The molar conductivity of 0.04 M solution of `MgCl_2` is 200 `Scm^2mol^(-1)` at 298 k. A cell with electrodes that are 2.0`cm^2` in surface area and 0.50cm apart is filled with `MgCl_2` solution
How much current will flow when the potential difference between the two electrodes is 5.0V?

Options

A. 156.25V
B. 0.16A
C. 160A
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Specific Conductance (k)
The relationship between molar conductivity (Λ) and specific conductance (k) is given by the formula:
\[ k = \frac{\Lambda \times C}{1000} \]
where:
- \(\Lambda = 200 \, \text{S cm}^2/\text{mol}\) (molar conductivity)
- \(C = 0.04 \, \text{mol/L}\) (molarity)

Substituting the values:
\[ k = \frac{200 \times 0.04}{1000} = 0.008 \, \text{S/cm} \]

### Step 2: Calculate the Resistance (R)
The resistance (R) of the solution can be calculated using the formula:
\[ R = \frac{\rho \cdot l}{A} \]
where:
- \(\rho = \frac{1}{k}\) (specific resistance)
- \(l = 0.5 \, \text{cm}\) (distance between electrodes)
- \(A = 2.0 \, \text{cm}^2\) (surface area of electrodes)

First, calculate \(\rho\):
\[ \rho = \frac{1}{0.008} = 125 \, \text{ohm cm} \]

Now, substitute into the resistance formula:
\[ R = \frac{125 \cdot 0.5}{2.0} = \frac{62.5}{2.0} = 31.25 \, \text{ohms} \]

### Step 3: Calculate the Current (I)
Using Ohm's law, the current (I) can be calculated using the formula:
\[ I = \frac{V}{R} \]
where:
- \(V = 5.0 \, \text{V}\) (potential difference)

Substituting the values:
\[ I = \frac{5.0}{31.25} \approx 0.16 \, \text{A} \]

### Final Answer
The current that will flow when the potential difference of 5.0 V is applied is approximately **0.16 A**.

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