The molar conductance of 0.05 M solution of `MgCl_(2)` is 194.5 `Omega^(-1) cm^(2) mol^(-1)` at `25^(@)C`. A cell with electrodes having 1.50 `cm^(2)` surface area and 0.50 cm apart is filled with 0.05 M solution of `MgCl_2`. How much current will flow when the potential difference between the electrodes is 5.0V?

Explanation

To solve the problem step by step, we will follow the outlined procedure to calculate the current flowing through the cell when a potential difference is applied.

### Step 1: Calculate the Conductivity (κ)
We know that the molar conductance (Λ_m) is related to the conductivity (κ) by the formula:
\[ \kappa = \Lambda_m \times C \]
where:
- \(\Lambda_m = 194.5 \, \Omega^{-1} \, cm^2 \, mol^{-1}\) (given)
- \(C = 0.05 \, mol/L\)

First, we need to convert the concentration from mol/L to mol/cm³:
\[ C = 0.05 \, mol/L = 0.05 \, mol/1000 \, cm^3 = 0.00005 \, mol/cm^3 \]

Now we can calculate κ:
\[ \kappa = 194.5 \, \Omega^{-1} \, cm^2 \, mol^{-1} \times 0.05 \, mol/cm^3 = 9.725 \, \Omega^{-1} \, cm^{-1} \]

### Step 2: Calculate the Cell Constant (L/A)
The cell constant (k) is given by:
\[ k = \frac{L}{A} \]
where:
- \(L = 0.50 \, cm\) (distance between electrodes)
- \(A = 1.50 \, cm^2\) (surface area of electrodes)

Calculating the cell constant:
\[ k = \frac{0.50 \, cm}{1.50 \, cm^2} = \frac{1}{3} \, cm^{-1} \approx 0.333 \, cm^{-1} \]

### Step 3: Calculate the Resistance (R)
The resistance (R) can be calculated using the formula:
\[ R = \frac{1}{\kappa \cdot k} \]
Substituting the values:
\[ R = \frac{1}{9.725 \, \Omega^{-1} \, cm^{-1} \cdot 0.333 \, cm^{-1}} \approx \frac{1}{3.241} \approx 0.308 \, \Omega \]

### Step 4: Calculate the Current (I)
Using Ohm's Law, we can find the current (I) using the formula:
\[ I = \frac{V}{R} \]
where:
- \(V = 5.0 \, V\) (potential difference)

Now substituting the values:
\[ I = \frac{5.0 \, V}{0.308 \, \Omega} \approx 16.23 \, A \]

### Final Answer
The current that will flow when the potential difference between the electrodes is 5.0 V is approximately **16.23 A**.

---