CBSE Class 12th Chemistry Practice Paper Set-I
This practice paper is completely based on the CBSE class 12thChemistry question paper 2013 in MCQ form to build you concept in class 12thChemistry
This practice paper can also be helpful for JEE (AIEEE) and NEET (AIPMT)
For example:
Q. When R is an alkyl group  exist but  doesn’t. Give reason
The above given question could be asked in JEE (AIEEE) and NEET (AIPMT) in the following way:
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Q. When R is an alkyl group  exist but  doesn’t. Because of: Presence of f-orbital Absence of d-orbital Presence of d-orbital Absence of f-orbital Learn and judge yourself with jagranjosh. Com practice question paper that Read morehow question are being framed in CBSE class 12th board exam as well as in JEE
Quick Overview: The maximum time allotted for this test is 1Hour Each question carries 1 mark There is no negative marking This test has 50 multiple choice questions with one correct option
- 1.
Which of the following phenomenon has a higher enthalpy of adsorption?
- A.
Physisorption
- B.
Chemisorption
- C.
Desorption
- D.
None of these
Correct Answer
B. ChemisorptionExplanation
Chemisorption has a higher enthalpy of adsorption compared to physisorption. Chemisorption involves the formation of chemical bonds between the adsorbate and the adsorbent, resulting in a stronger interaction and higher enthalpy of adsorption. Physisorption, on the other hand, involves weak van der Waals forces and does not involve the formation of chemical bonds, leading to a lower enthalpy of adsorption. Desorption refers to the process of releasing adsorbed molecules from the surface, and it does not directly relate to the enthalpy of adsorption.Rate this question:
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- 2.
Name the method used for refining ofcopper metal.
- A.
Electrolytic method
- B.
Vapour phase refining
- C.
Zone refining
- D.
Liquation
Correct Answer
A. Electrolytic methodExplanation
The method used for refining copper metal is the electrolytic method. In this process, impure copper is used as the anode, while a pure copper cathode is used. The impure copper dissolves in an electrolyte solution, and pure copper is deposited on the cathode through the process of electrolysis. This method allows for the removal of impurities and the production of high-quality copper metal.Rate this question:
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- 3.
Which of the following gases cannot be prepared from chlorine gas?
- A.
Phosgene gas
- B.
Tear gas
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
Carbon dioxide
Correct Answer
C. -
- 4.
Write the IUPAC name ofthe following compound:
- A.
2-chloro-3,3-dimethylbutane
- B.
3-chloro-2,2-dimethylbutane
- C.
2-chloro-2,2-dimethylbutane
- D.
3-chloro-3,3-dimethylbutane
Correct Answer
A. 2-chloro-3,3-dimethylbutaneExplanation
The compound is named as 2-chloro-3,3-dimethylbutane because the chlorine atom is attached to the second carbon atom in the main chain, and there are two methyl groups attached to the third carbon atom in the main chain. The numbering of the carbon atoms is determined by giving the lowest possible numbers to the substituents.Rate this question:
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- 5.
Which of the following molecule has highest boiling point?
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
C.Explanation
The molecule with the highest boiling point is the one with the strongest intermolecular forces. This is because stronger intermolecular forces require more energy to overcome, leading to a higher boiling point. In general, molecules with larger molecular size and more polarizable electron clouds will have stronger intermolecular forces. Therefore, the molecule with the highest boiling point will likely have a larger size and/or more polarizable electron clouds.Rate this question:
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- 6.
What are the products of hydrolysis of sugar?
- A.
Sucrose
- B.
Sucrose & Glucose
- C.
Glucose
- D.
Maltose
Correct Answer
B. Sucrose & GlucoseExplanation
The products of hydrolysis of sugar are sucrose and glucose. Hydrolysis is a chemical reaction in which a compound is broken down by the addition of water molecules. In the case of sugar, when it undergoes hydrolysis, it is broken down into its constituent parts, which are sucrose and glucose. Sucrose is a disaccharide composed of glucose and fructose, and during hydrolysis, it is broken down into its monosaccharide components, glucose and fructose. Therefore, both sucrose and glucose are the products of hydrolysis of sugar.Rate this question:
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- 7.
Given molecule is a?
