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Chemistry Test - 24

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Chemistry Test - 24
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  • Question 1
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    Which reagent is used for detecting \(\mathrm{Ni}^{2+}\) ions in solution?

    Solution

    When dimethylglyoxime \(( \text{DMG})\) is added to a solution along with ammonium hydroxide, the formation of a bright red precipitate indicates the presence of \(\mathrm{Ni}^{2+}\) ions in solution. This is due to the formation of nickel dimethylglyoxime complex.

    \(2 \mathrm{DMG}+\mathrm{NiCl}_{2}+2 \mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{NiDMG}_{2} \downarrow(\) red ppt. \()+2 \mathrm{NH}_{4} \mathrm{Cl}+2 \mathrm{H}_{2} \mathrm{O}\)

  • Question 2
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    Match correctly the functional group given in List-I with the Nomenclature of that functional group given in list-II:

    List IList II
    a. Aldehydei. Propanone
    b. Halogenii. Propanal
    c. Ketoneiii. Propanol
    d. Alcoholiv. Bromopropane
    Solution

    Correctly match functional group given in List-I with the Nomenclature of that functional group given in list-II:

    Functional

    Group

    Suffix

    Nomenclature

    Aldehyde

    -al

    Propanal

    Halogen

    -pane

    Bromopropane

    Ketone

    -one

    Propanone

    Alcohol

    -ol

    Propanol


  • Question 3
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    Which method can be used to separate mixture of 1°, 2° and 3° amines?

    Solution

    A mixture of 1°, 2° and 3° amines can be separated by Hinsberg’s method.

    Mixture of primary, secondary and tertiary amines can be separated by Hinsberg's method and fractional distillation.
    They give different products with Hinsberg's. These products differ in solubility. Hence, they can be separated with Hinsberg's method. They differ in boiling points. Hence they can be separated by fractional distillation.

  • Question 4
    1 / -0

    The correct order of increasing acidic strength is:

    Solution

    The correct order of increasing acidic strength is Ethanol < Phenol < Acetic acid < Chloroacetic acid.

    • Phenol is more acidic than ethanol because in phenol, the phenoxide ion obtained on deprotonation is stabilized by resonance which is not possible in case of ethanol.
    • Also carboxylic acids are more acidic than alcohols and phenols as the carboxylate ion is stabilized by resonance. Chloroacetic acid is more acidic than acetic acid due to inductive effect of chlorine atom which stabilizes the carboxylate anion.
  • Question 5
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    Which of the following is the chemical formula of halothane?

    Solution

    The chemical formula of halothane is \(\mathrm{CF}_{3} \mathrm{CHClBr}\).

    • Halothane, the brandname Fluothane, is a general anaesthetic.  
    • It can be used to induce or maintain anaesthesia.

    The molecular structure of halothane is as follows:

     

  • Question 6
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    To detect the reducing and non reducing sugars, which of the following test is used?

    Solution

    (A) Molisch's test is a chemical test which is used to check for the presence of carbohydrates in a given analyte.

    (B) The biuret test, also known as Piotrowski's test, is a chemical test used for detecting the presence of peptide bonds. In the presence of peptides, a copper (II) ion forms mauve-colored coordination complexes in an alkaline solution.

    (C) Fehling’s test is used for reducing sugars and non-reducing sugars, supplementary to the Tollens’ reagent test. The test was developed by German chemist Hermann von Fehling in1849.

    (D) Millon's reagent is an analytical reagent used to detect the presence of soluble proteins.

  • Question 7
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    The heats of adsorption (in kJ/mol) in physisorption or physical adsorption lie in the range of:

    Solution

    The heats of adsorption in physisorption or physical adsorption lies in the range of 10−40 kJ/mol.

    It is less as compared to chemisorption as there is no chemical reaction involved in it. The heat of chemisorption lie in the range of 80 to 240kJmol−1.

  • Question 8
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    Which of the following statements is correct regarding variations of molar conductivity with concentration?

    Solution

    We know the relation between molar conductivity and Molarity,

    \(\Lambda_{m}=\frac{\kappa \times 1000}{M}\)

    As seen in the above equation we can arrive at a conclusion that \(\Lambda_{m}^{0}\) is inversely proportional to Molarity.

    And therefore, if we increase concentration of solution molar conductivity decreases and vice-versa.

