If the cost of electricity required to deposit \(1 g\) of aluminum is \(\# 4.00\), how much would it cost to deposit \(24 g\) of copper' \([ A ]=27, Cu =64]\)
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Correct Answer: Option A
Explanation:
No of mole of \(\frac{1 g }{27 g / mol }\).
\(=0.037\) motl. deposited
The discharge of \(Al l^{3+}\) is:
\(Al _{(\text {oq })}^{3+}+3 e ^{-} \rightarrow Al _{(s)}\)
1 mole of \(Al\) is deposited by \(3 F\)
\(0.037\) mole of \(A /\) would be deposited by.
\((0.037 \times 3) F =0.111 F\)lso
The no of mole of Cato beideposited
\(=\frac{24 g }{64 g / mol }=0.375 mol\)
The discharge of \(Cu ^{2+}\) is:
\(Cu _{(a+)}^{2+}+2 e ^{-} \rightarrow Cu _{(s)}\)
1 mole of \(Cu\) is deposited by \(2 F\)
\(0.375 mole\) of \(Cu\) would be deposited by
\((0.375 \times 2) F =0.75 F\)ut the cost of \(0.111 F\) of electricity is \(\# 4.00\)
\(0.75 F\) would cost \(\left(\frac{0.75 F \times \# 4: 00}{0.111 F }\right)\)
\(=\# 27.02\)
No of mole of \(\frac{1 g }{27 g / mol }\).
\(=0.037\) motl. deposited
The discharge of \(Al l^{3+}\) is:
\(Al _{(\text {oq })}^{3+}+3 e ^{-} \rightarrow Al _{(s)}\)
1 mole of \(Al\) is deposited by \(3 F\)
\(0.037\) mole of \(A /\) would be deposited by.
\((0.037 \times 3) F =0.111 F\)lso
The no of mole of Cato beideposited
\(=\frac{24 g }{64 g / mol }=0.375 mol\)
The discharge of \(Cu ^{2+}\) is:
\(Cu _{(a+)}^{2+}+2 e ^{-} \rightarrow Cu _{(s)}\)
1 mole of \(Cu\) is deposited by \(2 F\)
\(0.375 mole\) of \(Cu\) would be deposited by
\((0.375 \times 2) F =0.75 F\)ut the cost of \(0.111 F\) of electricity is \(\# 4.00\)
\(0.75 F\) would cost \(\left(\frac{0.75 F \times \# 4: 00}{0.111 F }\right)\)
\(=\# 27.02\)