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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.
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QUESTION: 1

**Only One Option Correct Type**

Direction (Q. Nos. 1-19) This section contains 19 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})**

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 H_{2}O 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 (K_{a}) of weak acid is

Solution:

For a weak acid.

QUESTION: 6

Given limiting ionic conductance of H^{+ }ion : = 350 S cm^{2} equi^{-1} and of ion : = 80 S cm^{2} equi^{-1} Thus for H_{2}SO_{4}, limiting values of molar conductance and equivalent conductance are

Solution:

Since, equivalent mass of H_{2}SO_{4} = 1/2 molar mass

Hence, = 430 x 2 = 860 S cm^{2} 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 Scm^{2} 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 H_{2}SO_{4 }solution is diluted to 0.01 M H_{2}SO_{4 }?.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 (m^{2})

∴

S = k (Specific conductivity) mol m^{-2}

∴ k = Sm^{2 }mol^{-1}

QUESTION: 13

Given, (Scm^{2} mol^{-1})for different electrolytes

Thus, of CH_{3}COOH is

Solution:

(CH_{3}COOH) = (CH_{3}COONa) + (HCI) - (NaCI)

= 91.0+ 426.2 -126.5

= 517.2 - 126.5

= 390.7 Scm^{2}mol^{-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 λ_{c }and λ_{∞} respectively.The correct relationship λ_{c }and λ_{∞} between is given as (where constanr B is positive).

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

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 }cm^{2 }equiv^{-1 }

**Thus, solubility product (K _{Sp}) 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^{-6}Ω ^{-1}cm^{-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^{-}] = S^{2}

= (1.31 x 10^{-5})^{2}

= 1.72 x 10^{-10} mol^{2}L^{-2}

QUESTION: 17

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 AgNO_{3 }is

**[IIT JEE 2011]**

Solution:

When AgNO_{3} 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 AgNO_{3} 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 Sm^{2}mol^{-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} Sm^{2} mol^{-1}

Note 1 mol L^{-1} = 1 mol dm^{-3} = 10^{3} 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 CaCl _{2 }solution is 126.36 S cm^{2} 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 Scm^{2}/eq.into Sm2/eq.

we get (126.36×10^{-4} Sm^{2}/eq.)

which is equal to (252.72×10^{-4} Sm^{2}/mol)

[since n factor for CaCl^{2} 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 λ_{0}(H^{+}) = 350 Scm^{2 }equiv^{-1 },^{ }λ_{0}(OH^{+}) = 200 Scm^{2 }equiv^{-1 }

Thus,equivalent conductance of water is

Solution:

= (350 + 200) Scm^{2} equiv^{-1}

= 550 Scm^{2}equiv^{-1}

= 1.045 x 10^{-10} Sm^{2} 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 [(C_{2}H_{5})_{3}N] is added to y (CH_{3}COOH), 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 CH_{3}COOH (X) is added, OH^{-} is neutralised by CH_{3}COOH.

CH_{3}COO^{-} is hydrolysed at the same time.

CH_{3}COO^{-} + H_{2}O → CH_{3}COOH + OH^{-}

But being in small extent, conductance decreases. After complete reaction of KOH, further addition of CH_{3}COOH, (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), H_{2}O 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 cm^{2 }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

**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 NH_{4}OH solution at 298K = 3.0 x 10^{-3}Sm^{2}mol^{-1}.

Limiting molar conductance of

aq. NH_{4}CI = 1.50 x 10^{-2}Sm^{2}mol^{-1}

aq. NaCl = 1.26 x 10^{-2}Sm^{2} mol^{-1}

and aq. NaOH = 2.48 x 10^{-2} Sm^{2} mol^{-1}

**Q.**

**Degree of dissociation of NH _{4}OH 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} Sm^{2}mol^{-1}

QUESTION: 27

**Passage II**

Given molar conductance of 0.001 M NH_{4}OH solution at 298K = 3.0 x 10^{-3}Sm^{2}mol^{-1}.

Limiting molar conductance of

aq. NH_{4}CI = 1.50 x 10^{-2}Sm^{2}mol^{-1}

aq. NaCl = 1.26 x 10^{-2}Sm^{2} mol^{-1}

and aq. NaOH = 2.48 x 10^{-2} Sm^{2} mol^{-1}

**Q.**

**pK _{b} of NH_{4}OH 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} Sm^{2}mol^{-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 H_{2}A 2H^{+} + A^{2-}

∴ 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

(4) Faraday

(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 H_{2}O

(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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