QBManager

ELECTROCHEMISTRY

276151 The specific conductivity of $0.1 N KCl$ solution is $0.0129 {ohm}^{- 1} {cm}^{- 1}$ . The resistance of the solution in the cell is $100 ohm$ . The cell constant of the cell will be :

1 0.56

2 2.80

3 1.10

4 1.29

Explanation:

Given,
specific conductivity $(κ) = 0.0129 {ohm}^{- 1} {cm}^{- 1}$ and resistance $(R) = 100 ohm$
We know,
Specific conductivity $(κ) = \frac{1}{R} \times$ Cell constant
$\begin{aligned} Cell constant & = κ \times R \\ = 0.0129 \times 100 \\ = 1.29 \end{aligned}$

Manipal-2019

ELECTROCHEMISTRY

276152 The resistance of $\frac{1}{10} M$ solution is $2.5 \times 10^{3} ohm$ .
What is the molar conductivity of solution? (cell constant $= 1.25 {cm}^{- 1}$ )

1

3.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

5.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

2.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

2.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

Given, resistance (R) $2.5 \times 10^{3} ohm$
Cell constant $(G^{*}) = 1.25 {cm}^{- 1}$
We know,
Specific conductivity $(κ) = \frac{G^{*}}{R}$
$= \frac{1.25}{2.5 \times 10^{3}} = 5 \times 10^{- 4} {ohm}^{- 1} {cm}^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{m}$
$\begin{aligned} Λ_{m} = 5 \times 10^{- 4} \times 1000 \times 10 \\ Λ_{m} = 5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276154 Consider the statements $S_{1}$ and $S_{2}$ :
$S_{1}$ : Conductivity always increases with decrease in the concentration of electrolyte.
$S_{2}$ : Molar conductivity always increase with decrease in the concentration of electrolyte. The correct option among the following is

1

S_{1}

is correct and

S_{2}

is wrong

2

S_{1}

is wrong and

S_{2}

is correct

3 Both

S_{1}

and

S_{2}

are wrong

4 Both

S_{1}

and

S_{2}

are correct

Explanation:

On dilution number of ions per unit volume decrease so conductivity decreases with dilution and hence, $S_{1}$ is wrong.
Conductivity $(κ) \propto$ Concentration
Molar conductivity $(C) \propto \frac{1}{Concentration}$
On dilution $(C)$ and $(κ)$ both decrease but the effect of (C) is more dominating so $Λ_{m}$ increases hence $S_{2}$ is right.
$Λ_{m} = \frac{1000 \times κ}{C}$

JEE Main 2019

ELECTROCHEMISTRY

276156 $Λ_{m}^{o}$ for $NaCl, HCl$ and $NaA$ are 126.4, 425.9 and $100.5 S cm {mol}^{- 1}$ , respectively. If the conductivity of $0.001 M HA$ is $5 \times 10^{- 5} S {cm}^{- 1}$ , degree of dissociation of $HA$ is

1 0.25

2 0.50

3 0.75

4 0.125

Explanation:

Given that,
$\begin{aligned} Λ_{m} NaCl = 126.45 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 425.95 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} NaA = 100.55 {cm}^{2} 55 {cm}^{2} {mol}^{- 1} \end{aligned}$
According to Formula-
$\begin{aligned} Λ_{m}^{\circ} HA = Λ_{m}^{\circ} HCl + Λ_{m}^{\circ} NaA - Λ_{m}^{\circ} NaCl \\ = 425.9 + 100.5 - 126.4 \\ = 400 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} = \frac{1000 κ}{M} = \frac{1000 \times 5 \times 10^{- 5}}{10^{- 3}} \end{aligned}$
$\begin{aligned} = 5 \times 10^{- 5} \times \frac{1000}{0.001} = 50 S {cm}^{2} {mol}^{- 1} \\ a = \frac{Λ_{m}}{Λ_{m}} = \frac{50}{400} = 0.125 \end{aligned}$

JEE Main 2019

ELECTROCHEMISTRY

276157 The decreasing order of electrical conductivity of the following aqueous solution is
0.1 M formic acid (A),
$0.1 M$ acetic acid (B),
0.1 M benzoic acid (C).

