Question

The molar conductances at infinite dilution for electrolytes $BA$ and $CA$ are $140$ and $120oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively. If the molar conductance at infinite dilution of $BX$ is $198oh{m}^{-1}c{m}^{2}mo{l}^{-1}$, then at infinite dilution, the molar conductance of $CX$ is:

• A. $178oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• B. $198oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• C. $218oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• D. $130oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is A $178oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Let the limiting molar conductivity of ions $A,B,C$ and $X$ be
${\lambda }_A^0,{\lambda }_B^0,{\lambda }_C^0\text{and}{\lambda }_X^0$ respectively.

${\Lambda }_{BA}^0=140oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\lambda }_A^0+{\lambda }_B^0=\mathrm{140......}eqn.1$

${\Lambda }_{CA}^0=120oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\lambda }_A^0+{\lambda }_C^0=\mathrm{120......}eqn.2$

${\Lambda }_{BX}^0=198oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\lambda }_B^0+{\lambda }_X^0=\mathrm{198......}eqn.3$

Equation $\left(2\right)-\left(1\right)+\left(3\right)$, we get
${\lambda }_A^0+{\lambda }_C^0-{\lambda }_A^0-{\lambda }_B^0+{\lambda }_B^0+{\lambda }_X^0=120-140+198$
${\lambda }_C^0+{\lambda }_X^0=178$
$\therefore$
Molar conductance at infinite dilution of CX is,
${\Lambda }_{CX}^0=178oh{m}^{-1}c{m}^{2}mo{l}^{-1}$