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Chemistry Test - 21

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Chemistry Test - 21
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  • Question 1
    1 / -0

    How much electricity in terms of Faraday is required to produce \(100 {~g}\) of Ca from molten \(C a C l_{2} ?\)

    Solution

    We have to note that, one mole of electron charge is equivalent to oneFaraday.

    Given reaction,

    Ca is undergoing oxidation i.e., \({Ca}^{2+}+2 {e}^{-} \rightarrow {Ca}\)

    We can observe that, \(40 {~g}\) of \({Ca}\) takes \(2 e^{-}\)charge.

    So, if 1 mole of \(C a(40 g) \equiv 2 F\)

    Then,\(100 g\) of \(C a =\frac{2}{40}\times 100\)

    \(=5F\)

  • Question 2
    1 / -0

    In the reaction, \(\mathrm{Cl}_{2}+\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}+\mathrm{H}_{2} O\), chlorine is:

    Solution

    Given reaction is:

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

    The balanced reaction will be:

    \(3 \mathrm{Cl}_{2}+6 \mathrm{OH}^{-} \longrightarrow 5 \mathrm{Cl}^{-}+\mathrm{ClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\)

    Oxidation number of \(C l_{2}=0\)

    Oxidation number \(C l\) in \(C l^{-}=-1\)

    Oxidation number of\(\mathrm{Cl}\) in \(\mathrm{ClO}_{3}:\)

    \(\Rightarrow x+3 \times(-2) =-1\)

    \(\Rightarrow x=+5\)

    So, \(\mathrm{Cl}_{2}\) is oxidised as well as reduced.

  • Question 3
    1 / -0

    Which characteristic is not correct for physical adsorption?

    Solution

    Physical adsorption decreases with increase in temperature.

    Physical adsorption is non-directional, reversible, multilayers exothermic process where adsorbate molecules are held on surface of adsorbent by physical forces such as van der Waals forces.

    Van der Waals interactions cause these surface atoms to be reactive, causing them to attract liquids, vapors and gases in order to satisfy their atomic force imbalance. The attracted molecules fill in the pores on the surface of the solid when adsorption occurs.

  • Question 4
    1 / -0

    Which of the following complex species is not expected to exhibit optical isomerism?

    Solution

    Optical isomerism is exhibited by only those complexes which lack elements of symmetry.

    \(\left[ Co \left( NH _{3}\right)_{3} Cl _{3}\right]\) shows facial as well as meridional isomerism. But both the forms contain plane of symmetry.

    Thus, this complex does not exhibit optical isomerism.



  • Question 5
    1 / -0

    The compound that does not produce nitrogen has by the thermal decomposition is:

    Solution

    \(\mathrm{NH}_{4} \mathrm{NO}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2}\)

    \(2\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}\)

    \(\mathrm{Ba}\left(\mathrm{N}_{3}\right)_{2} \rightarrow \mathrm{Ba}+3 \mathrm{~N}_{2}\)

    \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{N}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

    The compound that does not produce nitrogen has by the thermal decomposition is\(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

  • Question 6
    1 / -0

    How long would it take to deposit 50 g of \(A l\) from an electrolytic cell containing \(\mathrm{Al}_{2} \mathrm{O}_{3}\) using a current of 105 ampere?

    Solution

    Faraday's First Law of Electrolysis states that only, according to this law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

    The reaction of Al is,

    \(A l^{3+}+3 e^{-} \rightarrow A l\)

    Given,

    Current, \(I= 105~amp\)

    Now as \(n=3\),

    Equivalent weight of \(\mathrm{Al}=\frac{27}{3}=9\)

    As per Faraday's first law.

    \(W=Z \times I \times t\)

    where, \( Z=\frac{E q \cdot w t}{96500}\)

    Then, \( t=\frac{W \times 96500}{E q \cdot w t \times I}\)

    \(=\frac{50 \times 96500}{9 \times 105}\)

    \(=5102.2\)second

    \(=1.42\) hours

  • Question 7
    1 / -0

    At a particular temperature, the vapour pressures of two liquids A and B are 120 mm and 180 mm of mercury respectively. If 2 moles of A and 3 moles of B are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be: (in mm of mercury)

    Solution

    Given: Vapour pressure of liquid A, \(P_{A} =120~ mm\)

    Vapour pressure of liquid B, \(P_{B} =180~ mm\)

