Units of conductivity and molar conductivity are, respectively

Read the passage carefully and answer.

The molar conductivity of a solution at any given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two platinum electrodes with unit area of cross-section and at a distance of unit length. Both conductivity and molar conductivity change with the concentration of the electrolyte. Kohlrausch examined $Ʌ°_m$ values for a number of strong electrolytes and observed certain regularities. He noted that the difference in $Ʌ°_m$ of the electrolytes NaX and KX for any X is nearly constant. On the basis of the above observations, he enunciated the Kohlrausch law of independent migration of ions.

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → $S/cm,\, S\, cm^2\, mol^{-1}$

Breakdown of Units

1. Conductivity ($\kappa$):

Conductivity is the inverse of resistivity. Its units are derived from:

$\kappa = \frac{1}{\rho} = \frac{l}{R \cdot A}$

$R$ (Resistance) is measured in Ohms ($\Omega$) or Siemens ($S$) as its inverse ($1/\Omega = S$).
$l$ (Length) is in cm.
$A$ (Area) is in $\text{cm}^2$.
Therefore, the unit is $S \cdot \text{cm} / \text{cm}^2 = \mathbf{S/cm}$ (or $S \cdot \text{cm}^{-1}$).

2. Molar Conductivity ($\Lambda_m$):

As shown on the y-axis of the provided graph, molar conductivity represents the conductivity of a solution containing one mole of electrolyte.

The formula is:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

$\kappa$ is in S/cm.
$C$ (Concentration) is in $\text{mol/cm}^3$.
This results in the unit: $\mathbf{S \text{ cm}^2 \text{mol}^{-1}}$.

Graph Analysis

The graph shows how molar conductivity ($\Lambda_m$) varies with the square root of concentration ($c^{1/2}$):

Electrolyte A (Weak Electrolyte): Shows a steep increase in conductivity upon dilution.
Electrolyte B (Strong Electrolyte): Shows a linear increase that can be extrapolated to point C, the limiting molar conductivity.

Correct Option:

S/cm, $\text{S cm}^2 \text{mol}^{-1}$

Read the passage carefully and answer.

Units of conductivity and molar conductivity are, respectively

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → $S/cm,\, S\, cm^2\, mol^{-1}$

Breakdown of Units

1. Conductivity ($\kappa$):

Conductivity is the inverse of resistivity. Its units are derived from:

$\kappa = \frac{1}{\rho} = \frac{l}{R \cdot A}$

$R$ (Resistance) is measured in Ohms ($\Omega$) or Siemens ($S$) as its inverse ($1/\Omega = S$).
$l$ (Length) is in cm.
$A$ (Area) is in $\text{cm}^2$.
Therefore, the unit is $S \cdot \text{cm} / \text{cm}^2 = \mathbf{S/cm}$ (or $S \cdot \text{cm}^{-1}$).

2. Molar Conductivity ($\Lambda_m$):

As shown on the y-axis of the provided graph, molar conductivity represents the conductivity of a solution containing one mole of electrolyte.

The formula is:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

$\kappa$ is in S/cm.
$C$ (Concentration) is in $\text{mol/cm}^3$.
This results in the unit: $\mathbf{S \text{ cm}^2 \text{mol}^{-1}}$.

Graph Analysis

The graph shows how molar conductivity ($\Lambda_m$) varies with the square root of concentration ($c^{1/2}$):

Electrolyte A (Weak Electrolyte): Shows a steep increase in conductivity upon dilution.
Electrolyte B (Strong Electrolyte): Shows a linear increase that can be extrapolated to point C, the limiting molar conductivity.

Correct Option:

S/cm, $\text{S cm}^2 \text{mol}^{-1}$