1

A > C > B

2

C > B > A

3

A > B > C

4

C > A > B

Explanation:

Electrical conductivity of the aqueous solution depends on the degree of ionisation. Degree of ionisation is directly proportional to the acidic strength. $\begin{array}{ccc} A & B & C \\ HCOOH & C_{6} H_{5} COOH & {CH}_{3} COOH \end{array}$
As the acidic strength decreases rate of dissociation decreases and hence conductivity decreases.
Order of acidic strength
$A > C > B$
Acidic strength $↑=$ degree of ionization $↑$

JEE Main 2019

ELECTROCHEMISTRY

276151 The specific conductivity of $0.1 N KCl$ solution is $0.0129 {ohm}^{- 1} {cm}^{- 1}$ . The resistance of the solution in the cell is $100 ohm$ . The cell constant of the cell will be :

1 0.56

2 2.80

3 1.10

4 1.29

Explanation:

Given,
specific conductivity $(κ) = 0.0129 {ohm}^{- 1} {cm}^{- 1}$ and resistance $(R) = 100 ohm$
We know,
Specific conductivity $(κ) = \frac{1}{R} \times$ Cell constant
$\begin{aligned} Cell constant & = κ \times R \\ = 0.0129 \times 100 \\ = 1.29 \end{aligned}$

Manipal-2019

ELECTROCHEMISTRY

276152 The resistance of $\frac{1}{10} M$ solution is $2.5 \times 10^{3} ohm$ .
What is the molar conductivity of solution? (cell constant $= 1.25 {cm}^{- 1}$ )

1

3.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

5.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

2.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

2.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

Given, resistance (R) $2.5 \times 10^{3} ohm$
Cell constant $(G^{*}) = 1.25 {cm}^{- 1}$
We know,
Specific conductivity $(κ) = \frac{G^{*}}{R}$
$= \frac{1.25}{2.5 \times 10^{3}} = 5 \times 10^{- 4} {ohm}^{- 1} {cm}^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{m}$
$\begin{aligned} Λ_{m} = 5 \times 10^{- 4} \times 1000 \times 10 \\ Λ_{m} = 5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276154 Consider the statements $S_{1}$ and $S_{2}$ :
$S_{1}$ : Conductivity always increases with decrease in the concentration of electrolyte.
$S_{2}$ : Molar conductivity always increase with decrease in the concentration of electrolyte. The correct option among the following is

1

S_{1}

is correct and

S_{2}

is wrong

2

S_{1}

is wrong and

S_{2}

is correct

3 Both

S_{1}

and

S_{2}

are wrong

4 Both

S_{1}

and

S_{2}

are correct

Explanation:

On dilution number of ions per unit volume decrease so conductivity decreases with dilution and hence, $S_{1}$ is wrong.
Conductivity $(κ) \propto$ Concentration
Molar conductivity $(C) \propto \frac{1}{Concentration}$
On dilution $(C)$ and $(κ)$ both decrease but the effect of (C) is more dominating so $Λ_{m}$ increases hence $S_{2}$ is right.
$Λ_{m} = \frac{1000 \times κ}{C}$

JEE Main 2019

ELECTROCHEMISTRY

276156 $Λ_{m}^{o}$ for $NaCl, HCl$ and $NaA$ are 126.4, 425.9 and $100.5 S cm {mol}^{- 1}$ , respectively. If the conductivity of $0.001 M HA$ is $5 \times 10^{- 5} S {cm}^{- 1}$ , degree of dissociation of $HA$ is

1 0.25

2 0.50

3 0.75

4 0.125

Explanation:

