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 can follow these steps:

### Step 1: Identify Given Values
- Resistance \( R = 2450 \, \Omega \)
- Distance between electrodes \( L = 4 \, \text{cm} \)
- Area of electrodes \( A = 7 \, \text{cm}^2 \)
- Normality of KCl solution \( N = \frac{1}{10} \, N \) or \( 0.1 \, N \)

### Step 2: Convert Normality to Molarity
Normality (N) is related to molarity (M) by the equation:
\[ \text{Normality} = \text{Molarity} \times n \]
For KCl, the \( n \) factor is 1 (since KCl dissociates into one K\(^+\) and one Cl\(^-\)). Therefore:
\[ \text{Molarity} = \frac{0.1 \, N}{1} = 0.1 \, M \]

### Step 3: Calculate Specific Conductance (κ)
The specific conductance (κ) can be calculated using the formula:
\[ \kappa = \frac{L}{A \cdot R} \]
Substituting the values:
\[ \kappa = \frac{4 \, \text{cm}}{7 \, \text{cm}^2 \cdot 2450 \, \Omega} \]
Calculating the denominator:
\[ \kappa = \frac{4}{7 \cdot 2450} \]
\[ \kappa = \frac{4}{17150} \approx 0.000233 \, \text{S/cm} \]

### Step 4: Calculate Molar Conductance (Λ)
Molar conductance (Λ) is given by the formula:
\[ \Lambda = \frac{1000 \cdot \kappa}{C} \]
Where \( C \) is the molarity. Substituting the values:
\[ \Lambda = \frac{1000 \cdot 0.000233}{0.1} \]
Calculating:
\[ \Lambda = \frac{0.233}{0.1} = 2.33 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer
The molar conductance of the \( \frac{N}{10} \) KCl aqueous solution is approximately:
\[ \Lambda \approx 2.33 \, \text{S cm}^2 \text{mol}^{-1} \]