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Q2.8:
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S (cm^{-1}). Calculate its molar conductivity.

Solution :

Step 1: Given Variables & Theoretical Foundation

To calculate the molar conductivity ($\Lambda_m$) of an electrolytic solution, we utilize the relationship between conductivity ($\kappa$) and molarity ($C$).

  • Conductivity ($\kappa$) = $0.0248 \text{ S cm}^{-1}$
  • Molar Concentration ($C$) = $0.20 \text{ M}$ (which is $\text{mol L}^{-1}$)
  • Temperature ($T$) = $298 \text{ K}$

[Definition: Molar conductivity is the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution.]

Step 2: Unit Conversion and Dimensional Consistency

The standard unit for molar conductivity is $\text{S cm}^2 \text{ mol}^{-1}$. Since the molarity is given in $\text{mol L}^{-1}$, we must convert it to $\text{mol cm}^{-3}$ to maintain dimensional consistency with $\kappa$ ($\text{S cm}^{-1}$).

We know that $1 \text{ L} = 1000 \text{ cm}^3$. Therefore:

$C (\text{mol cm}^{-3}) = \frac{0.20 \text{ mol}}{1000 \text{ cm}^3} = 2.0 \times 10^{-4} \text{ mol cm}^{-3}$

Step 3: Applying the Molar Conductivity Formula

The mathematical expression for molar conductivity is:

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

[Formula derivation: By dividing conductivity by the concentration in appropriate units, we obtain the conductance per mole of the solute.]

Substituting the values:

$\Lambda_m = \frac{0.0248 \text{ S cm}^{-1}}{2.0 \times 10^{-4} \text{ mol cm}^{-3}}$

Step 4: Final Calculation

Perform the arithmetic division:

$\Lambda_m = \frac{0.0248}{0.0002} \text{ S cm}^2 \text{ mol}^{-1}$

$\Lambda_m = 124 \text{ S cm}^2 \text{ mol}^{-1}$

Given: κ = 0.0248 S/cm, C = 0.20 M Formula: Λm = (κ × 1000) / M Result: Λm = 124 S cm²/mol

Final Solution: The molar conductivity of the 0.20 M KCl solution is 124 S cm² mol⁻¹.


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