    Number of moles of liquid A,  \({n}_{{A}}=2\)

    Number of moles of liquid B, \({n}_{{B}}=3\)

    Therefore,

    Mole fraction of liquid A,  \({X}_{{A}}=\frac{n_A}{n_A+n_B}\)\(=\frac{2}{5}\)

    Mole fraction of liquid B,  \({X}_{{B}}=\frac{n_B}{n_A+n_B}\)\(= \frac{3}{5}\)

    We know that: Vapor Pressure of solution, \({P}={X}_{{A}} {P}_{{A}}+{X}_{{B}} {P}_{{B}}\)

    \(=\frac{2}{5} \times 120+\frac{3}{5} \times 180\)\(=48+108\)

    \(P=156 {~mm} {~Hg}\)

  • Question 8
    1 / -0

    Which of the following is an example of Baeyer-Villiger rearrangement?

    Solution

    The treatment of ketones with peracids (RCOOOH) or hydrogen peroxide gives esters by insertion of oxygen. The reaction is called Baeyer-Villiger rearrangement or Baeyer-Villiger oxidation.

  • Question 9
    1 / -0

    Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

    Assertion : CCl4 is not a fire extinguisher.

    Reason : CCl4 is insoluble in water.

    Solution

    \(CCl_4\) is used as a fire extinguisher. The dense, non combustible vapours cover the burning substance and prevent the availability of oxygen around burning material.

  • Question 10
    1 / -0

    Benzoylation of phenol in alkaline medium is known as:

    Solution

    The process of benzoylation of compounds containing active hydrogen such as phenol, aniline, alcohol, etc, with benzoyl chloride in the presence of aqueous NaOH is called Schotten-Baumann reaction.

    The reaction are as follows:

  • Question 11
    1 / -0

    The freezing point of equimolal aqueous solution will be highest for:

    Solution

    We know that:

    Freezing point \(\propto \frac{1}{\text { Van't Hoff factor }}\)

    (A) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \rightarrow C_{6} \mathrm{H}_{5} N H_{3}^{\oplus}+C l^{\ominus}\)

    Here, Van't Hoff factor, i = 2

    (B)\(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{\ominus}\)

    Here, Van't Hoff factor, i = 3

    (C)\(\mathrm{} \mathrm{La}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{La}^{3+}+3 \mathrm{NO}_{3}^{\ominus}\)

    Here, Van't Hoff factor, i = 4

    (D) Glucose does not dissociate. So, it has the minimum number of particles, and therefore, it shows minimum depression in freezing point. So, it has the maximum freezing point.

  • Question 12
    1 / -0

    Which one of the following is correct regarding instantaneous rate of reactions?

    Solution

    The instantaneous rate of reaction is defined as the change in concentration of an infinitely small time interval. An instantaneous rate is a rate at some instant in time. The average rate is the average of the instantaneous reaction rate over a period of time during the reaction. It depicts a more accurate value.The instantaneous rate of reaction is the rate at which the reaction is proceeding at any given time.There are two ways to determine the instantaneous reaction rate:

    • By using data from experiments and finding the slope of the tangent from the concentration-time graph.
    • By using data from experiments by getting their average rate and then determining the reaction rate from the slope of the tangents.
  • Question 13
    1 / -0

    Read the following road map carefully

    Solution

    Both the ethers obtained by the two routes have opposite but equal optical rotation

    One of the ether is obtained as a racemic mixture

    Step II & III both are S2N  reaction and both have inversion Step II has inversion but step III has retention.

  • Question 14
    1 / -0

    The below shown carbohydrate is:

    Solution

    According to the question,

    It is a \(\beta\)-furanose, thus, a ketohexose.

    A ketohexose is a ketone-containing hexose (a six-carbon monosaccharide). The most common ketohexoses, each of which represents a pair of enantiomers (D-isomers and L-isomers), include psicose, fructose, sorbose, and tagatose.

  • Question 15
    1 / -0

    The oxidative rusting of iron under Earth's atmosphere is an example of:

    Solution

    The oxidative rusting of iron under Earth's atmosphere is slow reaction.

    The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a particular reaction is intuitively defined as how fast or slow a reaction takes place. For example, the oxidative rusting of iron under Earth's atmosphere is a slow reaction that can take many yearsbut the combustion of cellulose in a fire is a reaction that takes place in fractions of a second. For most reactions, the rate decreases as the reaction proceeds.

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