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*Multiple options can be correct

QUESTION: 1

Which of the following options are correct:

Solution:

*Multiple options can be correct

QUESTION: 2

The correct expressions between conc. mobility and molar conductivity are:

Solution:

*Multiple options can be correct

QUESTION: 3

Which of the following are correct:

Solution:

QUESTION: 4

Which of the following expressions are correct:

Solution:

**Correct Answer :- b**

**Explanation :** Al_{2}(SO_{4})_{3} --------> 2A(Al^{3+}) + 3SO_{4}^{2-}

We can calculate the equivalent conductance only for ions, so the equivalent conductance at infinite dilution,

∧∞eq = ∧^{o}Al^{3+} + ∧^{o}SO_{4}^{2-}

*Multiple options can be correct

QUESTION: 5

Which of the following plots are correct:

Solution:

*Multiple options can be correct

QUESTION: 6

The correct expressions of molar conductivity are:

Solution:

*Multiple options can be correct

QUESTION: 7

The correct expressions for dissociation constant of weak acid are:

Solution:

*Multiple options can be correct

QUESTION: 8

The correct expressions which explain the Debye-Huckel-Onsager Theory for strong electrolytes are

Solution:

QUESTION: 9

The correct order of molar conductivities at infinite dilution are:

Solution:

**Correct Answer :- d**

**Explanation : ** Li+ would be the smallest because it has only 2 electrons and therefore the least electron- electron repulsion.

K+ is larger than Na+ because the ionic radius increases in a particular group on moving from top to bottom due to increase in the principle energy shell though the number or electrons in the valence shell remain the same.

The conductivity of an ion depends mainly on its mobility. That is how fast it moves in an ionic solution. H+ being the smallest ion moves fast. So it has the highest conductivity.

*Multiple options can be correct

QUESTION: 10

The correct expressions for equivalent conductance are:

Solution:

**G is conductance and l is length.**

**We know that **

**Now, G(conductance) = KA/l => K = Gl/A and Veq = Al**

**So, (Gl/A)(Al) = Gl2 **

**Hence A and B are correct.**

*Multiple options can be correct

QUESTION: 11

Choose the correct statements:

Solution:

QUESTION: 12

Choose the correct options:

Solution:

**Correct Answer :- c**

**Explanation :** molar conductance = Valency × equivalent conductance.

Al_{2}(SO_{4})_{3} is commonly known as aluminium sulphate. There is two aluminium atom in aluminium sulphate, and because of this, the total valency of the formula is 6+

A_{M} = 6_{eq} for Al_{2}(SO_{4})_{3}

*Multiple options can be correct

QUESTION: 13

Which of the following will have same value of molar conductivity and equivalent conductivity:

Solution:

*Multiple options can be correct

QUESTION: 14

If = 73.5Ω^{-1}cm^{2} mol^{-1} = 189Ω^{-1}cm^{2} mol^{-1} and = 160Ω^{-1}cm^{2} mol^{-1} then:

Solution:

Potash alum is K_{2}SO_{4} · Al_{2}(SO_{4})_{3} · 24H_{2}O

On dilutionsions will be 2K^{+}, 2Al^{3+} and 4SO_{4}^{2–}

∧_{M} = (2x73.5)+ (2x189) + (4x160)

= **1165 S cm ^{2} mol^{-1}**

*Multiple options can be correct

QUESTION: 15

The unit of conductance is (are):

Solution:

*Multiple options can be correct

QUESTION: 16

The unit of conductivity is(are):

Solution:

**A and C are correct.**

**R = ρ l/A**

**ρ = RA/l**

**Units of ρ = Ω cm**

**σ = 1/ρ**

**Units of σ = Ω ^{-1} cm^{-1 }**

**Ω ^{-1} = S (Siemens)**

**Units of σ = S cm ^{-1 }**

QUESTION: 17

Choose the correct order of molar ionic conductivities of the following ions.

