By passing a certain amount of charge through NaCl solution, 9.08 L of chlorine gas were liberated a — Electrochemistry Chemistry Question
Question
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
💡 Solution & Explanation
Moles of $Cl_2 = 9.08 / 22.4 = 0.405$ mol. Since $2Cl^- \rightarrow Cl_2 + 2e^-$, the equivalents of $Cl_2 = 0.405 \times 2 = 0.81$ eq. By Faraday's second law, 0.81 eq of metal M is also deposited. Equivalent weight of M = $52.8 / 0.81 = 65.18$ g/eq. By Dulong-Petit law, approximate atomic mass = $6.4 / \text{specific heat} = 6.4 / 0.032 = 200$ g/mol. Valency = Atomic mass / Equivalent weight = $200 / 65.18 \approx 3.06 \approx 3$. Therefore, correct answer is 3.