Question

The difference in limiting molar conductivity between $KCl$ and $KN{O}_3$ is the same as the difference between :

• A. $NaCl$ and $KN{O}_3$
• B. $KCl$ and $NaCl$
• C. $NaCl$ and $NaN{O}_3$
• D. $KN{O}_3$ and $NaN{O}_3$

Answer 1

The correct option is C $NaCl$ and $NaN{O}_3$

Kohlrausch law:
At infinite dilution, when all the interionic effects disappear and dissociation of electrolyte is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated.

Kohlrausch law of independent migration of ions:
${\Lambda }_m^0={\lambda }_{+}^0+{\lambda }_{-}^0$

${\lambda }_{+}^0\to \text{Limiting molar conductivities of cation}$
${\lambda }_{-}^0\to \text{Limiting molar conductivities of anion}$

For $KCl$ and $KN{O}_3$
${\Lambda }_{KCl}^0={\lambda }_{K^{+}}^0+{\lambda }_{C{l}^{-}}^0$
${\Lambda }_{KN{O}_3}^0={\lambda }_{K^{+}}^0+{\lambda }_{N{O}_3^{-}}^0$

${\Lambda }_{KCl}^0-{\Lambda }_{KN{O}_3}^0={\lambda }_{C{l}^{-}}^0-{\lambda }_{N{O}_3^{-}}^0....\left(1\right)$
The obtained value is due to the difference in ${\lambda }_m\text{of}C{l}^{-}\text{and}N{O}_3^{-}$ ions.

For $NaCl$ and $NaN{O}_3$,
${\Lambda }_{NaCl}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{C{l}^{-}}^0$
${\Lambda }_{NaN{O}_3}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{N{O}_3^{-}}^0$

${\Lambda }_{NaCl}^0-{\Lambda }_{NaN{O}_3}^0={\lambda }_{C{l}^{-}}^0-{\lambda }_{N{O}_3^{-}}^0....\left(2\right)$
The obtained value is due to the difference in ${\lambda }_m\text{of}C{l}^{-}\text{and}N{O}_3^{-}$ ions.
This proves the Kohlrausch law of independent migration of ions.

Hence, correct option is (a)