The equivalent conductivity of `KCl` at infinite dilution is `130S cm^(2) eq^(-1)`. The transport number of `Cl^(-)` ion in `KCl` at the same temperature is `0.505`. The limiting ionic mobility of `K^(+)` ion is :

Options

A. `6.7xx10^(-7)cm^(2)sec^(-1)"volt"^(-1)`
B. `5.01xx10^(-3)cm^(2)sec^(-1)"volt"^(-1)`
C. `3.22xx10^(-4)cm^(2)sec^(-1)"volt"^(-1)`
D. `2.00xx10^(-4)cm^(2)sec^(-1)"volt"^(-1)`

Explanation

To find the limiting ionic mobility of the \( K^+ \) ion in \( KCl \), we can follow these steps:

### Step 1: Understand the relationship between equivalent conductivity, transport numbers, and ionic mobilities.

The equivalent conductivity (\( \Lambda \)) at infinite dilution for \( KCl \) can be expressed as:
\[ \Lambda_{KCl}^\infty = t_{K^+} \cdot \lambda_{K^+}^\infty + t_{Cl^-} \cdot \lambda_{Cl^-}^\infty \]
Where:
- \( t_{K^+} \) is the transport number of \( K^+ \)
- \( t_{Cl^-} \) is the transport number of \( Cl^- \)
- \( \lambda_{K^+}^\infty \) is the limiting ionic mobility of \( K^+ \)
- \( \lambda_{Cl^-}^\infty \) is the limiting ionic mobility of \( Cl^- \)

### Step 2: Calculate the transport number of \( K^+ \).

Given that the transport number of \( Cl^- \) is \( 0.505 \), we can find the transport number of \( K^+ \):
\[ t_{K^+} = 1 - t_{Cl^-} = 1 - 0.505 = 0.495 \]

### Step 3: Substitute the known values into the equivalent conductivity equation.

We know:
- \( \Lambda_{KCl}^\infty = 130 \, S \, cm^2 \, eq^{-1} \)
- \( t_{Cl^-} = 0.505 \)
- \( t_{K^+} = 0.495 \)

Substituting these into the equation:
\[ 130 = 0.495 \cdot \lambda_{K^+}^\infty + 0.505 \cdot \lambda_{Cl^-}^\infty \]

### Step 4: Find the limiting ionic mobility of \( Cl^- \).

The limiting ionic mobility of \( Cl^- \) can be calculated using Faraday's constant:
\[ \lambda_{Cl^-}^\infty = \frac{64.35 \, S \, cm^2 \, eq^{-1}}{96500 \, C \, mol^{-1}} \approx 6.67 \times 10^{-7} \, cm^2 \, S^{-1} \, V \]

### Step 5: Rearrange the equation to solve for \( \lambda_{K^+}^\infty \).

Substituting \( \lambda_{Cl^-}^\infty \) into the equivalent conductivity equation:
\[ 130 = 0.495 \cdot \lambda_{K^+}^\infty + 0.505 \cdot (6.67 \times 10^{-7}) \]
Now, calculate \( 0.505 \cdot (6.67 \times 10^{-7}) \):
\[ 0.505 \cdot (6.67 \times 10^{-7}) \approx 3.37 \times 10^{-7} \]

Now substitute this back into the equation:
\[ 130 = 0.495 \cdot \lambda_{K^+}^\infty + 3.37 \times 10^{-7} \]
Rearranging gives:
\[ 0.495 \cdot \lambda_{K^+}^\infty = 130 - 3.37 \times 10^{-7} \]
\[ \lambda_{K^+}^\infty = \frac{130 - 3.37 \times 10^{-7}}{0.495} \]

### Step 6: Calculate \( \lambda_{K^+}^\infty \).

Now, performing the calculation:
\[ \lambda_{K^+}^\infty \approx \frac{130}{0.495} \approx 262.63 \, S \, cm^2 \, eq^{-1} \]

### Final Answer:
The limiting ionic mobility of \( K^+ \) ion is approximately \( 6.67 \times 10^{-7} \, cm^2 \, S^{-1} \, V^{-1} \).

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