Λ°m values for electrolytes AB, HB and AD are 120.5, 300.9 and $80.0\, S\, cm^2\, mol^{-1}$, respectively. Λ°m value for HD is

$260.4\, S\, cm^2\, mol^{-1}$

The correct answer is Option (1) → $260.4\, S\, cm^2\, mol^{-1}$

The limiting molar conductivity ($\Lambda^\circ_m$) for the electrolyte $HD$ can be calculated using Kohlrausch's Law of Independent Migration of Ions.

According to this law, the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions.

Calculation Steps

Identify the relationship:

To find the value for $HD$, we need the contributions of $H^+$ and $D^-$. We can combine the given electrolytes such that the unwanted ions cancel out:

$\Lambda^\circ_m(HD) = \Lambda^\circ_m(HB) + \Lambda^\circ_m(AD) - \Lambda^\circ_m(AB)$

Verify with Ion Contributions:

• $\Lambda^\circ_m(HB) = \lambda^\circ(H^+) + \lambda^\circ(B^-)$

• $\Lambda^\circ_m(AD) = \lambda^\circ(A^+) + \lambda^\circ(D^-)$

• $\Lambda^\circ_m(AB) = \lambda^\circ(A^+) + \lambda^\circ(B^-)$

Substituting these into the equation:

$(\lambda^\circ(H^+) + \lambda^\circ(B^-)) + (\lambda^\circ(A^+) + \lambda^\circ(D^-)) - (\lambda^\circ(A^+) + \lambda^\circ(B^-)) = \lambda^\circ(H^+) + \lambda^\circ(D^-)$

Plug in the Values:

• $\Lambda^\circ_m(HB) = 300.9 \, \text{S cm}^2 \text{mol}^{-1}$

• $\Lambda^\circ_m(AD) = 80.0 \, \text{S cm}^2 \text{mol}^{-1}$

• $\Lambda^\circ_m(AB) = 120.5 \, \text{S cm}^2 \text{mol}^{-1}$

$\Lambda^\circ_m(HD) = 300.9 + 80.0 - 120.5$

$\Lambda^\circ_m(HD) = 380.9 - 120.5$

$\Lambda^\circ_m(HD) = 260.4 \, \text{S cm}^2 \text{mol}^{-1}$