`wedge _(eq)` of `0.10N` solution of `CaI_(2)` is `100.0 Scm^(2)eq^(-1)` at `298 K.G^(**)` of the cell `=0.25 cm^(-1)`. How much current will flow potential difference between the electrode is `5V` ?

Options

A. 0.4
B. 0.6
C. 0.2
D. 0.8

Explanation

To solve the problem step by step, we will use the given data to find the current flowing through the solution when a potential difference is applied.

### Step 1: Understand the Given Data
- Equivalent conductivity (Λ) of 0.10 N solution of CaI₂ = 100 S cm² eq⁻¹
- Cell constant (G*) = 0.25 cm⁻¹
- Potential difference (V) = 5 V

### Step 2: Calculate Conductivity (k)
The relationship between equivalent conductivity (Λ) and conductivity (k) is given by:

\[ k = \frac{\Lambda \times n}{1000} \]

Where:
- n = normality of the solution = 0.10 N

Substituting the values:

\[ k = \frac{100 \, \text{S cm}^2 \text{eq}^{-1} \times 0.10 \, \text{eq L}^{-1}}{1000} \]

Calculating k:

\[ k = \frac{10}{1000} = 0.01 \, \text{S cm}^{-1} \]

### Step 3: Calculate Resistance (R)
The relationship between conductivity (k), resistance (R), length (L), and area (A) is given by:

\[ k = \frac{L}{R \cdot A} \]

Rearranging the formula to find R:

\[ R = \frac{L}{k \cdot A} \]

Since the cell constant (G*) is defined as:

\[ G^* = \frac{L}{A} \]

We can substitute G* into the equation for R:

\[ R = \frac{G^*}{k} \]

Substituting the values:

\[ R = \frac{0.25 \, \text{cm}^{-1}}{0.01 \, \text{S cm}^{-1}} = 25 \, \Omega \]

### Step 4: Calculate Current (I)
Using Ohm's law, the current (I) can be calculated as:

\[ I = \frac{V}{R} \]

Substituting the values:

\[ I = \frac{5 \, \text{V}}{25 \, \Omega} = 0.2 \, \text{A} \]

### Final Answer
The current that will flow when a potential difference of 5 V is applied is **0.2 A**.

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