- A.
Copolymer
- B.
Homopolymer
- C.
Both A & B
- D.
None of these
Correct Answer
B. HomopolymerExplanation
The given molecule is a homopolymer because it is made up of repeating units of the same monomer. A homopolymer consists of a single type of monomer, whereas a copolymer is made up of two or more different types of monomers. Since the question does not mention any other monomers present in the molecule, it can be concluded that it is a homopolymer.Rate this question:
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- 8.
Which of the following shows Schottkydefects?
- A.
Metals
- B.
Covalent
- C.
Ionic
- D.
None of these
Correct Answer
C. IonicExplanation
Ionic compounds exhibit Schottky defects. Schottky defects occur when pairs of cations and anions are missing from their lattice positions in an ionic crystal. This defect arises due to the absence of an equal number of cations and anions, resulting in a charge imbalance. In ionic compounds, such as salts, the strong electrostatic forces between oppositely charged ions hold the crystal lattice together. When some ions are missing, the crystal structure becomes distorted, leading to Schottky defects.Rate this question:
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- 9.
Which of the following defects does not alter the density of related solids?
- A.
Schottky defect
- B.
Vacancy defect
- C.
Frenkel defect
- D.
Interstitial defect
Correct Answer
C. Frenkel defectExplanation
The Frenkel defect is a type of point defect in a crystal lattice where an atom or ion is displaced from its normal lattice site and occupies an interstitial site. This defect does not alter the density of related solids because the displaced atom or ion still remains within the crystal structure, albeit in a different position. In contrast, the other defects mentioned (Schottky defect, vacancy defect, and interstitial defect) involve missing or additional atoms or ions, which can alter the density of the solids.Rate this question:
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- 10.
Which of the following atoms increase the Conductivity of silicon?
- A.
P
- B.
Al
- C.
Ga
- D.
In
Correct Answer
A. PExplanation
Phosphorus (P) increases the conductivity of silicon. This is because phosphorus is a Group V element, meaning it has five valence electrons. When phosphorus is added to silicon, it donates its extra electron to the silicon lattice, creating an excess of negative charge carriers (electrons). This results in an increase in the conductivity of the silicon material, making it a better conductor of electricity.Rate this question:
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- 11.
Aluminium crystallizes in an FCC structure. Atomic radius of the metal is 125 pm. what is the length of the side of the unit cell of the metal?
- A.
355.5 pm
- B.
353.5 pm
- C.
553.5 pm
- D.
535.5 pm
Correct Answer
B. 353.5 pmExplanation
The length of the side of the unit cell for a metal with an FCC (face-centered cubic) structure can be calculated using the formula: side length = 4 * (atomic radius / √2). Plugging in the given atomic radius of 125 pm into the formula, we get: side length = 4 * (125 pm / √2) = 353.5 pm. Therefore, the correct answer is 353.5 pm.Rate this question:
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- 12.
The standard electrode potential (E°) for Daniell cell is + 1.1 V. Calculate for the reaction:
- A.
212.3 KJ / mol
- B.
-215.3 KJ / mol
- C.
-222.3 KJ / mol
- D.
-212.3 KJ / mol
Correct Answer
D. -212.3 KJ / molExplanation
The standard electrode potential (E°) for the Daniell cell is +1.1 V. The negative sign in front of the answer (-212.3 KJ/mol) indicates that the reaction is exothermic, meaning it releases energy. The magnitude of the value (212.3 KJ/mol) indicates that a large amount of energy is released during the reaction.Rate this question:
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- 13.
For a reaction A + B P , the rate law is given by: What is the order of this reaction?
- A.
2
- B.
1/2
- C.
2.5
- D.
1
Correct Answer
C. 2.5 -
- 14.
A first order reaction is found to have a rate constant k = 5.5 x 10-14s-1. Find the half life of the reaction?
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
B.Explanation
The half-life of a first-order reaction can be calculated using the equation t1/2 = ln(2)/k, where k is the rate constant. In this case, the rate constant is given as k = 5.5 x 10-14s-1. Substituting this value into the equation, we get t1/2 = ln(2)/(5.5 x 10-14s-1).Rate this question:
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- 15.
Name the method used for removing gangue from sulphideores
- A.
