A) \[5.49\times {{10}^{1}}\text{ }C\] of electricity
B) \[5.49\times {{10}^{4}}\text{ }C\] of electricity
C) \[1.83\times {{10}^{7}}C\] of electricity
D) \[5.49\times {{10}^{7}}C\] of electricity
Correct Answer: D
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] \[w=zQ\] where w = amount of metal \[=5.12\text{ }kg\] \[=5.12\times {{10}^{3}}g\] z = electrochemical equivalent \[=\frac{equivalent\text{ }weight}{96500}\] \[=\frac{atomic\,mass}{electron\times 96500}\] \[=\frac{27}{3\times 96500}\] \[5.12\times {{10}^{3}}=\frac{27}{3\times 96500}\times Q\] \[Q=\frac{5.12\times {{10}^{3}}\times 3\times 96500}{27}C\] \[=5.49\times {{10}^{7}}C\]
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