The `wedge_(m)^(@)` values for NaCl and KCl are 126.5 and 149.9`Omega^(-1)cm^(2)mol^(-1)` respectively. The ionic conductances of `Na^(+)` at infinite dilution is 50.1 `Omega^(-1)cm^(2)mol^(-1)`. Calculate the ionic conductance at infinite dilution for `K^(+)` ion.

Explanation

To calculate the ionic conductance at infinite dilution for the K⁺ ion, we can use the following relationship:

\[ \Lambda_{m}^\circ(\text{NaCl}) = \Lambda_{m}^\circ(\text{Na}^+) + \Lambda_{m}^\circ(\text{Cl}^-) \]

\[ \Lambda_{m}^\circ(\text{KCl}) = \Lambda_{m}^\circ(\text{K}^+) + \Lambda_{m}^\circ(\text{Cl}^-) \]

Where:
- \(\Lambda_{m}^\circ(\text{NaCl})\) = 126.5 \(\Omega^{-1}cm^2mol^{-1}\)
- \(\Lambda_{m}^\circ(\text{KCl})\) = 149.9 \(\Omega^{-1}cm^2mol^{-1}\)
- \(\Lambda_{m}^\circ(\text{Na}^+)\) = 50.1 \(\Omega^{-1}cm^2mol^{-1}\)

### Step 1: Write the equations for NaCl and KCl

From the given data, we can write:

\[ \Lambda_{m}^\circ(\text{NaCl}) = \Lambda_{m}^\circ(\text{Na}^+) + \Lambda_{m}^\circ(\text{Cl}^-) \]

\[ \Lambda_{m}^\circ(\text{KCl}) = \Lambda_{m}^\circ(\text{K}^+) + \Lambda_{m}^\circ(\text{Cl}^-) \]

### Step 2: Rearranging the equations

We can express \(\Lambda_{m}^\circ(\text{Cl}^-)\) from the NaCl equation:

\[ \Lambda_{m}^\circ(\text{Cl}^-) = \Lambda_{m}^\circ(\text{NaCl}) - \Lambda_{m}^\circ(\text{Na}^+) \]

Substituting the known values:

\[ \Lambda_{m}^\circ(\text{Cl}^-) = 126.5 - 50.1 = 76.4 \, \Omega^{-1}cm^2mol^{-1} \]

### Step 3: Substitute \(\Lambda_{m}^\circ(\text{Cl}^-)\) into the KCl equation

Now, substituting \(\Lambda_{m}^\circ(\text{Cl}^-)\) into the KCl equation:

\[ \Lambda_{m}^\circ(\text{KCl}) = \Lambda_{m}^\circ(\text{K}^+) + \Lambda_{m}^\circ(\text{Cl}^-) \]

Substituting the known values:

\[ 149.9 = \Lambda_{m}^\circ(\text{K}^+) + 76.4 \]

### Step 4: Solve for \(\Lambda_{m}^\circ(\text{K}^+)\)

Now, we can solve for \(\Lambda_{m}^\circ(\text{K}^+)\):

\[ \Lambda_{m}^\circ(\text{K}^+) = 149.9 - 76.4 = 73.5 \, \Omega^{-1}cm^2mol^{-1} \]

### Final Answer

The ionic conductance at infinite dilution for the K⁺ ion is:

\[ \Lambda_{m}^\circ(\text{K}^+) = 73.5 \, \Omega^{-1}cm^2mol^{-1} \]