A) 4.5g
B) 9.0g
C) 13.5g
D) 27g
Correct Answer: A
Solution :
According to Faradays second law of electrolysis, the amount of different substances deposited by the passage of same quantity of electricity are proportional to their chemical equivalent weights. \[W\propto E\] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\] \[{{W}_{KCl}}=19.5g;{{E}_{{{K}^{+}}}}=39\] \[{{W}_{AlC{{l}_{3}}}}=?;{{E}_{A{{l}^{3+}}}}=\frac{27}{3}=9\] then , \[\frac{19.5}{x}=\frac{39}{9}\] \[x=\frac{19.5\times 9}{39}=4.5\,g\] So, 4.5 g of aluminium is deposited.You need to login to perform this action.
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