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Chemistry Test 232
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Chemistry Test 232
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  • Question 1/10
    4 / -1

    What pressure (in bar) of ( O2 ) would be required to make the emf of an oxygen electrode zero in a 1.0 M ( H2SO4 ) solution at 25ºC?

    Solutions

    CONCEPT:

    EMF of Electrodes in Solutions

    • The electromotive force (EMF) of an electrode depends on the concentration (or pressure) of the reacting species and the temperature of the system.
    • For a hydrogen electrode, the potential is zero when the pressure of \text{H}_2 gas is 1 bar and the concentration of  H+ ions is 1 M.
    • In the case of an oxygen electrode, the Nernst equation can be used to determine the EMF:
      • The general Nernst equation is:


     

  • Question 2/10
    4 / -1

    Adsorption of gas on platinum surface is less if :

    Solutions

    CONCEPT:

    Adsorption of Gas on Solid Surface

    • Adsorption is a surface phenomenon where molecules of a gas adhere to the surface of a solid.
    • The extent of adsorption depends on factors such as the nature of the gas, nature of the adsorbent, temperature, and pressure.
    • Gases with a higher critical temperature are adsorbed more easily because they are more polarizable and have stronger intermolecular forces, which enhance adsorption.
    • Gases that are easily liquefiable (having high critical temperatures) are generally adsorbed more efficiently on solid surfaces.

    Explanation:-

    • Given that adsorption is less when:
      • The critical temperature of the gas is low because lower critical temperatures indicate weaker intermolecular forces and less polarizability, making adsorption less favorable.
      • Gases that are not easily liquefiable (indicating low critical temperatures) are less likely to be adsorbed efficiently.

    The correct answer is 2) the critical temperature of the gas is low

     

  • Question 3/10
    4 / -1

    One gram of a monobasic acid when dissolved in 100 g of water lowers the freezing point by 0.186°C. Now, 0.25 g of the same acid are dissolved and titrated with 15.1 mL of N/10 alkali. The degree of dissociation of the acid is (k(H2O) = 1.86)

    Solutions

    CONCEPT:

    Degree of Dissociation

    • The degree of dissociation ( α ) of an acid can be found using the relationship between the colligative property (freezing point depression), molar mass, and titration data.
    • Freezing point depression (ΔTf) is related to molality (m): ΔTf = k⋅ m
    • The titration data can be used to find the moles of the acid and relate this to the degree of dissociation.

    Explanation:-

     

  • Question 4/10
    4 / -1

    Major product of the following reaction is 

    Solutions

    CONCEPT:

    Reaction of Primary Amines with Nitrous Acid

    • Primary aliphatic amines react with nitrous acid (HNO2) to form alcohols through a diazotization and subsequent hydrolysis process.
    • The intermediate diazonium salt is unstable in the case of aliphatic amines and quickly decomposes to form an alcohol and nitrogen gas.
    • This reaction is commonly used to convert a primary amine group (-NH2) into a hydroxyl group (-OH) in the final product.

    • In the given reaction, cyclohexylamine (cyclohexane with a -NH2 group) reacts with nitrous acid (HNO2).
    • The primary amine (-NH2) gets converted into a diazonium salt intermediate.
    • The diazonium salt quickly decomposes, resulting in the formation of option 1.

    The correct answer is option 1

     

  • Question 5/10
    4 / -1

    The correct increasing order of extent of hydrolysis in the following is

    Solutions

    CONCEPT:

    Extent of Hydrolysis in Metal Chlorides

    • Hydrolysis of metal chlorides involves the reaction of the chloride with water to form its corresponding acid and base, affecting the extent of hydrolysis.
    • The extent of hydrolysis generally increases with the charge density of the cation. High charge density leads to a stronger attraction to water molecules, thus a higher extent of hydrolysis.
    • Factors that affect the extent of hydrolysis include:
      • Charge of the cation: Higher charges result in greater hydrolysis.
      • Size of the cation: Smaller cations with high charge density show higher hydrolysis.
      • Electronegativity: Cations with higher electronegativity show lower hydrolysis.

    EXPLANATION:

    • CCl4 does not hydrolyze appreciably due to the lack of lone pairs on the carbon atom.
    • MgCl2 undergoes some hydrolysis, but not as significantly as AlCl3 due to its lower charge density.
    • AlCl3 has a high extent of hydrolysis because Al3+ has a high charge density, causing strong interactions with water molecules.
    • SiCl4 undergoes significant hydrolysis to form silicic acid due to the ability of silicon to accommodate additional bonds with oxygen.

