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# Test: Conduction

## 30 Questions MCQ Test Chemistry for JEE | Test: Conduction

Description
This mock test of Test: Conduction for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Conduction (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Conduction quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Conduction exercise for a better result in the exam. You can find other Test: Conduction extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

### 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

### 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

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

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

Given limiting ionic conductance of  Hion :  = 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

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

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

Cell constant is maximum in case of a

Solution:

QUESTION: 10

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

0.1 M H2SOsolution is diluted to 0.01 M H2SO?.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

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 = Smmol-1

QUESTION: 13

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

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

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

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 cmequiv-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

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 AgNOis

[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

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

*Multiple options can be correct
QUESTION: 19

One or More than One Options Correct Type

This section contains 5 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

Equivalent conductance of 0.01 N CaClsolution is 126.36 S cm2 equiv-1. This is equal to

Solution:

The correct answer is Option B and C.
Equiv. conductance= 126.36×10-4 Scm2/eq.

If we convert the unit Scm2/eq.into Sm2/eq.

we get (126.36×10-4 Sm2/eq.)

which is equal to (252.72×10-4 Sm2/mol)

[since n factor for CaCl2 is 2]

*Multiple options can be correct
QUESTION: 20

Degree of dissociation of very pure water is 1.9 x 10-9. Also limiting molar conductance of λ0(H+) = 350 Scmequiv-1 ,  λ0(OH+) = 200 Scmequiv-1

Thus,equivalent conductance of water is

Solution:

= (350 + 200) Scm2 equiv-1
= 550 Scm2equiv-1

= 1.045 x 10-10 Sm2 equiv-1

*Multiple options can be correct
QUESTION: 21

Select the correct statement(s).

Solution:

For strong electrolyte, variation in graph is uniform, hence can be extrapolated to
In case of weak electrolyte, variation of is rapid in lower concentration range (higher dilution), hence graph is not uniform and cannot be obtained by extrapolation to
Thus, (b) is incorrect.

Thus, correct based on Kohlrausch’s law.

Thus, it is also correct based on Kohlrausch's law.

QUESTION: 22

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

Limiting equivalent conductance (in S cmequiv-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

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

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

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

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

*Answer can only contain numeric values
QUESTION: 28

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 H2 2H+ + A2-
∴ 0.01M = 0.02N
Equivalent conductance at 0.02 N

Equivalent conductance at 0.004 N

*Answer can only contain numeric values
QUESTION: 29

At what serial number reciprocal of resistance is given?

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

Solution:

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

QUESTION: 30

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: