A 0.05 M NaOH solution offered a resistance of 31.6 `Omega` in a conductivity cell at 298 K. If the cell constant of the conductivity cell is 0.367 `cm^(-1)`, find out the specific and molar conductance of the sodium hydroxide solution.

Options

A. 232 `S cm^(2) mol^(-1)`
B. 116 `S cm^(2) mol^(-1)`
C. 58 `S cm^(2) mol^(-1)`
D. 105 `S cm^(2) mol^(-1)`

Answer

To solve the problem, we need to calculate the specific conductance (κ) and the molar conductance (λm) of the sodium hydroxide (NaOH) solution. Here are the steps to find the solution:

### Step 1: Calculate the Specific Conductance (κ)
The formula to calculate the specific conductance (κ) is given by:

\[ \kappa = \frac{B}{R} \]

Where:
- \( B \) is the cell constant (in cm\(^{-1}\))
- \( R \) is the resistance (in ohms)

Given:
- \( B = 0.367 \, \text{cm}^{-1} \)
- \( R = 31.6 \, \Omega \)

Substituting the values into the formula:

\[ \kappa = \frac{0.367}{31.6} \]

Calculating this gives:

\[ \kappa = 0.0116 \, \text{S/cm} \]

### Step 2: Calculate the Molar Conductance (λm)
The formula to calculate the molar conductance (λm) is:

\[ \lambda_m = \frac{\kappa \times 1000}{C} \]

Where:
- \( C \) is the concentration of the solution (in mol/L)

Given:
- \( C = 0.05 \, \text{mol/L} \)

Substituting the values into the formula:

\[ \lambda_m = \frac{0.0116 \times 1000}{0.05} \]

Calculating this gives:

\[ \lambda_m = \frac{11.6}{0.05} = 232 \, \text{S cm}^2/\text{mol} \]

### Final Results
- Specific Conductance (κ) = 0.0116 S/cm or 116 S cm\(^{-1}\)
- Molar Conductance (λm) = 232 S cm²/mol

### Summary
The specific conductance of the sodium hydroxide solution is 0.0116 S/cm (or 116 S cm\(^{-1}\)), and the molar conductance is 232 S cm²/mol.

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