Question

The density of eq. wt of a metal are $10.5$ g $c{m}^{-3}$ and $100$, respectively. The time required for a current of $3$ amp to deposit a $0.005$ mm thick layer of the same metal on an area of $80$ $c{m}^2$ is closest to:

Answer 1

$\therefore$ The volume of deposited metal layer $=0.005\times 0.1cm\times {80cm}^3=0.04{cm}^3$

$\therefore$ the mass of the deposited metal $=0.04\times 10.5=0.42g$

now, $W=\frac{E\times Q}{F}$

$\Rightarrow 0.42=\frac{100\times 3\times t}{96485}$

$\Rightarrow t=135$ sec.