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Electrochemistry Test - 8

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Electrochemistry Test - 8
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  • Question 1
    4 / -1

    The correct expression in SI system relating the equivalent conductance  specific conductance (K) and equivalent concentration (C) is (Given k in S cm-1, C in equivalent dm-3)

     

    Solution

     

     


     

     

  • Question 2
    4 / -1

     

    Conductivity (K) of 0.01 M NaCI solution is 0.00145 Scm-1. What happens to the conductivity if extra 100 mL of H2O be added to the above solution?

     

    Solution

     

     


    On dilution, molarity decreases and molar conductance increases, Hence, specific conductance decreases.

     

     

  • Question 3
    4 / -1

     

    An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

     

    Solution

     

     

    Number of ions remains constant but due to dilution, ionic mobility increases (as ionic attraction decreases). Hence, equivalent conductance increases.

     

     

  • Question 4
    4 / -1

     

    Specific conductance of 0.01 N KCI solution is x Scm-1 having conductance y S. Thus, specific conductance of 0.01 N NaCI having conductance zS is (in S cm-1)

     

    Solution

     

     

    Specific conductance = Conductance x Cell constant
    ∴ x = k (0.01 N KCI) = y x cell constant
    ∴ k (0.01N NaCI) = z x cell constant 

     

     

  • Question 5
    4 / -1

     

    In terms of  (molar conductance), (molar conductance of very dilute solution) ionisation constant (Ka) of weak acid is

     

    Solution

     

     

    For a weak acid.

     

     

  • Question 6
    4 / -1

     

    Given limiting ionic conductance of  H+  ion :  = 350 S cm2 equi-1 and of  ion :  = 80 S cm2 equi-1 Thus for H2SO4, limiting values of molar conductance and equivalent conductance are

     

    Solution

     

     



    Since, equivalent mass of H2SO4 = 1/2 molar mass
    Hence, = 430 x 2 = 860 S cm2 mol-1

     

     

  • Question 7
    4 / -1

     

    500 mL of an aqueous solution contains 0.1 mole of KCl. If its specific conductance is x Scm-1, its molar conductance will be (in Scm2 mol-1)

     

    Solution

     

     

    Molar conductance 

     

     

  • Question 8
    4 / -1

     

    Resistivity of a metal is equal to resistance when cell is constant is

     

    Solution

     

     

    Resistance = (Resistivity) x Cell constant
    ∴ Resistance = Resistivity
    when cell constant = 1cm-1

     

     

  • Question 9
    4 / -1

     

    Cell constant is maximum in case of a

     

    Solution

     

     



     

     

  • Question 10
    4 / -1

     

    Which quantity is temperature independent?

     

    Solution

     

     

    When temperature increases, conductance increases. Thus, resistivity and conductivity both are temperature dependent. By Nernst equation,

    Thus, Emf of the cell is dependent on temperature.


    Thus, cell constant is temperature independent.

     

     

  • Question 11
    4 / -1

     

    0.1 M H2SO4 solution is diluted to 0.01 M H2SO4 ?.Hence its molar conductivity will be 

     

    Solution

     

     


    If solution is diluted to x times then new molarity will be 

    Then new molar conductance will be


    ∴ New molar conductance will be 10 times.

     

     

  • Question 12
    4 / -1

     

    Conductivity (Siemen's S) is directly proportional to the area of the vessel and the concentration of the solution in it, and is inversely proportional to the length of the vessel,then constant of proportionality is expressed in

     

    Solution

     

     

    S concentration in mol dm-3
    or  S mol m-3
    S area (m2)

    ∴    


    S = k (Specific conductivity) mol m-2
     ∴ k = Sm2 mol-1 

     

     

  • Question 13
    4 / -1

     

    Given, (Scm2 mol-1)for different electrolytes

    Thus, of CH3COOH is

     

    Solution

     

     

    (CH3COOH) = (CH3COONa) + (HCI) - (NaCI)
    = 91.0+ 426.2 -126.5 
    = 517.2 - 126.5 
    = 390.7 Scm2mol-1

     

     

  • Question 14
    4 / -1

     

    Resistance of 0.2 M soluton of an electrolyte is 50?. The specific conductance of solution is 1.3 Sm-1. If resistance of the 0.4 M solution of the same electrolyte is 260Ω, its molar conductivity is

    [AlEEE 2011]

     

    Solution

     

     

    Specific conductance (k) of 0.2 M solution
    = Conductance x Cell constant


    Specific conductance of 0.4M solution


     

     

  • Question 15
    4 / -1

     

    The equivalent conductance of NaCl at concentration C and at infinite dilution are λc and λ∞ respectively.The correct relationship λc and λ∞ between  is given as (where constanr B is positive).

