96.5 amperes current is passed through the molten $\mathrm{AlCl}_3$ for 100 seconds. The mass of aluminium deposited at the cathode is (atomic weight of $\mathrm{Al}=27 \mathrm{u}$ )

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Aicharya · Accepted Answer

Given, $I=96.5 \mathrm{~A}$ $t=100 \mathrm{~s}$ $\Rightarrow \quad m=?$ Atomic weight of $\mathrm{Al}=27 \mathrm{u}$ $Q=I 	imes t=96.5 	imes 100$ $=9650 \mathrm{C}$ $\begin{aligned} & \mathrm{AlCl}_3 \longrightarrow \mathrm{Al}^{3+}+3 \mathrm{Cl}^{-} \ & \mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al} \ & 96500 	imes 3 \mathrm{C} 	ext { of electricity deposits, } 1 	ext { mole } \mathrm{Al} .\end{aligned}$ $	herefore 9650 \mathrm{C}$ will deposit $\mathrm{Al}=\frac{9650}{96500 	imes 3}=0.0333 \mathrm{~mol}$. Weight of $\mathrm{Al}$ deposited $=n 	imes$ molar mass $\begin{aligned} & =0.0333 	imes 27 \ & \cong 0.90 \mathrm{~g}\end{aligned}$

96.5 amperes current is passed through the molten $\mathrm{AlCl}_3$ for 100 seconds. The mass of aluminium…

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