The resistance of a `0.01N` solution of an electroyte was found to 210 ohm at `298K` using a conductivity cell with a cell constant of `0.88 cm^(-1)`. Calculate specific conductance and equilvalent conductance of solution.

Options

A. `"400 mho cm"^(2)"g eq"^(-1)`
B. `"295 mho cm"^(2)"g eq"^(-1)`
C. `"419 mho cm"^(2)"g eq"^(-1)`
D. `"425 mho cm"^(2)"g eq"^(-1)`

Answer

To solve the problem step by step, we will calculate the specific conductance and equivalent conductance of the given electrolyte solution.

### Step 1: Calculate Specific Conductance (κ)

The formula for specific conductance (κ) is given by:

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

where:
- \(L\) is the cell constant (given as \(0.88 \, \text{cm}^{-1}\))
- \(R\) is the resistance (given as \(210 \, \Omega\))

Now substituting the values into the formula:

\[ \kappa = \frac{0.88 \, \text{cm}^{-1}}{210 \, \Omega} \]

Calculating this gives:

\[ \kappa = \frac{0.88}{210} \approx 0.00419 \, \text{S/cm} = 4.19 \times 10^{-3} \, \text{S/cm} \]

### Step 2: Calculate Equivalent Conductance (Λ)

The formula for equivalent conductance (Λ) is given by:

\[ \Lambda = \frac{\kappa \times 1000}{N} \]

where:
- \(N\) is the normality of the solution (given as \(0.01 \, \text{N}\))

Substituting the values into the formula:

\[ \Lambda = \frac{4.19 \times 10^{-3} \, \text{S/cm} \times 1000}{0.01} \]

Calculating this gives:

\[ \Lambda = \frac{4.19 \times 10^{-3} \times 1000}{0.01} = \frac{4.19}{0.01} = 419 \, \text{S cm}^2/\text{eq} \]

### Final Answers

- Specific Conductance (κ) = \(4.19 \times 10^{-3} \, \text{S/cm}\)
- Equivalent Conductance (Λ) = \(419 \, \text{S cm}^2/\text{eq}\)