     

    Solution

     

     

    By Debye-Huckel Onsager equation 

    when, λC = equivalent conductivity at concentration C and λ∞ = limiting equivalent conductivity at infinite dilution 
    Limit C → 0. λC = λ∞

     

     

  • Question 16
    4 / -1

     

    At 298 K, given specific conductance of saturated
    AgCl solution=3.41 x 10-6 Ω-1 cm-1 and that of water used =1.60 x 10-6 Ω-1 cm-1.
    Equivalent conductance of saturated AgCl solution= 138.3 Ω-1 cm2 equiv-1

    Thus, solubility product (KSp) of AgCl is

     

    Solution

     

     

    Water has also conductance, hence to obtain pure value of conductance of AgCI, conductance of water should be substracted from the conductance of AgCI given.

    Pure specific conductance of AgCI = (3.41 x 10-6 - 1.60 x 10-6)

    = 1.81 x 10-6Ω -1cm-1

    AgCI is sparingly soluble, hence its solubility may be taken its concentration.
    Also, in saturated solution sparingly soluble salt

    ∴   

    ∴   AgCl (s)  Ag+ + Cl-

    ∴ Ksp = [Ag+] [Cl-] = S2
    = (1.31 x 10-5)2
    = 1.72 x 10-10 mol2L-2

     

     

  • Question 17
    4 / -1

     

    AgNO3 (aq) was added to an aq. KCl solution gradually and the conductivity of the solution was measured.the plot of conductance vs the volume of AgNO3 is

    [IIT JEE 2011]

     

    Solution

     

     


    When AgNO3 is added to KCI solution, in soluble AgCI(s) is formed. Thus, conductance a remains constant. After KCI has been completely precipitated as AgCI, further addition of AgNO3 causes increases in conductance.

     

     

  • Question 18
    4 / -1

     

    Resistance of 0.2 M solution of an electrolyte is 50Ω.The specific conductance of this solution is 1.4 Sm-1. The resistance  0.5 M solution of the same electrolyte is 280Ω. The molar conductivity of 0.5 solution of the electrolyte in Sm2mol-1 is

    [JEE Main 2014]

     

    Solution

     

     

    Specific conductance (0.2 M) = Conductance x Cell constant



    Specific conductance (0.5M)
    = Conductance x Cell constant

    k = 0.25 S m-1


    = 5 x 10-4 Sm2 mol-1
    Note 1 mol L-1 = 1 mol dm-3 = 103 mol m-3

     

     

  • Question 19
    4 / -1

     

    Molar conductivity of solution is represented by:

     

    Solution

     

    Molar conductivity of a solution is the conductivity of that volume containing 1 mole of an electrolyte when placed between two sufficiently large electrodes, which are 1 cm. apart. It is represented by Λm.

  • Question 20
    4 / -1

     

    Which of the following statements is incorrect about cell constant?

     

    Solution

     

    The correct answer is Option C.

    Cell constant can be defined as the ratio of the distance between conductance-titration electrodes to the area of the electrodes, measured from the determined resistance of a solution of known specific conductance. It has dimension length-1

     

  • Question 21
    4 / -1

     

    The inverse of resistance is called:

     

    Solution

     

    The correct answer is option D

    The inverse quantity is electrical conductance, and is the ease with which an electric current passes. The SI unit of electrical resistance is the ohm (Ω), while electrical conductance is measured in siemens (S).

  • Question 22
    4 / -1

     

    Matching List Type

    Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

    Q.

    An aqueous solution of X is added slowly to an aqueous solution of Y as Shown in column I.The variation in conductivity of these reactions is given in Column II.Match Column I with Column II and select the correct answer using the codes given below the lists.

     

    Solution

     

     

    (i) X and V both are weak electrolytes, hence have minimum conductance. When X [(C2H5)3N] is added to y (CH3COOH), ions are formed hence conductance increases.


    After complete reaction further addition of Y ,makes no change in conductance.
    Thus, (i) → (r)
    (ii) Net reaction is

    where X(I-) is added to Y(Ag+) insoluble Agl is formed. Hence, there is no change in conductance. After complete reaction of Ag+, further addition of I- causes increase in conductance.

    Thus, (ii) → (s)
    (iii) KOH is a strong electrolyte. As CH3COOH (X) is added, OH- is neutralised by CH3COOH.

    CH3COO- is hydrolysed at the same time.
    CH3COO- + H2O → CH3COOH + OH-
    But being in small extent, conductance decreases. After complete reaction of KOH, further addition of CH3COOH, (weak) there is no change in conductance.

    Thus, (iii) → (q)
    (iv) NaOH(X) and HI(Y) both are strong electrolytes, Thus, net reaction is

    When (X) is added to (Y), H2O is formed hence conductance decreases. After complete reaction of OH-, further addition of NaOH (X), causes increases in conductance.

    Thus, (iv) → (p)

     

     

  • Question 23
    4 / -1

     

    Limiting equivalent conductance (in S cm2 equiv-1) of a weak monobasic acid (HA) at infinite dilution is 100 and that of its 0.01 M solution is 5 at 298K.Match the parameters given in Column I with their values in Column II and select the answer from the codes given.           

     

    Solution

     

     




    ∴ [A-] = Cx, x = degree of ionisation 


    [A-] = Cx = 0.01 x 0.05 = 5 x 10-4 M
    [H+] = Cx = 5 x 10-4 M
    pH = - log [H+]
    = 4.log 5 = 3.3
    By Ostwald's dilution law

     

     

  • Question 24
    4 / -1

     

    Comprehension Type

    Direction : This section contains 2 paragraphs, each describing theory, experiments, data, etc. Four questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

    Passage I

    Consider the following solutions of an electrolyte 

    Q. 

    Conductivity.of 0.2 M solution is

     

    Solution

     

     

    For 0.1 M solution, k = 1.30 Sm-1
    Conductivity (k) = Conductance x Cell constant



     

     

  • Question 25
    4 / -1

     

    Passage I

    Consider the following solutions of an electrolyte 

    Q.

    Molar conductivity of 0.2 M solution is

     

    Solution

     

     

    For 0.1 M solution, k = 1.30 Sm-1
    Conductivity (k) = Conductance x Cell constant


     

     

  • Question 26
    4 / -1

     

    Passage II

    Given molar conductance of 0.001 M NH4OH solution at 298K = 3.0 x 10-3Sm2mol-1.
    Limiting molar conductance of

    aq. NH4CI = 1.50 x 10-2Sm2mol-1

    aq. NaCl = 1.26 x 10-2Sm2 mol-1

    and aq. NaOH = 2.48 x 10-2 Sm2 mol-1

    Q.

    Degree of dissociation of NH4OH at 298 K is 

     

    Solution

     

     

    By Kohlrausch’s law,

    = 1.50 x 10-2 + 2.48 x 10-2 - 1.26 x 10-2
    = 2.72 x 10-2 Sm2mol-1
     


     

     

  • Question 27
    4 / -1

     

    Passage II

    Given molar conductance of 0.001 M NH4OH solution at 298K = 3.0 x 10-3Sm2mol-1.
    Limiting molar conductance of

    aq. NH4CI = 1.50 x 10-2Sm2mol-1

    aq. NaCl = 1.26 x 10-2Sm2 mol-1

    and aq. NaOH = 2.48 x 10-2 Sm2 mol-1

    Q.

    pKb of NH4OH  is

     

    Solution

     

     

    By Kohlrausch’s law,

    = 1.50 x 10-2 + 2.48 x 10-2 - 1.26 x 10-2
    = 2.72 x 10-2 Sm2mol-1
     


     

     

  • Question 28
    4 / -1

     

    One Integer Value Correct Type

    This section contains 2 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

    Q.

    0.01 M aqueous solution of a dibasic acid is diluted to 0.004N such that equivalent conductance is x times.What is the value of x?

     

    Solution

     

    Dibasic acid is H2A   2H+ + A2- 
    ∴ 0.01M = 0.02N
    Equivalent conductance at 0.02 N

    Equivalent conductance at 0.004 N 

  • Question 29
    4 / -1

     

    At what serial number reciprocal of resistance is given?

    (1) Volt
    (2) Ampere
    (3) Coulomb
    (4) Faraday
    (5) Conductivity
    (6) Resistivity
    (7) Siemen
    (8) Debye

     

    Solution

     

    Reciprocal of resistance is Siemen (Conductance Ω-1).

     

  • Question 30
    4 / -1

     

    Only One Option Correct Type

    This section contains 1 multiple choice question. A question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

     Q.

    The equivalent conductance of two strong electrolytes at infinite dilution in H2O
    (where ions move freely through a solution) at 25°C are given below: 


    Q.

    What additional information/quantity are needed to calculate of an aqueous solution of acetic acid?

     

    Solution

     

     

     

     

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