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CHXII03:ELECTROCHEMISTRY

330392 The electrical conductance of unit volume $(1 {cm}^{3})$ of solution is called as

1 Electrical resistance

2 Resistivity

3 Molar conductivity

4 Conductivity

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330393 The molar conductivity of $0.4 M KCl$ solution is $2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . What is the resistivity of solution?

1

2.1 \times 10^{2}

2

2.5 \times 10^{2}

3

1 \times 10^{- 2}

4

2.8 \times 10^{- 2}

Explanation:

$c = 0.4 M, \land = 2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ , $ρ$ ?
Step-1 Finding conductivity :
$\begin{aligned} \land = \frac{1000 k}{C} ∴ k = \frac{\land C}{1000} \\ ∴ k = \frac{2.5 \times 10^{5} \times 0.4}{1000} = 100 Ω^{-} {cm}^{-} \end{aligned}$
Step-2 Finding resistivity :
$k = \frac{1}{ρ} ∴ ρ = \frac{1}{k} = \frac{1}{100} = 1 \times 10^{- 2} Ω cm$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330394 The molar conductivity of $0 .1M BaC l_{2}$ solution is $106 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ at $25^{\circ} C$ . What is it's conductivity?

1

1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

2

5.3 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

3

3.66 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

4

2.6 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

Explanation:

$\land = \frac{1000 k}{c} ∴ k = \frac{\land c}{1000}$
Given $\land = 106 Ω^{- 1} c m^{2} mo l^{- 1}, c = 0.1 mol L^{- 1}$
$∴ k = \frac{106 Ω^{- 1} c m^{2} mo l^{- 1} \times 0.1 mol L^{- 1}}{1000 c m^{3} L^{- 1}}$
$∴ k = 1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330395 Conductance of 0.1 M KCl (conductivity $= X o h m^{- 1} c m^{- 1}$ ) filled in a conductivity cell is $Y o h m^{- 1}$ . If the conductance of 0.1 M NaOH filled in the same cell is $Z o h m^{- 1}$ , the molar conductance of NaOH will be

1

1 0^{3} X Z / Y

2

1 0^{4} X Z / Y

3

10 X Z / Y

4

0 .1 X Z / Y

Explanation:

Conductivity (X) = conductance $\times$ cell constant
Cell constant = X/Y
Conductivity of NaOH = XZ/Y
$Λ_{m} (N a O H) = 1 0^{4} X Z / Y$ .

CHXII03:ELECTROCHEMISTRY

330392 The electrical conductance of unit volume $(1 {cm}^{3})$ of solution is called as

1 Electrical resistance

2 Resistivity

3 Molar conductivity

4 Conductivity

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330393 The molar conductivity of $0.4 M KCl$ solution is $2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . What is the resistivity of solution?

1

2.1 \times 10^{2}

2

2.5 \times 10^{2}

3

1 \times 10^{- 2}

4

2.8 \times 10^{- 2}

Explanation:

$c = 0.4 M, \land = 2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ , $ρ$ ?
Step-1 Finding conductivity :
$\begin{aligned} \land = \frac{1000 k}{C} ∴ k = \frac{\land C}{1000} \\ ∴ k = \frac{2.5 \times 10^{5} \times 0.4}{1000} = 100 Ω^{-} {cm}^{-} \end{aligned}$
Step-2 Finding resistivity :
$k = \frac{1}{ρ} ∴ ρ = \frac{1}{k} = \frac{1}{100} = 1 \times 10^{- 2} Ω cm$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330394 The molar conductivity of $0 .1M BaC l_{2}$ solution is $106 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ at $25^{\circ} C$ . What is it's conductivity?

1

1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

2

5.3 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

3

3.66 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

4

2.6 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

Explanation:

$\land = \frac{1000 k}{c} ∴ k = \frac{\land c}{1000}$
Given $\land = 106 Ω^{- 1} c m^{2} mo l^{- 1}, c = 0.1 mol L^{- 1}$
$∴ k = \frac{106 Ω^{- 1} c m^{2} mo l^{- 1} \times 0.1 mol L^{- 1}}{1000 c m^{3} L^{- 1}}$
$∴ k = 1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330395 Conductance of 0.1 M KCl (conductivity $= X o h m^{- 1} c m^{- 1}$ ) filled in a conductivity cell is $Y o h m^{- 1}$ . If the conductance of 0.1 M NaOH filled in the same cell is $Z o h m^{- 1}$ , the molar conductance of NaOH will be

1

1 0^{3} X Z / Y

2

1 0^{4} X Z / Y

3

10 X Z / Y

4

0 .1 X Z / Y

Explanation:

Conductivity (X) = conductance $\times$ cell constant
Cell constant = X/Y
Conductivity of NaOH = XZ/Y
$Λ_{m} (N a O H) = 1 0^{4} X Z / Y$ .