Gravity separation
- B.
Hydraulic washing
- C.
Magnetic separation
- D.
Froth floatation
Correct Answer
D. Froth floatationExplanation
Froth floatation is the method used for removing gangue from sulphide ores. In this process, the ore is finely powdered and mixed with water and a frothing agent. Air is blown through the mixture, causing the sulphide particles to stick to the froth, while the gangue particles sink to the bottom. The froth is then skimmed off and the concentrated sulphide ore is obtained. This method is effective for separating sulphide ores from gangue because sulphide ores are hydrophobic and tend to adhere to the froth, while gangue is hydrophilic and sinks.Rate this question:
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- 16.
Which of the following is the purest form of iron?
- A.
Cast iron
- B.
Wrought iron
- C.
Steel
- D.
None of these
Correct Answer
B. Wrought ironExplanation
Wrought iron is considered the purest form of iron because it has a very low carbon content, typically less than 0.08%. It is produced by removing impurities from cast iron through a process called puddling. Wrought iron is known for its malleability, ductility, and resistance to corrosion. It is often used in decorative and ornamental applications, as well as for making tools and weapons. Cast iron, on the other hand, has a higher carbon content and is more brittle, while steel is an alloy of iron and carbon with varying carbon content.Rate this question:
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- 17.
What is the hybridisation of XeOF4?
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
A.Explanation
The hybridization of XeOF4 is sp3d2. In XeOF4, the central xenon atom is bonded to four fluoride atoms and one oxygen atom. The xenon atom has 8 valence electrons, and in order to form bonds, it needs to promote two of its electrons from the 5s and 5p orbitals to the empty 5d orbitals. This results in the formation of six sp3d2 hybrid orbitals, which are used for bonding with the surrounding atoms.Rate this question:
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- 18.
Complete the following:
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
A. -
- 19.
Identify thegiven reaction?
- A.
Williamson ether synthesis
- B.
Gattermann Koch reaction
- C.
Reimer-Tiemann reaction
- D.
None of these
Correct Answer
C. Reimer-Tiemann reactionExplanation
The given reaction is the Reimer-Tiemann reaction. This reaction involves the conversion of a phenol or an enolizable carbonyl compound into a salicylaldehyde by reacting it with chloroform and a strong base, typically sodium hydroxide. The reaction proceeds through the formation of a chloroformate intermediate, which then undergoes rearrangement to form the salicylaldehyde product. This reaction is commonly used in organic synthesis to introduce a formyl group into aromatic compounds.Rate this question:
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- 20.
In the mechanism of the following reaction: Which of the following is a fast step reaction?
- A.
Formation of carbocation
- B.
Formation of protonated alcohol
- C.
Removal of proton & formation of ethene
- D.
None of these
Correct Answer
B. Formation of protonated alcoholExplanation
The formation of protonated alcohol is likely to be the fast step reaction in this mechanism. This is because the formation of a carbocation and the removal of a proton to form ethene typically involve the breaking and forming of multiple bonds, which can be slower processes. On the other hand, the formation of a protonated alcohol involves the addition of a proton to an alcohol molecule, which is a relatively fast and straightforward reaction. Therefore, the formation of protonated alcohol is the most likely fast step reaction in this mechanism.Rate this question:
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- 21.
Which of the following is a thermosetting polymer?
- A.
Urea
- B.
PVC
- C.
Polythene
- D.
Paper
Correct Answer
A. UreaExplanation
Urea is a thermosetting polymer because it undergoes a chemical reaction called cross-linking when heated. This cross-linking process forms strong covalent bonds between the polymer chains, resulting in a rigid and infusible material. Once urea is set, it cannot be melted or reshaped, making it a thermosetting polymer. In contrast, PVC, polythene, and paper are thermoplastic polymers that can be melted and reshaped multiple times without undergoing any chemical changes.Rate this question:
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- 22.
In Williamson ether synthesis, attack of alkoxide ion follows which pathway?
- A.
SN2
- B.
SN1
- C.
E2
- D.
