 82.38
Explanation: This problem involves stoichiometry and conservation of mass in reactions of halide Salts with sulfuric acid. Both sodium iodide and sodium chloride react to ultimately form sodium sulfate, while releasing different byproducts depending on the halide present. The key idea is that the sodium content from both Salts ends up entirely in Na₂SO₄, so comparing initial mixture mass with final sodium sulfate mass gives a constraint. Since iodide and chloride Salts have different molar masses, their proportions affect total mass balance differently. By expressing both Salts in terms of moles and linking them through sodium balance, a solvable equation is formed. This type of problem tests understanding of mass conservation and reaction stoichiometry in mixed salt systems.
Option b – 28.38
Two samples of a compound of hydrogen and oxygen obtained by two different processes were analyzed. One sample contained 0.2 g of hydrogen and 3.2 g of oxygen. The other sample contained 5.88 % hydrogen and 94.12% oxygen. The experimental data obtained ……..
(A) supports the law of conservation of mass
(B) supports the law of definite proportions
(C) does not support the law of definite proportions
(D) shows that the composition of elements present is different for both the samples
Explanation: This question is based on the law of definite proportions, which states that a chemical compound always contains the same elements in a fixed mass ratio regardless of its source or method of preparation. By comparing the percentage composition of hydrogen and oxygen in both samples, we check whether the ratio remains constant. If both samples show identical mass ratios of hydrogen to oxygen, it confirms that the compound has a fixed composition. This principle is fundamental in chemical classification and supports the idea that compounds have definite molecular structures rather than variable mixtures.
Option b – supports the law of definite proportions
A gaseous hydrocarbon upon combustion gives 0.72 g of water and 3.08 g of CO₂. The empirical formula of the hydrocarbon is ………..
(A) C₂H4
(B) C₂H4
(C) C6Hs
(D) C7H8
Explanation: Combustion analysis is used to determine empirical formulas of Hydrocarbons by measuring products formed. Carbon in the hydrocarbon converts to carbon dioxide, while hydrogen converts to water. From the mass of CO₂, the amount of carbon is calculated using molar mass relationships. Similarly, hydrogen is determined from the mass of water. Once moles of carbon and hydrogen are found, their simplest whole number ratio gives the empirical formula. This method is widely used in Organic Chemistry to identify unknown compounds based on combustion data.
Option d – C7H8
81.4 g sample of ethyl Alcohol contains 0.002 g. The amount of pure ethyl Alcohol in water. sample up to the proper number of present in the significant figures is ………..
(A) 81.4 g
(B) 71.40 g
(C) 81.398 g
(D) 81 g
Explanation: This question involves significant figures and purity calculation in chemical samples. When a mixture contains a very small impurity or specified amount of substance, the actual pure content must be determined carefully using precision rules. The concept of significant figures ensures that reported values reflect the accuracy of measurements. In such calculations, extremely small quantities often do not affect the overall value significantly but must be handled according to rounding rules. This is important in analytical chemistry where precision and accuracy determine reliability of experimental data and results.
Option a – 81.4 g
Total number of electrons in 1 mole of CO₂ (in terms of N₁) is ………
(A) 3NA
(B) NA
(C) 16NA
(D) 22NA
Explanation: To determine total electrons in a mole of molecules, first count electrons in a single Molecule based on atomic numbers. Carbon dioxide consists of one carbon Atom and two oxygen atoms. The total electrons in one Molecule are obtained by summing atomic numbers of all atoms. Then this value is multiplied by Avogadro’s number (Nₐ) to get electrons in one mole. This method connects Atomic Structure with macroscopic quantities and is commonly used in mole concept problems involving particles, atoms, or electrons in bulk Matter.
Option d – 22NA
Equal masses of H₂, O₂, and CH have been taken in a container of volume (V) at a temperature of 27 °C in identical conditions. The ratio of the volumes of gases H₂ : O₂ : CH4 would be ………..
(A) 8 : 16 : 1
(B) 16 : 8 : 1
(C) 16 : 1 : 2
(D) 8 : 1 : 2
Explanation: At the same temperature and pressure, gas volume is directly proportional to the number of moles according to Avogadro’s law. Since equal masses of different gases are given, their mole amounts differ because molar masses are different. To find volume ratio, each mass is divided by its molar mass to calculate moles. These mole values are then compared directly because volume is proportional to moles. This approach is commonly used in gas mixture problems where mass is given instead of moles, requiring conversion before applying gas laws.
Option c – 16 : 1 : 2
The amount of HCl in gastric juice is about 3.0 g per liter. If a person produces 2.5 liters of gastric juice per day then the number of antacid tablets each containing 300 mg of Al(OH)3 needed to neutralize the HC1l produced in one day is approximately ………(Atomic mass of Al = 27, CI = 35.5, O = 16, H= 1)
(A) 12
(B) 15
(C) 18
(D) 25
Explanation: This is a neutralization stoichiometry problem involving acid–base reaction between hydrochloric acid and aluminium hydroxide. First, total mass of HCl produced per day is calculated from concentration and volume, then converted into moles using molar mass. The balanced reaction between HCl and Al(OH)₃ gives a fixed mole ratio, which determines how much base is required to neutralize the acid. Finally, the required mass of antacid is divided by tablet mass to estimate number of tablets. Such problems are important in Pharmaceutical chemistry and medical dosage calculations involving acid neutralization.
