The resistance of 0.05 M NaOH solution is `31.6 Omega` and its cell constant is `0.357 cm^(-1)` . Calculate its conductivity and molar conductivity .

Explanation

To solve the problem, we need to calculate the conductivity (κ) and the molar conductivity (λm) of the 0.05 M NaOH solution using the given resistance and cell constant.

### Step 1: Calculate Conductance (G)
Conductance (G) is the reciprocal of resistance (R). The formula is:
\[ G = \frac{1}{R} \]
Given that the resistance \( R = 31.6 \, \Omega \), we can calculate conductance:
\[ G = \frac{1}{31.6} \approx 0.031645 \, \text{S} \, (\text{Siemens}) \]

### Step 2: Calculate Conductivity (κ)
The conductivity (κ) can be calculated using the formula:
\[ κ = G \times \text{Cell Constant} \]
The cell constant is given as \( 0.357 \, \text{cm}^{-1} \). Now substituting the values:
\[ κ = 0.031645 \, \text{S} \times 0.357 \, \text{cm}^{-1} \approx 0.0113 \, \text{S cm}^{-1} \]

### Step 3: Calculate Molar Conductivity (λm)
Molar conductivity (λm) can be calculated using the formula:
\[ λ_m = \frac{κ \times 1000}{C} \]
where \( C \) is the concentration in mol/L. Given that the concentration \( C = 0.05 \, \text{mol/L} \):
\[ λ_m = \frac{0.0113 \, \text{S cm}^{-1} \times 1000}{0.05} \approx 0.226 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Results
- Conductivity (κ) = \( 0.0113 \, \text{S cm}^{-1} \)
- Molar Conductivity (λm) = \( 0.226 \, \text{S cm}^2 \text{mol}^{-1} \)