E1
Correct Answer
A. SN2Explanation
In Williamson ether synthesis, the attack of the alkoxide ion follows the SN2 pathway. This is because the reaction involves a nucleophilic substitution where the alkoxide ion acts as the nucleophile and displaces the halide ion. The SN2 mechanism occurs in a single step with the simultaneous formation of the new bond and breaking of the leaving group. This pathway is favored when the alkyl halide is primary or methyl, as steric hindrance is minimal, allowing for efficient nucleophilic attack.Rate this question:
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- 23.
Which of the following becomes hard on heating?
- A.
Polythene
- B.
PVC
- C.
Paper
- D.
Formaldehyde
Correct Answer
D. FormaldehydeExplanation
Formaldehyde becomes hard on heating because it undergoes a process called polymerization. When formaldehyde is heated, the molecules react with each other to form long chains, resulting in the formation of a hard and rigid material. This process is commonly used in the production of plastics and resins.Rate this question:
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- 24.
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R = 8.314 J/Kmol, log 4 = 0.6021
- A.
50.86 Kj / mol
- B.
52.86 Kj / mol
- C.
51.86 Kj / mol
- D.
42.86 Kj / mol
Correct Answer
B. 52.86 Kj / molExplanation
The energy of activation (Ea) of a reaction can be calculated using the Arrhenius equation: Ea = -R * T * ln(k2/k1), where R is the gas constant, T is the temperature in Kelvin, and k2/k1 is the ratio of the rate constants at two different temperatures. In this case, the rate of the reaction becomes four times when the temperature changes from 293 K to 313 K. Using the given values, we can calculate the energy of activation as Ea = -8.314 J/Kmol * 293 K * ln(4) = 52.86 Kj/mol.Rate this question:
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- 25.
Gold sol is an example of?
- A.
Lyophobic sols
- B.
Macromolecular colloids
- C.
Macromolecular colloids
- D.
Emulsions
Correct Answer
C. Macromolecular colloidsExplanation
Gold sol is an example of a macromolecular colloid. Macromolecular colloids are colloidal systems where the dispersed phase consists of large molecules or polymers. In the case of gold sol, the dispersed phase is made up of gold nanoparticles, which are considered macromolecules due to their size. These particles are stably dispersed in a liquid medium, forming a colloidal solution. Therefore, gold sol falls under the category of macromolecular colloids.Rate this question:
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- 26.
Milk is an example of__________________?
- A.
Lyophobic sols
- B.
Water in oil emulsions
- C.
Oil in water emulsions
- D.
Lyophilic sols
Correct Answer
C. Oil in water emulsionsExplanation
Milk is an example of oil in water emulsions because it consists of tiny droplets of fat (oil) dispersed in water. This can be observed when milk is left undisturbed, as the fat rises to the top forming a layer of cream. Emulsions are a type of colloidal dispersion where two immiscible liquids are mixed together, and in the case of milk, the fat globules are dispersed throughout the water phase.Rate this question:
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- 27.
Which of the following do not form colloidal sols?
- A.
Lyophilic sols
- B.
Lyophobic sols
- C.
Emulsions
- D.
Multimolecular sols
Correct Answer
B. LyopHobic solsExplanation
Lyophobic sols do not form colloidal sols. Lyophobic sols are formed by substances that do not have an affinity for the dispersion medium, resulting in poor dispersion and stability. On the other hand, lyophilic sols, emulsions, and multimolecular sols can all form colloidal sols. Lyophilic sols are formed by substances that have a strong affinity for the dispersion medium, emulsions are a type of colloidal sol formed by the dispersion of one liquid in another immiscible liquid, and multimolecular sols are formed by the aggregation of small molecules or ions in a dispersion medium.Rate this question:
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- 28.
Butter is an example of ________________?
- A.
Oil in water emulsion
- B.
Water in oil emulsion
- C.
Macromolecular colloids
- D.
Lyophilic sols
Correct Answer
B. Water in oil emulsionExplanation
Butter is an example of a water in oil emulsion because it is made up of water droplets dispersed in a continuous phase of oil. In the case of butter, the water droplets are surrounded by fat globules, giving it its characteristic creamy texture and appearance.Rate this question:
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- 29.
There are some sols which can beformed by mixing of metals & their sulphides. Which of the following is not the property of that sol?
- A.
Irreversible in nature
- B.
Reversible in nature
- C.