    ORDER OF EXTENT OF HYDROLYSIS is  CCl4 < MgCl2 < AlCl3 < SiCl4

    So, the correct is answer  is CCl4 < MgCl2 < AlCl3 < SiCl4

     

  • Question 6/10
    4 / -1

    Which of the following is an example of ketohexose?

    Solutions

    CONCEPT:

    Ketohexose

    • A ketohexose is a six-carbon sugar (hexose) that contains a ketone functional group (carbonyl group at C-2).
    • Ketohexoses have the general formula C6H12O6 and are important in metabolism.
    • Common examples of ketohexoses include:
      • Fructose: Often found in fruits and honey, it's one of the most common ketohexoses.


     

     

  • Question 7/10
    4 / -1

    In Lassaigne’s test, the organic compound is fused with a piece of sodium metal in order to

    Solutions

    CONCEPT:

    Lassaigne’s Test

    • Lassaigne’s test is a qualitative analysis used to detect the presence of halogens, nitrogen, and sulfur in an organic compound.
    • In this test, the organic compound is fused with sodium metal to convert covalently bonded elements (halogens, nitrogen, and sulfur) into their ionic forms.
    • The conversion into ionic compounds facilitates their detection in subsequent steps.

    EXPLANATION:

    • Fusion with sodium helps break down the organic compound and form ionic salts such as sodium halide (NaX), sodium cyanide (NaCN), and sodium sulfide (Na2S).
    • These ionic salts are soluble in water and can be easily tested using specific reagents for identification of halogens, nitrogen, and sulfur.
    • The primary purpose of this fusion process is not to increase ionisation or change melting points but to facilitate the detection process by converting covalent bonds to ionic ones.

    So, the correct answer is Convert the covalent compound into a mixture of ionic compounds

     

  • Question 8/10
    4 / -1

    Which of the following has the highest value of pKb?

    Solutions

    CONCEPT:

    Basicity and pKB of Amines

    • pKB is the logarithmic measure of the base dissociation constant (KB) of a compound; the higher the pKB, the weaker the base.
    • In amines, the availability of the lone pair of electrons on the nitrogen atom determines the basicity. The more available the lone pair, the stronger the base, and the lower the pKB value.
    • Factors like inductive effects, resonance, and steric hindrance can affect the availability of the lone pair, thereby influencing the basicity and pKB value.

    EXPLANATION:

    • Option 1 (Cyclohexylamine) is a simple aliphatic amine, where the lone pair on nitrogen is readily available, leading to a relatively lower pKB.
    • Option 2 (Piperdin) has a nitrogen atom not involved in the aromatic ring system, where the lone pair is localized, making it more available for protonation, hence a lower pKB.
    • Option 3 (Aniline) has a lone pair that is delocalized into the benzene ring, reducing its availability for protonation. This significant delocalization effect leads to a higher pKB compared to aliphatic amines.
    • Option 4 (N-Methylcyclohexylamine) is similar to cyclohexylamine but has a methyl group attached, which does not significantly affect the availability of the lone pair compared to the resonance effect in aniline.

    The correct answer is option 3: Aniline (highest pKB due to strong resonance effect reducing the lone pair availability).

     

  • Question 9/10
    4 / -1

    In which case van’t Hoff factor is maximum?

    Solutions

    CONCEPT:

    Van’t Hoff Factor (i)

    • The van’t Hoff factor (i) represents the number of particles into which a solute dissociates in solution.
    • It is used for calculating colligative properties and understanding the extent of ionisation or dissociation of a compound in solution.
    • The van’t Hoff factor can be calculated using the formula:
      i = 1 + α(n - 1), where α is the degree of ionisation, and n is the number of particles the compound dissociates into.

    EXPLANATION:

    So the correct answer is Aqueous potassium ferricyanide, 30% ionised

     

  • Question 10/10
    4 / -1

    The correct hybridisation state of iodine atom in  molecules is , respectively

    Solutions

    CONCEPT:

    Hybridization of Iodine in Different Molecules

    • Hybridization is determined by the steric number, which is the sum of the number of atoms bonded to a central atom and the number of lone pairs on the central atom.
    • The steric number is used to determine the hybridization state:
      • sp hybridization: Steric number 2
      • sp2 hybridization: Steric number 3
      • sp3 hybridization: Steric number 4
      • sp3d hybridization: Steric number 5
      • sp3d2 hybridization: Steric number 6
      • sp3d3 hybridization: Steric number 7

    EXPLANATION:

    1. I3:

    So, the correct 

    Option 1: sp3d, sp3d3, sp3d2

     

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