A conductivity cell find with 0.1 M KCI gives at `25^(@) C` a resistance of 85.5 other . The conductivity of 0.1 M KCI at `25^(@) C` is 0.01286 `"ohm"^(-1) cm^(-1) `. The same cell filled with 0.005 M HCI gives a resistance of 529 ohms. What is the molar conductivity of HCI solution at `25^(@) C`?

Explanation

To find the molar conductivity of the HCl solution at 25°C, we can follow these steps:

### Step 1: Calculate the Cell Constant (B)
We know that the conductivity (K) is related to the cell constant (B) and resistance (R) by the formula:

\[ K = \frac{B}{R} \]

From the data given for the 0.1 M KCl solution:
- Resistance (R) = 85.5 ohms
- Conductivity (K) = 0.01286 ohm\(^{-1}\) cm\(^{-1}\)

Rearranging the formula to find B:

\[ B = K \times R \]

Substituting the values:

\[ B = 0.01286 \, \text{ohm}^{-1} \text{cm}^{-1} \times 85.5 \, \text{ohms} \]

Calculating B:

\[ B = 1.0993 \, \text{cm}^{-1} \]

### Step 2: Calculate the Conductivity of the HCl Solution
Now, we will use the same cell constant for the 0.005 M HCl solution, which has a resistance of 529 ohms.

Using the formula again:

\[ K = \frac{B}{R} \]

Substituting the values:

\[ K = \frac{1.0993 \, \text{cm}^{-1}}{529 \, \text{ohms}} \]

Calculating K:

\[ K = 0.00208 \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Calculate the Molar Conductivity of the HCl Solution
The molar conductivity (\(\Lambda_m\)) is given by the formula:

\[ \Lambda_m = \frac{K \times 1000}{C} \]

Where:
- K = conductivity of the HCl solution = 0.00208 ohm\(^{-1}\) cm\(^{-1}\)
- C = concentration of the HCl solution = 0.005 M

Substituting the values:

\[ \Lambda_m = \frac{0.00208 \, \text{ohm}^{-1} \text{cm}^{-1} \times 1000}{0.005} \]

Calculating \(\Lambda_m\):

\[ \Lambda_m = \frac{2.08}{0.005} = 416 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Final Answer
The molar conductivity of the HCl solution at 25°C is:

\[ \Lambda_m = 416 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]

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