Solution:

*Multiple options can be correct

QUESTION: 18

For which of the following relation is correct:

Solution:

*Multiple options can be correct

QUESTION: 19

Choose the correct statements:

Solution:

QUESTION: 20

On dilution, the:

Solution:

*Answer can only contain numeric values

QUESTION: 21

The equivalent conductance of 0.01 M solution of K_{2}SO_{4} whose specific conductance is 1.26 × 10^{–3} Ω^{-1} cm^{-1} are ________ Ω^{-1}cm^{2} eq^{-1 }

Solution:

*Answer can only contain numeric values

QUESTION: 22

The conductivity of 0.1 M NaOH solution is 0.0221 Ω^{-1} cm^{-1}. When an equal volume of 0.1 M HCl is added, the conductivity decreases to 0.0056 Ω^{-1} cm^{-1}. A further addit ion HCl solution, the volume of which of which is equal to that of the first portion added, the conductivity increases to 0.017 Ω^{-1} cm^{-1}.

Solution:

*Answer can only contain numeric values

QUESTION: 23

The conductivity of 0.1 M NaOH solution is 0.0221 Ω^{-1} cm^{-1}. When an equal volume of 0.1 M HCl is added, the conductivity decreases to 0.0056 Ω^{-1} cm^{-1}. A further addit ion HCl solution, the volume of which of which is equal to that of the first portion added, the conductivity increases to 0.017 Ω^{-1} cm^{-1}. ________, _____ and _______ Ω^{-1} cm^{-1} respectively:

Solution:

*Answer can only contain numeric values

QUESTION: 24

The specific conductance of a saturated solution of AgCl at 25ºC after subtracting the specific conductance of water is 2.28 × 10^{–4} Sm^{–1}. The molar ionic conductance of Ag^{+} and Cl^{–} ions are 73.3 × 10^{–4} and 65 × 10^{–4} Sm^{2} mol^{–1}. The solubility of AgCl at 25ºC is _______ g/dm^{3}.

Solution:

= (73.3 + 65) ×10^{-4} S m^{2} mol^{-1} = 0.01383 S m^{2} mol^{-1} = 0.01383 ×10^{4} S cm^{2} mol^{-1} K = 2.28 ×10^{-4} S m^{-1} = 2.28 ×10^{-6} S cm^{-1}

1mol of AgCl = 108 + 35.5 = 143.5g

B = 1.648 ×10^{-5} ×143.5g / L or g / dm^{3}

= 2.36 ×10^{-3} g dm^{-3}

*Answer can only contain numeric values

QUESTION: 25

The ionic conductivities at infinite dilution of O_{x}^{2–}, K^{+} and Na^{+} are 148.2, 50.1 and 73.5 Ω^{-1} cm^{2} mol^{-1} respectively. The equivalent conductivities at infinite dilution of the salt KOOC. COONa is _______ Ω^{-1} cm^{2} mol^{-1}

Solution:

= 50.1 + 148.2 + 73.5 = 271.8Ω^{-1}cm ^{2} mol^{-1}

*Answer can only contain numeric values

QUESTION: 26

The molar conductivity of NH4Cl at infinite dilution is 149.7 Ω^{-1} cm^{2} mol^{-1} and the ionic conductivities of OH^{-} and Cl^{-} ions are 198 and 76.3 Ω^{-1} cm^{2} mol^{-1} respectively. The molar conductivity of NH_{4}OH at the dilution is _______ Ω^{-1} cm^{2} mol^{-1}:

Solution:

= (149.7 - 76.3 + 198)Ω^{-1}cm^{2} mol^{-1}

= 271.4Ω^{-1}cm^{2} mol^{-1}

*Answer can only contain numeric values

QUESTION: 27

A conductivity cell whose cell constant is 2cm^{–1} is filled with 0.1 M acetic acid solution. Its resistance is found to be 3765 Ω. The degree of dissociation of acetic acid are ______ . Given that and

Solution:

*Answer can only contain numeric values

QUESTION: 28

The conductivity of a saturated solution of silver okalate is 4.5 × 10^{–5} Ω^{-1} cm^{-1}. If its K_{sp} = 1.35 × 10^{-11} M^{3}, the molar conductivity of the saturated solution would be _______ Ω^{-1} cm^{2} mol^{-1}.

Solution:

*Answer can only contain numeric values

QUESTION: 29

The molar conductivity of 0.05 M of solution of an electrolyte is 200 Ω^{-1} cm^{2} mol^{-1}. The resistance offered by a conductivity cell with cell constant (1/3 cm^{-1}) would be _______Ω.

Solution:

*Answer can only contain numeric values

QUESTION: 30

A conductivity cell whose cell constant is 3 cm^{–1} is filled with 0.1 M solution of weak acids. Its resistance is found to be 3000 Ω. the degree of dissociation of weak acid is _______.

Solution:

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