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]

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

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

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

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

On passing \[C\] ampere of electricity through a electrolyte solution

1.	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.	\[E=\frac{C\times t\times 96500}{m}\]
Answer» D. \[E=\frac{C\times t\times 96500}{m}\]

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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]

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