  • Question 9
    1 / -0

    Ether is formed when alkyl halide is heated with sodium alkoxides. This method is called:

    Solution

    Ether is formed when alkyl halide is heated with sodium alkoxides. This method is calledWilliamson's Synthesis.

    Ethers can be prepared in the laboratory byWilliamson's synthesis.

    In this method, when an alkyl halide is treated withsodium alkoxide, thenucleophilic substitutionof the halogen atom by the alkoxide group takes place.

    The general example is given below.


    Williamson's synthesis is used to preparesymmetrical as well as unsymmetrical ethers.

    As the reaction is a nucleophilic substitution reaction, thealkyl halideshould beprimary alkyl halide, using secondary and tertiary alkyl halide will yieldalkenesdue toelimination reactionand not ethers.

    It is preferable to takesecondary or tertiary alkoxide ionandprimary alkyl halide.

  • Question 10
    1 / -0

    In which of the following conditions, the potential for the following half-cell reaction is maximum?

    \(2 \mathrm{H}^{+}+2 e \rightarrow \mathrm{H}_{2}\)

    Solution

    In the given options,

    (A) Concentration of \(H^{+}\)in \(1.0 \mathrm{M} \mathrm{~HCl}=1 M\)

    (B) Since, \(p H\) of solution is 4 .

    So, concentration of \(H^{+}=10^{-4} M\)

    (C) \(p H\) of pure water \(=7\)

    So, concetration of \(H^{+}=10^{-7} M\)

    (D) Concentration of \(H^{+}\)in \(1.0 \mathrm{M~NaOH}\) is much less than the all the above options.

    Since, concentration of \(H^{+}\)is highest in option (A). So, potential of option (A) is maximum.

  • Question 11
    1 / -0

    Chemical kinetics is a branch of physical chemistry. It deals with _________________.

    Solution

    Chemical kinetics is branch of physical chemistry that is concerned with understanding the rates of chemical reactions.

    Structures and physical changes are deal in solid state and while studying liquids. Thermochemistry deals with energy changes in chemical reactions.

  • Question 12
    1 / -0

    From which of the following tertiary butyl alcohol is obtained by the action of methyl magnesium bromide?

    Solution

    Tertiary butyl alcohol is obtained by the action of methyl magnesium bromide on acetone CH3COCH3.

    Methyl magnesium bromide is CH3MgX and falls in the category of Grignard Reagents. The methyl group bears a partial negative charge and the metal bears a partial positive charge. Ketones react with a Grignard reagent to give tertiary alcohols as addition products.

    Aldehydes give secondary alcohol and formaldehyde gives primary alcohol. The reaction is a step-up reaction as the number of carbon atoms increases in the product.

    The reaction of acetone with methyl magnesium halide is given as follows:




    The product of the reaction is tertiary butyl alcohol.

  • Question 13
    1 / -0

    \(E_{\text {cell }}^{o}\) for the reaction, \(2 {H}_{2} {O} \rightarrow {H}_{3} {O}^{+}+{OH}^{-}\)at \(25^{\circ} {C}\) is \(-0.8277 {~V}\). The equilibrium constant for the reaction is:

    Solution

    Given reaction,

    \(2 {H}_{2} {O} \rightarrow {H}_{3} {O}^{+}+{OH}^{-}\)

    And, \(E_{c e l l}^0 = -0.8277 ~V\)

    It is evident from the cell reaction that it involves the transfer of one electron so that \(\mathrm{n}=1\)

    We know a relation:

    \(\log K=\frac{E_{c e l l}^{0} \times n}{0.0591}\)

    \(\log K=\frac{-0.8277 \times 1}{0.0591}=-14\)

    \( \therefore K=10^{-14}\)

  • Question 14
    1 / -0

    The ionization isomer of \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}\) is

    Solution

    lonisation isomers are the complexes that produces different ions in solution, i.e., they have ions interchanged inside and outside the coordination sphere. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\mathrm{NO}_{2}\) have different ions inside and outside the coordinate sphere and they are isomers.

    Therefore, they are ionisation isomers.

  • Question 15
    1 / -0

    Presence of unsaturation in organic compounds can be tested with:

    Solution

    Baeyer's reagent (alk. \(\mathrm{KMnO}_{4}\) ) which is pink in colour decolourises due to the presence of unsaturation.

    Thus, it shows the presence of unsaturation in an organic compound.

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