The resistance of decinormal solution is found to be `2.5 xx 10^(3) Omega`. The equivalent conductance of the solution is (cell constant `= 1.25 cm^(–1)`)

Options

A. `2.5 Omega^(-1) cm^(2)" equiv"^(-1)`
B. `5.0 Omega^(-1) cm^(2)" equiv"^(-1)`
C. `25 Omega^(-1) cm^(2)" equiv"^(-1)`
D. `50 Omega^(-1) cm^(2)" equiv"^(-1)`

Answer

To find the equivalent conductance of a decinormal solution given its resistance and cell constant, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance (R) = \(2.5 \times 10^3 \, \Omega\)
- Cell constant (L/A) = \(1.25 \, \text{cm}^{-1}\)
- Normality (N) = 0.1 N (decinormal solution)

2. **Calculate Specific Conductance (κ):**
The specific conductance (κ) can be calculated using the formula:
\[ \kappa = \frac{L}{R} \]
Where:
- L = Cell constant
- R = Resistance

Plugging in the values:
\[ \kappa = \frac{1.25 \, \text{cm}^{-1}}{2.5 \times 10^3 \, \Omega} \]

3. **Calculate κ:**
\[ \kappa = \frac{1.25}{2.5 \times 10^3} = 0.0005 \, \text{S/cm} \]

4. **Calculate Equivalent Conductance (Λ):**
The equivalent conductance (Λ) can be calculated using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{N} \]
Where:
- κ = Specific conductance
- N = Normality

Plugging in the values:
\[ \Lambda = \frac{0.0005 \, \text{S/cm} \times 1000}{0.1} \]

5. **Calculate Λ:**
\[ \Lambda = \frac{0.5}{0.1} = 5 \, \text{S cm}^{-1} \text{ equivalent}^{-1} \]

### Final Answer:
The equivalent conductance of the decinormal solution is \(5 \, \text{S cm}^{-1} \text{ equivalent}^{-1}\).

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