Given that,
$\begin{aligned} Λ_{m} NaCl = 126.45 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 425.95 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} NaA = 100.55 {cm}^{2} 55 {cm}^{2} {mol}^{- 1} \end{aligned}$
According to Formula-
$\begin{aligned} Λ_{m}^{\circ} HA = Λ_{m}^{\circ} HCl + Λ_{m}^{\circ} NaA - Λ_{m}^{\circ} NaCl \\ = 425.9 + 100.5 - 126.4 \\ = 400 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} = \frac{1000 κ}{M} = \frac{1000 \times 5 \times 10^{- 5}}{10^{- 3}} \end{aligned}$
$\begin{aligned} = 5 \times 10^{- 5} \times \frac{1000}{0.001} = 50 S {cm}^{2} {mol}^{- 1} \\ a = \frac{Λ_{m}}{Λ_{m}} = \frac{50}{400} = 0.125 \end{aligned}$

JEE Main 2019

ELECTROCHEMISTRY

276157 The decreasing order of electrical conductivity of the following aqueous solution is
0.1 M formic acid (A),
$0.1 M$ acetic acid (B),
0.1 M benzoic acid (C).

1

A > C > B

2

C > B > A

3

A > B > C

4

C > A > B

Explanation:

Electrical conductivity of the aqueous solution depends on the degree of ionisation. Degree of ionisation is directly proportional to the acidic strength. $\begin{array}{ccc} A & B & C \\ HCOOH & C_{6} H_{5} COOH & {CH}_{3} COOH \end{array}$
As the acidic strength decreases rate of dissociation decreases and hence conductivity decreases.
Order of acidic strength
$A > C > B$
Acidic strength $↑=$ degree of ionization $↑$

JEE Main 2019

ELECTROCHEMISTRY

276151 The specific conductivity of $0.1 N KCl$ solution is $0.0129 {ohm}^{- 1} {cm}^{- 1}$ . The resistance of the solution in the cell is $100 ohm$ . The cell constant of the cell will be :

1 0.56

2 2.80

3 1.10

4 1.29

Explanation:

Given,
specific conductivity $(κ) = 0.0129 {ohm}^{- 1} {cm}^{- 1}$ and resistance $(R) = 100 ohm$
We know,
Specific conductivity $(κ) = \frac{1}{R} \times$ Cell constant
$\begin{aligned} Cell constant & = κ \times R \\ = 0.0129 \times 100 \\ = 1.29 \end{aligned}$

Manipal-2019

ELECTROCHEMISTRY

276152 The resistance of $\frac{1}{10} M$ solution is $2.5 \times 10^{3} ohm$ .
What is the molar conductivity of solution? (cell constant $= 1.25 {cm}^{- 1}$ )

1

3.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

5.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

2.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

2.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

Given, resistance (R) $2.5 \times 10^{3} ohm$
Cell constant $(G^{*}) = 1.25 {cm}^{- 1}$
We know,
Specific conductivity $(κ) = \frac{G^{*}}{R}$
$= \frac{1.25}{2.5 \times 10^{3}} = 5 \times 10^{- 4} {ohm}^{- 1} {cm}^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{m}$
$\begin{aligned} Λ_{m} = 5 \times 10^{- 4} \times 1000 \times 10 \\ Λ_{m} = 5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276154 Consider the statements $S_{1}$ and $S_{2}$ :
$S_{1}$ : Conductivity always increases with decrease in the concentration of electrolyte.
$S_{2}$ : Molar conductivity always increase with decrease in the concentration of electrolyte. The correct option among the following is

1

S_{1}

is correct and

S_{2}

is wrong

2

S_{1}

is wrong and

S_{2}

is correct

3 Both

S_{1}

and

S_{2}

are wrong

4 Both

S_{1}

and

S_{2}

are correct

Explanation:

On dilution number of ions per unit volume decrease so conductivity decreases with dilution and hence, $S_{1}$ is wrong.
Conductivity $(κ) \propto$ Concentration
Molar conductivity $(C) \propto \frac{1}{Concentration}$
On dilution $(C)$ and $(κ)$ both decrease but the effect of (C) is more dominating so $Λ_{m}$ increases hence $S_{2}$ is right.
$Λ_{m} = \frac{1000 \times κ}{C}$