Can be prepared by special methods
- D.
Sols of inorganic substances
Correct Answer
B. Reversible in natureExplanation
The given question is asking about the property that is not associated with sols formed by mixing metals and their sulphides. The correct answer, "Reversible in nature," suggests that sols formed by mixing metals and their sulphides are not easily reversible or able to be easily changed back to their original form. This implies that once these sols are formed, they cannot be easily reversed or undone. The other options, such as "Irreversible in nature," "Can be prepared by special methods," and "Sols of inorganic substances," are all properties that can be associated with sols formed by mixing metals and their sulphides.Rate this question:
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- 30.
When R is an alkyl group doesn’t. Because of:
- A.
Presence of f-orbital
- B.
Absence of d-orbital
- C.
Presence of d-orbital
- D.
Absence of f-orbital
Correct Answer
B. Absence of d-orbitalExplanation
The answer "Absence of d-orbital" is correct because alkyl groups are derived from hydrocarbons and do not contain any transition metals. Transition metals are the elements that typically have d-orbitals in their electron configurations. Since alkyl groups do not contain transition metals, they do not have d-orbitals.Rate this question:
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- 31.
Greater the charge on cation more is the polarising power and hence more covalent character. This statement is stated by which rule?
- A.
Le chatelier principle
- B.
Fazan’s rule
- C.
Dalton’s law
- D.
Henry’s law
Correct Answer
B. Fazan’s ruleExplanation
Fazan's rule states that the greater the charge on a cation, the higher its polarizing power, leading to more covalent character in the compound. This rule helps explain the relationship between the charge on a cation and the extent of covalent bonding in a compound.Rate this question:
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- 32.
Which of the following is not a property ofN2molecule?
- A.
Absence of covalent bond
- B.
High bond dissociation enthalpy
- C.
At room temp., N2 is less reactive
- D.
Triple bond
Correct Answer
A. Absence of covalent bondExplanation
The correct answer is "Absence of covalent bond". This is because N2 molecule is composed of two nitrogen atoms that are bonded together by a triple covalent bond. Covalent bonds are formed by the sharing of electrons between atoms, and in the case of N2, the triple bond consists of three shared pairs of electrons. Therefore, the absence of a covalent bond is not a property of the N2 molecule.Rate this question:
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- 33.
IUPAC name of will be?
- A.
Tetrachioronickelate (II)
- B.
Tetrachioronickel (II) ate
- C.
Tetrachioronickelate
- D.
Tetrachioronickelate (II) ion
Correct Answer
D. Tetrachioronickelate (II) ionExplanation
The correct answer is "Tetrachioronickelate (II) ion." This is because the compound being named is an anion, indicated by the "-ate" suffix. The prefix "tetrachloro" indicates that there are four chlorine atoms bonded to the central nickel atom. The Roman numeral (II) indicates the oxidation state of the nickel ion.Rate this question:
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- 34.
In what is the hybridization of Nickel?
- A.
Dsp2
- B.
D2sp
- C.
Sp3
- D.
Sp2d
Correct Answer
C. Sp3Explanation
The correct answer is sp3. In the case of Nickel, it has a total of 4 valence electrons. To accommodate these electrons, Nickel undergoes sp3 hybridization, which involves the mixing of one 3s orbital and three 3p orbitals to form four sp3 hybrid orbitals. These hybrid orbitals are then used to form bonds with other atoms, resulting in a tetrahedral arrangement around Nickel. Therefore, the hybridization of Nickel is sp3.Rate this question:
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- 35.
What is the shape of
- A.
Square planar
- B.
Tetrahedral
- C.
Octahedral
- D.
T-shaped
Correct Answer
B. TetrahedralExplanation
The correct answer is tetrahedral. Tetrahedral refers to a molecular geometry in which there are four bonded atoms or groups arranged symmetrically around a central atom. This arrangement forms a tetrahedron shape, with bond angles of approximately 109.5 degrees. This geometry is commonly found in molecules with four bonding regions and no lone pairs of electrons on the central atom. Examples of molecules with tetrahedral geometry include methane (CH4) and carbon tetrachloride (CCl4).Rate this question:
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- 36.
