TS EAMCET · Chemistry · Electrochemistry
The number of moles of electrons required to deposit \(36 \mathrm{~g}\) of \(\mathrm{Al}\) from an aqueous solution of \(\mathrm{Al}\left(\mathrm{NO}_3\right)_3\) is (At. wt. of \(\mathrm{Al}=27\) )
- A \(4\)
- B \(2\)
- C \(3\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Al}^{3+}+\underset{3 \mathrm{~mol}}{3 e^{-}} \longrightarrow \underset{1 \mathrm{~mol}=27 \mathrm{~g}}{\mathrm{Al}}\) \(\because 27 \mathrm{~g}\) of \(\mathrm{Al}\) is deposited by 3 moles of electrons \(\therefore 36 \mathrm{~g} \mathrm{Al}\) will be deposited by…
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