MCQ
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
- A$E = \frac{{C \times t}}{{m \times 96500}}$
- B$E = \frac{{C \times m}}{{t \times 96500}}$
- ✓$E = \frac{{96500 \times m}}{{C \times t}}$
- D$E = \frac{{C \times t \times 96500}}{m}$
