In a galvanic cell, after running the cell for sometimes, the concentration of the electrolyte is automatically raised to 3 M HCl. Molar conductivity of the 3 M HCl is about `"240 S cm"^(2)" mol"^(-1)` and limiting molar conductivity of HCl is about `"420 S cm"^(2)" mol"^(-1)`. If `K_(b)` of water is `"0.52 K kg mol"^(-1)`, calculate the boiling point of the electrolyte at the end of the experiment.

Options

A. 375.6 K
B. 376.3 K
C. 378.1 K
D. 380.3 K

Answer

To calculate the boiling point of the electrolyte (3 M HCl) at the end of the experiment, we will follow these steps:

### Step 1: Identify the given data
- Molar conductivity of 3 M HCl, \( \Lambda_m = 240 \, \text{S cm}^{-2} \text{mol}^{-1} \)
- Limiting molar conductivity of HCl, \( \Lambda_0 = 420 \, \text{S cm}^{-2} \text{mol}^{-1} \)
- \( K_b \) of water = \( 0.52 \, \text{K kg mol}^{-1} \)
- Concentration (molality) of HCl = 3 M

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor \( i \) can be calculated using the formula:
\[ i = 1 + \frac{\alpha}{1 - \alpha} \]
Where \( \alpha \) is the degree of dissociation.

To find \( \alpha \), we can use the relation:
\[ \alpha = \frac{\Lambda_m}{\Lambda_0} \]
Substituting the values:
\[ \alpha = \frac{240}{420} = \frac{4}{7} \]

Now substituting \( \alpha \) back into the equation for \( i \):
\[ i = 1 + \frac{\frac{4}{7}}{1 - \frac{4}{7}} = 1 + \frac{\frac{4}{7}}{\frac{3}{7}} = 1 + \frac{4}{3} = \frac{7}{3} \]

### Step 3: Calculate the elevation in boiling point (\( \Delta T_b \))
The formula for the elevation in boiling point is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \( m \) is the molality (which is 3 in this case).

Substituting the values:
\[ \Delta T_b = \left(\frac{7}{3}\right) \cdot 0.52 \cdot 3 \]
Calculating this gives:
\[ \Delta T_b = \frac{7}{3} \cdot 0.52 \cdot 3 = 7 \cdot 0.52 = 3.64 \, \text{K} \]

### Step 4: Calculate the boiling point of the electrolyte
The boiling point of the solution can be calculated using:
\[ T_b = 273.15 + \Delta T_b \]
Substituting the value of \( \Delta T_b \):
\[ T_b = 273.15 + 3.64 = 276.79 \, \text{K} \]

### Step 5: Final Calculation
Since the boiling point is typically expressed in degrees Celsius, we can convert it:
\[ T_b = 276.79 \, \text{K} \approx 3.64 \, \text{°C} \]

### Conclusion
The boiling point of the electrolyte at the end of the experiment is approximately **276.79 K**.

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