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CHXII03:ELECTROCHEMISTRY

330396 Match the terms given in Column I with the units given in Column II and choose the correct
Column I
Column II
A
$Λ_{m}$
P
$l / A$
B
$G^{*}$
Q
$ρ l / A$
C
$κ$
R
$κ / C$
D
$R$
S
$G^{*} / R$

1 A - P, B - R, C - Q, D - S

2 A - R, B - P, C - S, D - Q

3 A - Q, B - S, C - R, D - P

4 A - S, B - Q, C - P, D - R

CHXII03:ELECTROCHEMISTRY

330397 The conductivity of $0.3 M$ solution of $KCl$ at $298 K$ is $0.0627 S {cm}^{- 1}$ . What is its molar conductivity?

1

104 S c m^{2} mo l^{- 1}

2

188 S c m^{2} mo l^{- 1}

3

209 S c m^{2} mo l^{- 1}

4

109 S c m^{2} mo l^{- 1}

Explanation:

$k = 0.0627 S c m^{- 1}, c = 0.3 mol L^{- 1}$
$\land = \frac{1000 k}{c}$
$= \frac{1000 c m^{3} L^{- 1} \times 0.0627 S c m^{- 1}}{0.3 mol L^{- 1}}$
$∴ \land = 209 S c m^{2} mo l^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330398 The resistance of $\frac{1}{10} M$ solution is $2 .5 \times 1 0^{3}$ ohm. What is the molar conductivity of solution? (cell constant $= 1 .25 c m^{- 1}$ )

1

3 .5 o h m^{- 1} c m^{2} m o l^{- 1}

2

5 .0 o h m^{- 1} c m^{2} m o l^{- 1}

3

2 .5 o h m^{- 1} c m^{2} m o l^{- 1}

4

2 .0 o h m^{- 1} c m^{2} m o l^{- 1}

Explanation:

Given, Molarity $= \frac{1}{10} M$
$R = 2 .5 \times 1 0^{3} Ω$
Cell constant $(\frac{l}{A}) = 1 .25 c m^{- 1}$
Now, $κ (c o n d u c t i v i t y) = \frac{1}{κ} \times \frac{l}{A}$
$∴ κ = \frac{1}{2.5 \times 10^{3}} \times 1.25 = 0.5 \times 10^{- 3} Ω^{- 1} c m^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{M}$
$∴ Λ_{m} = \frac{0.5 \times 10^{- 3} \times 1000}{1 / 10}$
$Λ_{m} = 5 Ω^{- 1} c m^{2} m o l^{- 1}$

MHTCET - 2019-1

CHXII03:ELECTROCHEMISTRY

330399 The conductivity of 0.20 M solution of KCl at 298 K is $0 .0248 S c m^{- 1}$ . Its molar conductivity is

1

124 S m o l^{- 1} c m^{2}

2

124 S m o l^{- 1} m^{2}

3

12 .4 S m o l^{- 1} c m^{2}

4

12 .4 S m o l^{- 1} m^{2}

Explanation:

Conductivity $(κ) = 0 .0248 S c m^{- 1}$
$= 0 .0248 o h m^{- 1} c m^{- 1}$
Molar concentration $(C) = 0 .20 m o l L^{- 1}$
$= \frac{(0 .2 m o l)}{(1000 c m^{3})} = 2 .0 \times 1 0^{- 4} m o l c m^{- 3}$
$M o l a r c o n d u c t i v i t y (Λ_{m}) =$
$\frac{κ}{C} = \frac{(0 .0248 o h m^{- 1} c m^{- 1})}{(2 .0 \times 1 0^{- 4} m o l c m^{- 3})}$
$= 124 o h m^{- 1} m o l^{- 1} c m^{2} o r 124 S m o l^{- 1} c m^{2}$
Molar conductivity $= 124 S m o l^{- 1} c m^{2}$ .

CHXII03:ELECTROCHEMISTRY

330396 Match the terms given in Column I with the units given in Column II and choose the correct
Column I
Column II
A
$Λ_{m}$
P
$l / A$
B
$G^{*}$
Q
$ρ l / A$
C
$κ$
R
$κ / C$
D
$R$
S
$G^{*} / R$

1 A - P, B - R, C - Q, D - S

2 A - R, B - P, C - S, D - Q

3 A - Q, B - S, C - R, D - P

4 A - S, B - Q, C - P, D - R

CHXII03:ELECTROCHEMISTRY

330397 The conductivity of $0.3 M$ solution of $KCl$ at $298 K$ is $0.0627 S {cm}^{- 1}$ . What is its molar conductivity?

1

104 S c m^{2} mo l^{- 1}

2

188 S c m^{2} mo l^{- 1}

3

209 S c m^{2} mo l^{- 1}

4

109 S c m^{2} mo l^{- 1}

Explanation:

$k = 0.0627 S c m^{- 1}, c = 0.3 mol L^{- 1}$
$\land = \frac{1000 k}{c}$
$= \frac{1000 c m^{3} L^{- 1} \times 0.0627 S c m^{- 1}}{0.3 mol L^{- 1}}$
$∴ \land = 209 S c m^{2} mo l^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330398 The resistance of $\frac{1}{10} M$ solution is $2 .5 \times 1 0^{3}$ ohm. What is the molar conductivity of solution? (cell constant $= 1 .25 c m^{- 1}$ )

1

3 .5 o h m^{- 1} c m^{2} m o l^{- 1}

2

5 .0 o h m^{- 1} c m^{2} m o l^{- 1}

3

2 .5 o h m^{- 1} c m^{2} m o l^{- 1}

4

2 .0 o h m^{- 1} c m^{2} m o l^{- 1}

Explanation:

Given, Molarity $= \frac{1}{10} M$
$R = 2 .5 \times 1 0^{3} Ω$
Cell constant $(\frac{l}{A}) = 1 .25 c m^{- 1}$
Now, $κ (c o n d u c t i v i t y) = \frac{1}{κ} \times \frac{l}{A}$
$∴ κ = \frac{1}{2.5 \times 10^{3}} \times 1.25 = 0.5 \times 10^{- 3} Ω^{- 1} c m^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{M}$
$∴ Λ_{m} = \frac{0.5 \times 10^{- 3} \times 1000}{1 / 10}$
$Λ_{m} = 5 Ω^{- 1} c m^{2} m o l^{- 1}$

MHTCET - 2019-1

CHXII03:ELECTROCHEMISTRY

330399 The conductivity of 0.20 M solution of KCl at 298 K is $0 .0248 S c m^{- 1}$ . Its molar conductivity is

1

124 S m o l^{- 1} c m^{2}

2

124 S m o l^{- 1} m^{2}

3

12 .4 S m o l^{- 1} c m^{2}

4

12 .4 S m o l^{- 1} m^{2}

Explanation:

Conductivity $(κ) = 0 .0248 S c m^{- 1}$
$= 0 .0248 o h m^{- 1} c m^{- 1}$
Molar concentration $(C) = 0 .20 m o l L^{- 1}$
$= \frac{(0 .2 m o l)}{(1000 c m^{3})} = 2 .0 \times 1 0^{- 4} m o l c m^{- 3}$
$M o l a r c o n d u c t i v i t y (Λ_{m}) =$
$\frac{κ}{C} = \frac{(0 .0248 o h m^{- 1} c m^{- 1})}{(2 .0 \times 1 0^{- 4} m o l c m^{- 3})}$
$= 124 o h m^{- 1} m o l^{- 1} c m^{2} o r 124 S m o l^{- 1} c m^{2}$
Molar conductivity $= 124 S m o l^{- 1} c m^{2}$ .

CHXII03:ELECTROCHEMISTRY

330396 Match the terms given in Column I with the units given in Column II and choose the correct
Column I
Column II
A
$Λ_{m}$
P
$l / A$
B
$G^{*}$
Q
$ρ l / A$
C
$κ$
R
$κ / C$
D
$R$
S
$G^{*} / R$

1 A - P, B - R, C - Q, D - S

2 A - R, B - P, C - S, D - Q

3 A - Q, B - S, C - R, D - P

4 A - S, B - Q, C - P, D - R

CHXII03:ELECTROCHEMISTRY

330397 The conductivity of $0.3 M$ solution of $KCl$ at $298 K$ is $0.0627 S {cm}^{- 1}$ . What is its molar conductivity?

1

104 S c m^{2} mo l^{- 1}

2

188 S c m^{2} mo l^{- 1}

3

209 S c m^{2} mo l^{- 1}

4

109 S c m^{2} mo l^{- 1}

Explanation:

$k = 0.0627 S c m^{- 1}, c = 0.3 mol L^{- 1}$
$\land = \frac{1000 k}{c}$
$= \frac{1000 c m^{3} L^{- 1} \times 0.0627 S c m^{- 1}}{0.3 mol L^{- 1}}$
$∴ \land = 209 S c m^{2} mo l^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330398 The resistance of $\frac{1}{10} M$ solution is $2 .5 \times 1 0^{3}$ ohm. What is the molar conductivity of solution? (cell constant $= 1 .25 c m^{- 1}$ )

1

3 .5 o h m^{- 1} c m^{2} m o l^{- 1}

2

5 .0 o h m^{- 1} c m^{2} m o l^{- 1}

3

2 .5 o h m^{- 1} c m^{2} m o l^{- 1}

4

2 .0 o h m^{- 1} c m^{2} m o l^{- 1}

Explanation:

Given, Molarity $= \frac{1}{10} M$
$R = 2 .5 \times 1 0^{3} Ω$
Cell constant $(\frac{l}{A}) = 1 .25 c m^{- 1}$
Now, $κ (c o n d u c t i v i t y) = \frac{1}{κ} \times \frac{l}{A}$
$∴ κ = \frac{1}{2.5 \times 10^{3}} \times 1.25 = 0.5 \times 10^{- 3} Ω^{- 1} c m^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{M}$
$∴ Λ_{m} = \frac{0.5 \times 10^{- 3} \times 1000}{1 / 10}$
$Λ_{m} = 5 Ω^{- 1} c m^{2} m o l^{- 1}$

MHTCET - 2019-1

CHXII03:ELECTROCHEMISTRY

330399 The conductivity of 0.20 M solution of KCl at 298 K is $0 .0248 S c m^{- 1}$ . Its molar conductivity is

1

124 S m o l^{- 1} c m^{2}

2

124 S m o l^{- 1} m^{2}

3

12 .4 S m o l^{- 1} c m^{2}

4

12 .4 S m o l^{- 1} m^{2}

Explanation:

Conductivity $(κ) = 0 .0248 S c m^{- 1}$
$= 0 .0248 o h m^{- 1} c m^{- 1}$
Molar concentration $(C) = 0 .20 m o l L^{- 1}$
$= \frac{(0 .2 m o l)}{(1000 c m^{3})} = 2 .0 \times 1 0^{- 4} m o l c m^{- 3}$
$M o l a r c o n d u c t i v i t y (Λ_{m}) =$
$\frac{κ}{C} = \frac{(0 .0248 o h m^{- 1} c m^{- 1})}{(2 .0 \times 1 0^{- 4} m o l c m^{- 3})}$
$= 124 o h m^{- 1} m o l^{- 1} c m^{2} o r 124 S m o l^{- 1} c m^{2}$
Molar conductivity $= 124 S m o l^{- 1} c m^{2}$ .

CHXII03:ELECTROCHEMISTRY

330396 Match the terms given in Column I with the units given in Column II and choose the correct
Column I
Column II
A
$Λ_{m}$
P
$l / A$
B
$G^{*}$
Q
$ρ l / A$
C
$κ$
R
$κ / C$
D
$R$
S
$G^{*} / R$

1 A - P, B - R, C - Q, D - S

2 A - R, B - P, C - S, D - Q

3 A - Q, B - S, C - R, D - P

4 A - S, B - Q, C - P, D - R

CHXII03:ELECTROCHEMISTRY

330397 The conductivity of $0.3 M$ solution of $KCl$ at $298 K$ is $0.0627 S {cm}^{- 1}$ . What is its molar conductivity?

1

104 S c m^{2} mo l^{- 1}

2

188 S c m^{2} mo l^{- 1}

3

209 S c m^{2} mo l^{- 1}

4

109 S c m^{2} mo l^{- 1}

Explanation:

$k = 0.0627 S c m^{- 1}, c = 0.3 mol L^{- 1}$
$\land = \frac{1000 k}{c}$
$= \frac{1000 c m^{3} L^{- 1} \times 0.0627 S c m^{- 1}}{0.3 mol L^{- 1}}$
$∴ \land = 209 S c m^{2} mo l^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330398 The resistance of $\frac{1}{10} M$ solution is $2 .5 \times 1 0^{3}$ ohm. What is the molar conductivity of solution? (cell constant $= 1 .25 c m^{- 1}$ )

1

3 .5 o h m^{- 1} c m^{2} m o l^{- 1}

2

5 .0 o h m^{- 1} c m^{2} m o l^{- 1}

3

2 .5 o h m^{- 1} c m^{2} m o l^{- 1}

4

2 .0 o h m^{- 1} c m^{2} m o l^{- 1}

Explanation:

Given, Molarity $= \frac{1}{10} M$
$R = 2 .5 \times 1 0^{3} Ω$
Cell constant $(\frac{l}{A}) = 1 .25 c m^{- 1}$
Now, $κ (c o n d u c t i v i t y) = \frac{1}{κ} \times \frac{l}{A}$
$∴ κ = \frac{1}{2.5 \times 10^{3}} \times 1.25 = 0.5 \times 10^{- 3} Ω^{- 1} c m^{- 1}$
Molar conductivity $(Λ_{m}) = \frac{κ \times 1000}{M}$
$∴ Λ_{m} = \frac{0.5 \times 10^{- 3} \times 1000}{1 / 10}$
$Λ_{m} = 5 Ω^{- 1} c m^{2} m o l^{- 1}$

MHTCET - 2019-1

CHXII03:ELECTROCHEMISTRY

330399 The conductivity of 0.20 M solution of KCl at 298 K is $0 .0248 S c m^{- 1}$ . Its molar conductivity is

1

124 S m o l^{- 1} c m^{2}

2

124 S m o l^{- 1} m^{2}

3

12 .4 S m o l^{- 1} c m^{2}

4

12 .4 S m o l^{- 1} m^{2}

Explanation:

Conductivity $(κ) = 0 .0248 S c m^{- 1}$
$= 0 .0248 o h m^{- 1} c m^{- 1}$
Molar concentration $(C) = 0 .20 m o l L^{- 1}$
$= \frac{(0 .2 m o l)}{(1000 c m^{3})} = 2 .0 \times 1 0^{- 4} m o l c m^{- 3}$
$M o l a r c o n d u c t i v i t y (Λ_{m}) =$
$\frac{κ}{C} = \frac{(0 .0248 o h m^{- 1} c m^{- 1})}{(2 .0 \times 1 0^{- 4} m o l c m^{- 3})}$
$= 124 o h m^{- 1} m o l^{- 1} c m^{2} o r 124 S m o l^{- 1} c m^{2}$
Molar conductivity $= 124 S m o l^{- 1} c m^{2}$ .

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