Correct Answer: Option D

Explanation

Mg\(^{2+}\)  + 2e\(^-\) → Mg

If        2F produces 24g Mg

then   x  will produce 1g Mg

 24x = 2F

     x = \(\frac{2F}{24}\)

Similarly, Al\(^{3+}\)  +  3e\(^-\) → Al

If        3F produces 27g Al

then   x  will produce 10g Al

27x = 3F

     x = \(\frac{{3F}\times{10}}{27}\)

     x =  \(\frac{30F}{27}\)

Now, if  \(\frac{2F}{24}\) of Mg costs N5

then    \(\frac{30F}{27}\) of Al will cost y

By cross- multiplying to make y the subject of formula.

    \(\frac{30F}{27}\times 5\) = \(\frac{2F}{24}\times y\)

      y =  \(\frac{30F}{27}\times 5\) ÷  \(\frac{2F}{24}\)

      y =  \(\frac{30F}{27}\times 5\) x \(\frac{24}{2F}\)

      y =  \(\frac{{30F}\times{24}\times{5}}{{27}{2F}}\)

      y = N 66.67

Therefore, it will cost N66.67 to deposit 10g of Aluminium.

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