A $5ampere$ current is passed through a solution of zinc sulphate for $40minutes$. The amount of zinc deposited at the cathode is:

The correct option is D $4.065g$
Current ($I$) = $5ampere$ and time ($t$) = $40minutes=2400seconds$.
Amount of electricity passed ($Q$) = $I\times$t
= $5\times 2400=12000C$
Now, $Z{n}^{2+}$ + $2e^{–}$ $\to$ $Zn$ ($1mole=65.39g$)
Since, two mole electronic charges (i.e., $2\times 96500C$) deposits $65.39g$ of zinc, therefore $12000C$ will deposit
= ($\frac{65.39\times 12000}{2\times 96500}$)
= $4.065g$ of zinc