The resistance of a `(N)/(10)KCl` aqueous solution is `2450Omega`. If the electrodes in the cell are 4cm apart abd area having `7 cm^(2)` each, the molar conductance of the solution will be

Options

A. 233
B. 2.33
C. 23.32
D. 0.233

Answer

To find the molar conductance of the \( \frac{N}{10} \) KCl aqueous solution, we will follow these steps:

### Step 1: Calculate the Cell Constant
The cell constant (k) can be calculated using the formula:
\[ k = \frac{L}{A} \]
where:
- \( L \) is the distance between the electrodes (in cm),
- \( A \) is the area of the electrodes (in cm²).

Given:
- \( L = 4 \, \text{cm} \)
- \( A = 7 \, \text{cm}^2 \)

Substituting the values:
\[ k = \frac{4 \, \text{cm}}{7 \, \text{cm}^2} = \frac{4}{7} \, \text{cm}^{-1} \]

### Step 2: Calculate the Conductance (G)
Conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} \]
Given:
- \( R = 2450 \, \Omega \)

Substituting the value:
\[ G = \frac{1}{2450} \, \text{S} = 0.00040816 \, \text{S} \]

### Step 3: Calculate the Conductivity (κ)
The conductivity (κ) can be calculated using the formula:
\[ \kappa = G \cdot k \]

Substituting the values:
\[ \kappa = 0.00040816 \, \text{S} \cdot \frac{4}{7} \, \text{cm}^{-1} \]
Calculating:
\[ \kappa = 0.00040816 \cdot 0.5714 = 0.000233 \, \text{S/cm} \]

### Step 4: Calculate the Molarity (M)
The normality of the solution is given as \( \frac{N}{10} \), which means:
\[ \text{Normality} = 0.1 \, N \]
For KCl, which dissociates into two ions (K⁺ and Cl⁻), the molarity (M) is equal to the normality:
\[ M = 0.1 \, M \]

### Step 5: Calculate Molar Conductance (Λ)
Molar conductance (Λ) can be calculated using the formula:
\[ \Lambda = \frac{\kappa \cdot 1000}{M} \]
Substituting the values:
\[ \Lambda = \frac{0.000233 \cdot 1000}{0.1} \]
Calculating:
\[ \Lambda = \frac{0.233}{0.1} = 2.33 \, \text{S cm}^2/\text{mol} \]

### Final Answer
The molar conductance of the \( \frac{N}{10} \) KCl aqueous solution is \( 2.33 \, \text{S cm}^2/\text{mol} \).

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