\(96.5\) amperes current is passed through the molten \(AlCl_3\) for \(100\) seconds. The mass of aluminium deposited at the cathode is
Atomic weight of \(Al=27\,u\)
Atomic weight of \(Al=27\,u\)
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- \(0.90\,g\)
- \(0.45\,g\)
- \(1.35\,g\)
- \(1.8\,g\)
The Correct Option is A
Solution and Explanation
In molten \(AlCl_3\), aluminium ions are reduced at the cathode: \[ Al^{3+}+3e^-\longrightarrow Al \] Thus, \(3\) moles of electrons are required to deposit \(1\) mole of aluminium.
Step 2: Use Faraday's law of electrolysis.
The mass deposited is given by \[ m=\frac{EIt}{F} \] where \[ E=\frac{\text{atomic mass}}{\text{valency}} \] For aluminium, \[ E=\frac{27}{3}=9 \]
Step 3: Substitute the given values.
Given, \[ I=96.5\,A \] \[ t=100\,s \] \[ F=96500\,C\,mol^{-1} \] So, \[ m=\frac{9\times 96.5\times 100}{96500} \] \[ m=\frac{9\times 9650}{96500} \] \[ m=0.90\,g \]
Step 4: Final conclusion.
Therefore, the mass of aluminium deposited at the cathode is \[ \boxed{0.90\,g} \]
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