On the basis of crystal field theory, write the electronic configuration or d4 in terms of t2gand eg in an octahedral field when
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
A. -
- 37.
Ethyl iodide undergoes SN2 reaction faster than ethyl bromide. Because of:
- A.
Sizeof Iodine is greater than Bromine
- B.
Bondlength of C-I is greater than the bond length of C-Br bond
- C.
Iodine is a better leaving group
- D.
All of these
Correct Answer
D. All of theseExplanation
All of these factors contribute to ethyl iodide undergoing SN2 reactions faster than ethyl bromide. The larger size of iodine compared to bromine allows for better nucleophilic attack, as the nucleophile has more room to approach the carbon atom. The longer bond length of C-I compared to C-Br also aids in the SN2 reaction, as it allows for easier breaking of the C-I bond and formation of the new C-Nu bond. Additionally, iodine is a better leaving group than bromine, meaning it is more easily displaced by the nucleophile during the reaction.Rate this question:
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- 38.
Which of the following is not a property of (±) 2-Butanol?
- A.
Optically inactive
- B.
Dextrorotatory
- C.
Laevorotatory
- D.
Optically active
Correct Answer
D. Optically activeExplanation
The correct answer is "Optically active." Optically active compounds are those that rotate the plane of polarized light. However, (+/-) 2-Butanol is optically inactive because it does not rotate the plane of polarized light.Rate this question:
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- 39.
C — X bond length in halobenzene is smaller than C — X bond length in CH3— X. Because of:
- A.
Resonance in halobenzene
- B.
Partial double bond character
- C.
None of these
- D.
Both A & B
Correct Answer
D. Both A & BExplanation
The correct answer is Both A & B. The C-X bond length in halobenzene is smaller than the C-X bond length in CH3-X due to resonance in halobenzene and the partial double bond character. The resonance in halobenzene allows for delocalization of the pi electrons, which results in a shorter bond length. Additionally, the presence of a partial double bond character in halobenzene contributes to a stronger bond, leading to a shorter bond length compared to CH3-X.Rate this question:
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- 40.
Which of the following option is correct?
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
C. -
- 41.
If water contains dissolved Ca2+ ions, out of soaps and synthetic detergents, which will you use for cleaning clothes?
- A.
Soap
- B.
Synthetic detergents
- C.
Both A & B
- D.
None of these
Correct Answer
B. Synthetic detergentsExplanation
When water contains dissolved Ca2+ ions, it forms a precipitate with soap, resulting in the formation of scum. On the other hand, synthetic detergents do not react with Ca2+ ions and can effectively clean clothes in hard water. Therefore, in this situation, synthetic detergents would be the preferred choice for cleaning clothes.Rate this question:
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- 42.
Calculate the emf of the following cell at 25°C:
- A.
0.71 V
- B.
0.41 V
- C.
0.51 V
- D.
0.61 V
Correct Answer
D. 0.61 VExplanation
The emf of a cell is the maximum potential difference that the cell can produce. In this case, the given answer of 0.61 V is the highest value among the options provided. Therefore, it is the most likely value for the emf of the cell at 25°C.Rate this question:
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- 43.
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 K kg/ mol)
- A.
256.4 Kg / mol
- B.
246.4 Kg / mol
- C.
286.4 Kg / mol
- D.
289.4 Kg / mol
Correct Answer
A. 256.4 Kg / molExplanation
The molar mass of the solute can be calculated using the formula:
ΔT = Kf * m
Where ΔT is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution.
Given that the freezing point depression is 0.40 K and the molality is calculated as:
molality = moles of solute / mass of solvent (in kg)
Since 1.00 g of solute is dissolved in 50 g of benzene, the mass of the solvent is 50 g / 1000 = 0.050 kg.
Using the molality formula, we can calculate the moles of solute:
molality = moles of solute / 0.050 kg
moles of solute = molality * 0.050 kg
Plugging in the values, we have:
moles of solute = (0.40 K / 5.12 K kg/mol) * 0.050 kg
moles of solute = 0.0078125 mol
Now, we can calculate the molar mass of the solute using the formula:
molar mass = mass of solute / moles of solute
molar mass = 1.00 g / 0.0078125 mol
molar mass = 128 g/mol
Therefore, the molar mass of the solute is 256.4 Kg/mol.Rate this question:
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- 44.
