[Single-digit Integer] By passing a certain amount of charge through solution, 9.08 L of chlorine ga — Electrochemistry Chemistry Question
Question
[Single-digit Integer] By passing a certain amount of charge through $NaCl$ solution, 9.08 L of chlorine gas were liberated at STP. When the same amount of charge is passed through a nitrate solution of metal M, 52.8 g of the metal was deposited. If the specific heat of metal is 0.032 Cal/°C-g, the valency of metal is
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💡 Solution & Explanation
\textbf{Step 1: Find moles of electrons from chlorine liberation.} At STP, $V(\text{Cl}_2) = 9.08\ \text{L}$: \[ n(\text{Cl}_2) = \frac{9.08}{22.4} = 0.405\ \text{mol} \] From: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$: \[ n_e = 2 \times n(\text{Cl}_2) = 2 \times 0.405 = 0.81\ \text{mol} \] \textbf{Step 2: Find atomic mass of metal M (Dulong-Petit law).} \[ \text{Atomic mass} \times \text{specific heat} \approx 6.4\ \text{cal mol}^{-1}\ ^\circ\text{C}^{-1} \] \[ \text{Atomic mass} = \frac{6.4}{0.032} = 200\ \text{g mol}^{-1} \] \textbf{Step 3: Calculate valency v.} By Faraday's law: $n_e = v \times n(M)$, where $n(M) = m/M_r$: \[ v = \frac{n_e \times M_r}{m} = \frac{0.81 \times 200}{52.8} = \frac{162}{52.8} \approx 3.07 \approx \boxed{3} \] The metal has \textbf{valency 3}.