JEE Main 2019

ELECTROCHEMISTRY

276156 $Λ_{m}^{o}$ for $NaCl, HCl$ and $NaA$ are 126.4, 425.9 and $100.5 S cm {mol}^{- 1}$ , respectively. If the conductivity of $0.001 M HA$ is $5 \times 10^{- 5} S {cm}^{- 1}$ , degree of dissociation of $HA$ is

1 0.25

2 0.50

3 0.75

4 0.125

Explanation:

Given that,
$\begin{aligned} Λ_{m} NaCl = 126.45 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 425.95 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} NaA = 100.55 {cm}^{2} 55 {cm}^{2} {mol}^{- 1} \end{aligned}$
According to Formula-
$\begin{aligned} Λ_{m}^{\circ} HA = Λ_{m}^{\circ} HCl + Λ_{m}^{\circ} NaA - Λ_{m}^{\circ} NaCl \\ = 425.9 + 100.5 - 126.4 \\ = 400 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} = \frac{1000 κ}{M} = \frac{1000 \times 5 \times 10^{- 5}}{10^{- 3}} \end{aligned}$
$\begin{aligned} = 5 \times 10^{- 5} \times \frac{1000}{0.001} = 50 S {cm}^{2} {mol}^{- 1} \\ a = \frac{Λ_{m}}{Λ_{m}} = \frac{50}{400} = 0.125 \end{aligned}$

JEE Main 2019

ELECTROCHEMISTRY

276157 The decreasing order of electrical conductivity of the following aqueous solution is
0.1 M formic acid (A),
$0.1 M$ acetic acid (B),
0.1 M benzoic acid (C).

1

A > C > B

2

C > B > A

3

A > B > C

4

C > A > B

Explanation:

Electrical conductivity of the aqueous solution depends on the degree of ionisation. Degree of ionisation is directly proportional to the acidic strength. $\begin{array}{ccc} A & B & C \\ HCOOH & C_{6} H_{5} COOH & {CH}_{3} COOH \end{array}$
As the acidic strength decreases rate of dissociation decreases and hence conductivity decreases.
Order of acidic strength
$A > C > B$
Acidic strength $↑=$ degree of ionization $↑$

JEE Main 2019

ELECTROCHEMISTRY

276151 The specific conductivity of $0.1 N KCl$ solution is $0.0129 {ohm}^{- 1} {cm}^{- 1}$ . The resistance of the solution in the cell is $100 ohm$ . The cell constant of the cell will be :

1 0.56

2 2.80

3 1.10

4 1.29

Explanation:

Given,
specific conductivity $(κ) = 0.0129 {ohm}^{- 1} {cm}^{- 1}$ and resistance $(R) = 100 ohm$
We know,
Specific conductivity $(κ) = \frac{1}{R} \times$ Cell constant
$\begin{aligned} Cell constant & = κ \times R \\ = 0.0129 \times 100 \\ = 1.29 \end{aligned}$

Manipal-2019

ELECTROCHEMISTRY

276152 The resistance of $\frac{1}{10} M$ solution is $2.5 \times 10^{3} ohm$ .
What is the molar conductivity of solution? (cell constant $= 1.25 {cm}^{- 1}$ )

1

3.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

5.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

2.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

2.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

Given, resistance (R) $2.5 \times 10^{3} ohm$
Cell constant $(G^{*}) = 1.25 {cm}^{- 1}$
We know,
Specific conductivity $(κ) = \frac{G^{*}}{R}$
$= \frac{1.25}{2.5 \times 10^{3}} = 5 \times 10^{- 4} {ohm}^{- 1} {cm}^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{m}$
$\begin{aligned} Λ_{m} = 5 \times 10^{- 4} \times 1000 \times 10 \\ Λ_{m} = 5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276154 Consider the statements $S_{1}$ and $S_{2}$ :
$S_{1}$ : Conductivity always increases with decrease in the concentration of electrolyte.
$S_{2}$ : Molar conductivity always increase with decrease in the concentration of electrolyte. The correct option among the following is

