question_answer

On passing \[C\] ampere of electricity through a electrolyte solution for \[t\] second, \[m\]gram metal deposits on cathode. The equivalent weight \[E\] of the metal is       [MP PMT 1990]

A)                 \[E=\frac{C\times t}{m\times 96500}\]  

B)                 \[E=\frac{C\times m}{t\times 96500}\]

C)                 \[E=\frac{96500\times m}{C\times t}\]  

D)                 \[E=\frac{C\times t\times 96500}{m}\]