CHXII03:ELECTROCHEMISTRY

330392 The electrical conductance of unit volume $(1 {cm}^{3})$ of solution is called as

1 Electrical resistance

2 Resistivity

3 Molar conductivity

4 Conductivity

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330393 The molar conductivity of $0.4 M KCl$ solution is $2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . What is the resistivity of solution?

1

2.1 \times 10^{2}

2

2.5 \times 10^{2}

3

1 \times 10^{- 2}

4

2.8 \times 10^{- 2}

Explanation:

$c = 0.4 M, \land = 2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ , $ρ$ ?
Step-1 Finding conductivity :
$\begin{aligned} \land = \frac{1000 k}{C} ∴ k = \frac{\land C}{1000} \\ ∴ k = \frac{2.5 \times 10^{5} \times 0.4}{1000} = 100 Ω^{-} {cm}^{-} \end{aligned}$
Step-2 Finding resistivity :
$k = \frac{1}{ρ} ∴ ρ = \frac{1}{k} = \frac{1}{100} = 1 \times 10^{- 2} Ω cm$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330394 The molar conductivity of $0 .1M BaC l_{2}$ solution is $106 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ at $25^{\circ} C$ . What is it's conductivity?

1

1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

2

5.3 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

3

3.66 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

4

2.6 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

Explanation:

$\land = \frac{1000 k}{c} ∴ k = \frac{\land c}{1000}$
Given $\land = 106 Ω^{- 1} c m^{2} mo l^{- 1}, c = 0.1 mol L^{- 1}$
$∴ k = \frac{106 Ω^{- 1} c m^{2} mo l^{- 1} \times 0.1 mol L^{- 1}}{1000 c m^{3} L^{- 1}}$
$∴ k = 1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330395 Conductance of 0.1 M KCl (conductivity $= X o h m^{- 1} c m^{- 1}$ ) filled in a conductivity cell is $Y o h m^{- 1}$ . If the conductance of 0.1 M NaOH filled in the same cell is $Z o h m^{- 1}$ , the molar conductance of NaOH will be

1

1 0^{3} X Z / Y

2

1 0^{4} X Z / Y

3

10 X Z / Y

4

0 .1 X Z / Y

Explanation:

Conductivity (X) = conductance $\times$ cell constant
Cell constant = X/Y
Conductivity of NaOH = XZ/Y
$Λ_{m} (N a O H) = 1 0^{4} X Z / Y$ .

CHXII03:ELECTROCHEMISTRY

330392 The electrical conductance of unit volume $(1 {cm}^{3})$ of solution is called as

1 Electrical resistance

2 Resistivity

3 Molar conductivity

4 Conductivity

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330393 The molar conductivity of $0.4 M KCl$ solution is $2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . What is the resistivity of solution?

1

2.1 \times 10^{2}

2

2.5 \times 10^{2}

3

1 \times 10^{- 2}

4

2.8 \times 10^{- 2}

Explanation:

$c = 0.4 M, \land = 2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ , $ρ$ ?
Step-1 Finding conductivity :
$\begin{aligned} \land = \frac{1000 k}{C} ∴ k = \frac{\land C}{1000} \\ ∴ k = \frac{2.5 \times 10^{5} \times 0.4}{1000} = 100 Ω^{-} {cm}^{-} \end{aligned}$
Step-2 Finding resistivity :
$k = \frac{1}{ρ} ∴ ρ = \frac{1}{k} = \frac{1}{100} = 1 \times 10^{- 2} Ω cm$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330394 The molar conductivity of $0 .1M BaC l_{2}$ solution is $106 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ at $25^{\circ} C$ . What is it's conductivity?

1

1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

2

5.3 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

3

3.66 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}

4

2.6 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}

Explanation:

$\land = \frac{1000 k}{c} ∴ k = \frac{\land c}{1000}$
Given $\land = 106 Ω^{- 1} c m^{2} mo l^{- 1}, c = 0.1 mol L^{- 1}$
$∴ k = \frac{106 Ω^{- 1} c m^{2} mo l^{- 1} \times 0.1 mol L^{- 1}}{1000 c m^{3} L^{- 1}}$
$∴ k = 1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330395 Conductance of 0.1 M KCl (conductivity $= X o h m^{- 1} c m^{- 1}$ ) filled in a conductivity cell is $Y o h m^{- 1}$ . If the conductance of 0.1 M NaOH filled in the same cell is $Z o h m^{- 1}$ , the molar conductance of NaOH will be

1

1 0^{3} X Z / Y

2

1 0^{4} X Z / Y

3

10 X Z / Y

4

0 .1 X Z / Y

Explanation:

Conductivity (X) = conductance $\times$ cell constant
Cell constant = X/Y
Conductivity of NaOH = XZ/Y
$Λ_{m} (N a O H) = 1 0^{4} X Z / Y$ .

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