1

S_{1}

is correct and

S_{2}

is wrong

2

S_{1}

is wrong and

S_{2}

is correct

3 Both

S_{1}

and

S_{2}

are wrong

4 Both

S_{1}

and

S_{2}

are correct

Explanation:

On dilution number of ions per unit volume decrease so conductivity decreases with dilution and hence, $S_{1}$ is wrong.
Conductivity $(κ) \propto$ Concentration
Molar conductivity $(C) \propto \frac{1}{Concentration}$
On dilution $(C)$ and $(κ)$ both decrease but the effect of (C) is more dominating so $Λ_{m}$ increases hence $S_{2}$ is right.
$Λ_{m} = \frac{1000 \times κ}{C}$

JEE Main 2019

ELECTROCHEMISTRY

276156 $Λ_{m}^{o}$ for $NaCl, HCl$ and $NaA$ are 126.4, 425.9 and $100.5 S cm {mol}^{- 1}$ , respectively. If the conductivity of $0.001 M HA$ is $5 \times 10^{- 5} S {cm}^{- 1}$ , degree of dissociation of $HA$ is

1 0.25

2 0.50

3 0.75

4 0.125

Explanation:

Given that,
$\begin{aligned} Λ_{m} NaCl = 126.45 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 425.95 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} NaA = 100.55 {cm}^{2} 55 {cm}^{2} {mol}^{- 1} \end{aligned}$
According to Formula-
$\begin{aligned} Λ_{m}^{\circ} HA = Λ_{m}^{\circ} HCl + Λ_{m}^{\circ} NaA - Λ_{m}^{\circ} NaCl \\ = 425.9 + 100.5 - 126.4 \\ = 400 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} = \frac{1000 κ}{M} = \frac{1000 \times 5 \times 10^{- 5}}{10^{- 3}} \end{aligned}$
$\begin{aligned} = 5 \times 10^{- 5} \times \frac{1000}{0.001} = 50 S {cm}^{2} {mol}^{- 1} \\ a = \frac{Λ_{m}}{Λ_{m}} = \frac{50}{400} = 0.125 \end{aligned}$

JEE Main 2019

ELECTROCHEMISTRY

276157 The decreasing order of electrical conductivity of the following aqueous solution is
0.1 M formic acid (A),
$0.1 M$ acetic acid (B),
0.1 M benzoic acid (C).

1

A > C > B

2

C > B > A

3

A > B > C

4

C > A > B

Explanation:

Electrical conductivity of the aqueous solution depends on the degree of ionisation. Degree of ionisation is directly proportional to the acidic strength. $\begin{array}{ccc} A & B & C \\ HCOOH & C_{6} H_{5} COOH & {CH}_{3} COOH \end{array}$
As the acidic strength decreases rate of dissociation decreases and hence conductivity decreases.
Order of acidic strength
$A > C > B$
Acidic strength $↑=$ degree of ionization $↑$

JEE Main 2019

ELECTROCHEMISTRY

276151 The specific conductivity of $0.1 N KCl$ solution is $0.0129 {ohm}^{- 1} {cm}^{- 1}$ . The resistance of the solution in the cell is $100 ohm$ . The cell constant of the cell will be :

1 0.56

2 2.80

3 1.10

4 1.29

Explanation:

Given,
specific conductivity $(κ) = 0.0129 {ohm}^{- 1} {cm}^{- 1}$ and resistance $(R) = 100 ohm$
We know,
Specific conductivity $(κ) = \frac{1}{R} \times$ Cell constant
$\begin{aligned} Cell constant & = κ \times R \\ = 0.0129 \times 100 \\ = 1.29 \end{aligned}$

Manipal-2019

ELECTROCHEMISTRY

276152 The resistance of $\frac{1}{10} M$ solution is $2.5 \times 10^{3} ohm$ .
What is the molar conductivity of solution? (cell constant $= 1.25 {cm}^{- 1}$ )

