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 with a resistance of \(2.5 \times 10^{3} \, \Omega\) and a cell constant of \(1.25 \, \text{cm}^{-1}\), we can follow these steps:

### Step 1: Calculate the Conductance
Conductance (\(G\)) is the reciprocal of resistance (\(R\)).

\[ G = \frac{1}{R} \]

Given \(R = 2.5 \times 10^{3} \, \Omega\):

\[ G = \frac{1}{2.5 \times 10^{3}} = 0.4 \times 10^{-3} \, \text{S} \, (\text{Siemens}) \]

### Step 2: Calculate the Specific Conductivity (\(\kappa\))
The specific conductivity (\(\kappa\)) can be calculated using the formula:

\[ \kappa = G \times \text{cell constant} \]

Substituting the values we have:

\[ \kappa = 0.4 \times 10^{-3} \, \text{S} \times 1.25 \, \text{cm}^{-1} \]

Calculating this gives:

\[ \kappa = 0.5 \times 10^{-3} \, \text{S/cm} \]

### Step 3: Calculate the Equivalent Conductance (\(\Lambda\))
The equivalent conductance (\(\Lambda\)) can be calculated using the formula:

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

Where \(N\) is the normality of the solution. For a decinormal solution, \(N = 0.1 \, \text{N}\).

Substituting the values:

\[ \Lambda = \frac{0.5 \times 10^{-3} \, \text{S/cm} \times 1000}{0.1} \]

Calculating this gives:

\[ \Lambda = \frac{0.5 \times 10^{-3} \times 1000}{0.1} = 5.0 \, \text{S cm}^{-1} \text{ equivalent}^{-1} \]

### Final Answer
The equivalent conductance of the decinormal solution is:

\[ \Lambda = 5.0 \, \text{S cm}^{-1} \text{ equivalent}^{-1} \]