Option c – 18
A sample of hydrated barium chloride weighing 61 g was heated until all the water of hydration is removed. The weight of the dried sample was found to be 52 g. What is the formula of the hydrated salt? (Atomic mass of Ba = 137, Cl = 35.5)
(A) BaCl₂.H₂O
(B) BaCl₂.2H₂O
(C) BaCl₂.3H₂O
(D) BaCl₂.4H₂O
Explanation: Hydrated salts contain water molecules bound within their crystal structure, called water of crystallization. When heated, this water is removed, leaving behind the anhydrous salt. The difference in mass between hydrated and anhydrous forms gives the mass of water lost. By converting these masses into moles, the ratio of salt to water molecules can be determined. This ratio provides the empirical formula of the hydrate. Such calculations are important in gravimetric analysis and help in determining composition of crystalline salts.
Option b – BaCl₂.2H₂O
Which of the following statements about a compound is INCORRECT?
(A) A Molecule of a compound has atoms of different elements.
(B) A compound cannot be separated into its constituent elements by physical methods of separation.
(C) A compound retains the physical properties of its constituent elements.
(D) The ratio of atoms of different elements in a compound is fixed.
Explanation: Compounds are substances formed when two or more elements chemically combine in a fixed ratio. They have distinct properties different from their constituent elements. Physical separation methods cannot separate compounds into elements; chemical methods are required. Compounds also have fixed composition and do not retain individual properties of elements. Understanding these characteristics helps distinguish compounds from mixtures. Incorrect statements often arise when properties of mixtures are confused with those of compounds. This concept is fundamental in basic chemistry classification and material identification.
Option c – A compound retains the physical properties of its constituent elements.
If the temperature (X°) of a substance is the same when expressed in °F and °C, then the value of Xº will be ………
(A) 180°
(B) – 273°
(C) – 40°
(D) – 32°
Explanation: Temperature conversion between Celsius and Fahrenheit follows a linear relationship. When both scales show the same numerical value, a specific equation can be formed equating the two expressions. Solving this relationship gives the temperature at which both scales coincide. This involves algebraic manipulation of the conversion formula. Such problems demonstrate the connection between different measurement systems and are commonly used to test understanding of unit conversions and linear equations in Physics and chemistry contexts.
Option c – – 40°
An Organic compound made up of C, H and N contain 20% nitrogen. What will be its molar mass in grams if it contains only one nitrogen Atom per Molecule of the compound?
(A) 70
(B) 140
(C) 100
(D) 65
Explanation: This problem is based on percentage composition and molecular formula determination. If nitrogen makes up a fixed percentage of the compound and each Molecule contains exactly one nitrogen atom, the molar mass can be deduced. The mass of nitrogen in one mole of compound is calculated from its atomic mass and percentage contribution. By setting up a proportion between nitrogen mass and total molar mass, the unknown molar mass is determined. This method is widely used in analytical chemistry for identifying Organic compounds from elemental analysis data.
Option a – 70
The amount of sodium sulphate formed when 1 L of 0.1 M sulphuric acid is allowed to react with 1 L of 0.1 M sodium hydroxide is …………..
(A) 7.10 g
(B) 3.55 g
(C) 28.4 g
(D) 142 g
Explanation: This problem is based on acid–base neutralization and stoichiometry. Sulfuric acid is a diprotic acid, meaning it can release two hydrogen ions per molecule, while sodium hydroxide provides one hydroxide ion per molecule. When solutions of known molarity and volume are mixed, the first step is to calculate the number of moles of each reactant. The reaction proceeds according to a balanced chemical equation forming sodium sulfate and water. The limiting reactant determines how much product is formed. Such calculations are important in quantitative chemistry to understand how reactants convert into products in exact proportions under complete neutralization conditions.
Option a – 7.10 g
The volume occupied by 2 g of the gas at STP having vapor density 11.2 is ……….. ( Class 11 Chemistry Chapter 1 MCQ )
(A) 2 L
(B) 4 L
(C) 11.2 L
(D) 22.4 L
Explanation: Vapor density is related to molar mass of a gas, where molar mass is twice the vapor density. Once the molar mass is known, the number of moles in a given mass can be calculated. At standard temperature and pressure, one mole of any gas occupies a fixed volume. Using this principle, the volume occupied by a given mass of gas can be determined. This connects mass, molar mass, moles, and gas volume in a single relationship. Such problems are fundamental in gas laws and help in understanding behavior of gases under standard conditions.
Option a – 2 L
Excess carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralized with 0.1 M hydrochloric acid. What will be the volume of hydrochloric acid required for the complete neutralization of CaCO3? (Atomic mass of Ca = 40)
(A) 200 mL
(B) 300 mL
(C) 400 mL
(D) 500 mL
Explanation: This is a multi-step stoichiometry problem involving precipitation and acid–base neutralization. Carbon dioxide reacts with calcium hydroxide to form calcium carbonate, a solid precipitate. The amount of calcium carbonate formed depends on the initial moles of calcium hydroxide, assuming excess CO₂ ensures complete reaction. The precipitate is then treated with hydrochloric acid, which reacts in a fixed mole ratio to form soluble calcium chloride. By tracking moles through each step, the final required volume of acid can be determined. Such problems test understanding of sequential reactions and conservation of Matter in solution chemistry.
Option d – 500 mL
We covered all the Simplified ICSE Chemistry Class 9 MCQ by Viraf J Dalal PDF above in this post for free so that you can practice well for the exam.
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