1

3.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

5.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

2.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

2.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

Given, resistance (R) $2.5 \times 10^{3} ohm$
Cell constant $(G^{*}) = 1.25 {cm}^{- 1}$
We know,
Specific conductivity $(κ) = \frac{G^{*}}{R}$
$= \frac{1.25}{2.5 \times 10^{3}} = 5 \times 10^{- 4} {ohm}^{- 1} {cm}^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{m}$
$\begin{aligned} Λ_{m} = 5 \times 10^{- 4} \times 1000 \times 10 \\ Λ_{m} = 5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276154 Consider the statements $S_{1}$ and $S_{2}$ :
$S_{1}$ : Conductivity always increases with decrease in the concentration of electrolyte.
$S_{2}$ : Molar conductivity always increase with decrease in the concentration of electrolyte. The correct option among the following is

1

S_{1}

is correct and

S_{2}

is wrong

2

S_{1}

is wrong and

S_{2}

is correct

3 Both

S_{1}

and

S_{2}

are wrong

4 Both

S_{1}

and

S_{2}

are correct

Explanation:

On dilution number of ions per unit volume decrease so conductivity decreases with dilution and hence, $S_{1}$ is wrong.
Conductivity $(κ) \propto$ Concentration
Molar conductivity $(C) \propto \frac{1}{Concentration}$
On dilution $(C)$ and $(κ)$ both decrease but the effect of (C) is more dominating so $Λ_{m}$ increases hence $S_{2}$ is right.
$Λ_{m} = \frac{1000 \times κ}{C}$

JEE Main 2019

ELECTROCHEMISTRY

276156 $Λ_{m}^{o}$ for $NaCl, HCl$ and $NaA$ are 126.4, 425.9 and $100.5 S cm {mol}^{- 1}$ , respectively. If the conductivity of $0.001 M HA$ is $5 \times 10^{- 5} S {cm}^{- 1}$ , degree of dissociation of $HA$ is

1 0.25

2 0.50

3 0.75

4 0.125

Explanation:

Given that,
$\begin{aligned} Λ_{m} NaCl = 126.45 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} HCl = 425.95 {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} NaA = 100.55 {cm}^{2} 55 {cm}^{2} {mol}^{- 1} \end{aligned}$
According to Formula-
$\begin{aligned} Λ_{m}^{\circ} HA = Λ_{m}^{\circ} HCl + Λ_{m}^{\circ} NaA - Λ_{m}^{\circ} NaCl \\ = 425.9 + 100.5 - 126.4 \\ = 400 S {cm}^{2} {mol}^{- 1} \\ Λ_{m}^{\circ} = \frac{1000 κ}{M} = \frac{1000 \times 5 \times 10^{- 5}}{10^{- 3}} \end{aligned}$
$\begin{aligned} = 5 \times 10^{- 5} \times \frac{1000}{0.001} = 50 S {cm}^{2} {mol}^{- 1} \\ a = \frac{Λ_{m}}{Λ_{m}} = \frac{50}{400} = 0.125 \end{aligned}$

JEE Main 2019

ELECTROCHEMISTRY

276157 The decreasing order of electrical conductivity of the following aqueous solution is
0.1 M formic acid (A),
$0.1 M$ acetic acid (B),
0.1 M benzoic acid (C).

1

A > C > B

2

C > B > A

3

A > B > C

4

C > A > B

Explanation:

Electrical conductivity of the aqueous solution depends on the degree of ionisation. Degree of ionisation is directly proportional to the acidic strength. $\begin{array}{ccc} A & B & C \\ HCOOH & C_{6} H_{5} COOH & {CH}_{3} COOH \end{array}$
As the acidic strength decreases rate of dissociation decreases and hence conductivity decreases.
Order of acidic strength
$A > C > B$
Acidic strength $↑=$ degree of ionization $↑$

JEE Main 2019

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: