Electrochemistry Test

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Electrochemistry Test - 22

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Weekly Quiz Competition

Question 1
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Only One Option Correct Type
Direction (Q. Nos. 1-18) This section contains 18 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
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)

A

B

C

D

Solution
Question 2
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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?

A
Increases

B
Decreases

C
Remains Unchanged

D
First increases and then decreases

Solution

On dilution, molarity decreases and molar conductance increases, Hence, specific conductance decreases.
Question 3
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An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

A
increase in number of ions

B
increase in ionic mobility of ions

C
100% ionisation of electrolyte at normal dilution.

D
increase in both, i.e number of ions and ionic mobility of ions.

Solution

Number of ions remains constant but due to dilution, ionic mobility increases (as ionic attraction decreases). Hence, equivalent conductance increases.
Question 4
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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)

A

B

C

D
None of these

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

A

B

C

D
None Of these

Solution

For a weak acid.
Question 6
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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

A
a

B
b

C
c

D
d

Solution

Since, equivalent mass of H₂SO₄ = 1/2 molar mass
Hence, = 430 x 2 = 860 S cm² mol^-1
Question 7
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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)

A
10x

B
200x

C
x

D
5000x

Solution

Molar conductance
Question 8
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Resistivity of a metal is equal to resistance when cell is constant is

A
1 cm^-1

B
0.1 cm^-1

C
10 cm^-1

D
100 cm^-1

Solution

Resistance = (Resistivity) x Cell constant
∴ Resistance = Resistivity
when cell constant = 1cm^-1
Question 9
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Cell constant is maximum in case of a

A
Wire of length 100 cm and area 100 cm²

B
Wire of length 10 cm and area 10 cm²

C
One centimetre cube of material

D
Equal in all cases

Solution
Question 10
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Which quantity is temperature independent?

A
Resistivity

B
Conductivity

C
Cell constant

D
Cell EMF

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
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0.1 M H₂SO₄solution is diluted to 0.01 M H₂SO₄?.Hence its molar conductivity will be

A
100 times

B
1/10th times

C
10 times

D
1000 times

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
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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

A
S m mol^-1

B
S² m² mol^-1

C
S m² mol^-1

D
S² m² mol

Solution

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

S = k (Specific conductivity) mol m^-2
∴ k = Sm²mol^-1
Question 13
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Given, (Scm² mol^-1)for different electrolytes
Thus, of CH₃COOH is

A
516.5

B
420.3

C
172.9

D
390.7

Solution

(CH₃COOH) = (CH₃COONa) + (HCI) - (NaCI)
= 91.0+ 426.2 -126.5
= 517.2 - 126.5
= 390.7 Scm²mol^-1
Question 14
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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]

A
6250 S m² mol^-1

B
6.25 x 10^-4 S m² mol^-1

C
62.5 x 10^-4 S m² mol^-1

D
62.5 S m² mol^-1

Solution

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

Specific conductance of 0.4M solution
Question 15
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The equivalent conductance of NaCl at concentration C and at infinite dilution are λ_cand λ_∞ respectively.The correct relationship λ_cand λ_∞ between is given as (where constanr B is positive).

A

B

C

D

Solution

By Debye-Huckel Onsager equation

when, λ_C = equivalent conductivity at concentration C and λ_∞ = limiting equivalent conductivity at infinite dilution
Limit C → 0. λ_C = λ_∞
Question 16
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At 298 K, given specific conductance of saturated
AgCl solution=3.41 x 10^-6Ω^-1cm^-1and that of water used =1.60 x 10^-6Ω^-1cm^-1.
Equivalent conductance of saturated AgCl solution= 138.3 Ω^-1cm²equiv^-1
Thus, solubility product (K_Sp) of AgCl is

A
1.81 x 10^-6

B
1.31 x 10^-5

C
1.72 x 10^-10

D
1.72 x 10^-8

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
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AgNO_{3 (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 AgNO₃is
[IIT JEE 2011]

A

B

C

D

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
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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^-1is
[JEE Main 2014]

A
5 x 10^-4

B
5 x 10^-3

C
5 x 10³

D
5 x 10²

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
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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.

A
a

B
b

C
c

D
d

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 20
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Limiting equivalent conductance (in S cm²equiv^-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.

A
a

B
b

C
c

D
d

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