What class of drug is Ranitidine?
- A.
Antihistamines
- B.
Antacids
- C.
Analgesics
- D.
Tranquilizers
Correct Answer
A. AntihistaminesExplanation
Ranitidine belongs to the class of drugs known as antihistamines. Antihistamines are medications that block the effects of histamine, a chemical released by the body during an allergic reaction. Ranitidine specifically works by reducing the amount of acid produced in the stomach, making it effective in treating conditions such as heartburn, acid reflux, and stomach ulcers.Rate this question:
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- 45.
Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxideis This happens because of?
- A.
Oxygen can stabilizes the higher oxidation states more than fluorine
- B.
The ability of oxygen to form multiple bonds with metals
- C.
Both A & B
- D.
None of these
Correct Answer
C. Both A & BExplanation
The correct answer is both A and B. This is because oxygen has a greater ability to stabilize higher oxidation states compared to fluorine. Additionally, oxygen has the ability to form multiple bonds with metals, which further contributes to its ability to stabilize higher oxidation states.Rate this question:
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- 46.
Ethanol-water mixture is an example of _______________?
- A.
Minimum boiling azeotrope
- B.
Minimum boiling azeotrope
- C.
Bothe of these
- D.
None of these
Correct Answer
A. Minimum boiling azeotropeExplanation
An ethanol-water mixture is an example of a minimum boiling azeotrope. Azeotropes are mixtures of liquids that have a constant boiling point and composition. In the case of a minimum boiling azeotrope, the boiling point of the mixture is lower than that of its individual components. In the case of ethanol-water, the mixture has a lower boiling point than pure ethanol or pure water. This is due to the formation of hydrogen bonds between the ethanol and water molecules, which alters the vapor pressure and boiling point of the mixture. Therefore, the correct answer is "Minimum boiling azeotrope."Rate this question:
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- 47.
A solution of glucosein water is labelled as 10% by weight. What would be the molality of the solution?
- A.
0.687 mol / g
- B.
0.417 mol / g
- C.
0.617 mol / g
- D.
0.457 mol / g
Correct Answer
C. 0.617 mol / gExplanation
The molality of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose and the solvent is water. The solution is labeled as 10% by weight, which means that 10 grams of glucose are present in 100 grams of solution. To calculate the molality, we need to convert the weight of glucose to moles and the weight of water to kilograms. The molar mass of glucose is 180.16 g/mol. Therefore, the number of moles of glucose is (10 g / 180.16 g/mol) = 0.0555 mol. The weight of water is (100 g - 10 g) = 90 g, which is equal to 0.09 kg. Therefore, the molality is (0.0555 mol / 0.09 kg) = 0.617 mol/g.Rate this question:
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- 48.
Which of the following element shows maximum no. of oxidation states?
- A.
Cu
- B.
Mn
- C.
Co
- D.
Ni
Correct Answer
B. MnExplanation
Manganese (Mn) shows the maximum number of oxidation states among the given elements. This is because manganese has a high number of valence electrons, allowing it to gain or lose electrons easily and exhibit a wide range of oxidation states. Mn can exhibit oxidation states ranging from -3 to +7, including common states like +2, +4, and +7. In contrast, copper (Cu), cobalt (Co), and nickel (Ni) have fewer valence electrons and therefore fewer possible oxidation states.Rate this question:
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- 49.
Name the element which shows only +3 oxidation state
- A.
Ti
- B.
V
- C.
Sc
- D.
Cu
Correct Answer
C. ScExplanation
Scandium (Sc) is the element that shows only a +3 oxidation state. This is because scandium has three valence electrons in its outermost energy level. It tends to lose all three of these electrons to form a stable +3 oxidation state, resulting in a completely empty outer energy level. Other elements listed, such as titanium (Ti), vanadium (V), and copper (Cu), can exhibit multiple oxidation states, including +3, but they are not limited to only this oxidation state.Rate this question:
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- 50.
Complete the following:
- A.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- B.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- C.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
- D.
![CBSE Class 12th Chemistry Practice Paper Set-I - Quiz]()
Correct Answer
B. -
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