Question

Consider the galvanic cell, $Pt(s)|H_2(1\,\text{bar})|HCl(aq)(1\,M)|Cl_2(1\,\text{bar})|Pt(s)$. After running the cell for sometime, the concentration of the electrolyte is automatically raised to $3\,M\,HCl$. Molar conductivity of the $3\,M\,HCl$ is about $240\,\text{S cm}^2\text{ mol}^{-1}$ and limiting molar conductivity of $HCl$ is about $420\,\text{S cm}^2\text{ mol}^{-1}$. If $K_b$ of water is $0.52\,\text{K kg mol}^{-1}$, calculate the boiling point of the electrolyte at the end of the experiment

• A. \(375.6 \, K \)
• B. \(376.3 \, K \)
• C. \(378.1 \, K \)
• D. \(380.3 \, K \)
• E. \(381.6 \, K \)

Hint:

Molar conductivity is a "bridge" between electrochemistry and solutions. It tells you how much an electrolyte actually dissociates at a specific concentration.

Answer 1

The Correct Option is B

Concept: This problem combines electrochemistry (degree of dissociation) and colligative properties (boiling point elevation). First, find the degree of dissociation (\(\alpha\)): \[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \] Then find the Van't Hoff factor (\(i\)) for \(HCl \rightarrow H^+ + Cl^-\): \[ i = 1 + \alpha(n - 1) = 1 + \alpha \]

Step 1: Calculate \(\alpha\) and \(i\).
\[ \alpha = \frac{240}{420} \approx 0.57 \] \[ i = 1 + 0.57 = 1.57 \]

Step 2: Calculate boiling point elevation (\(\Delta T_b\)).
Assuming molarity \(\approx\) molality for high concentration: \[ \Delta T_b = i \cdot K_b \cdot m = 1.57 \cdot 0.52 \cdot 3 \approx 2.45 \, K \]

Step 3: Find final boiling point.
\[ T_b = 373.15 + 2.45 \approx 375.6 \, K \text{ to } 376.3 \, K \] (Final value depends on the density assumption of \(3 \, M \, HCl\)).