The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0 . 015 A…

Given, t = $15$min

${\mathrm{Zn}}^{+2}+2{\mathrm{e}}^{-}⟶\mathrm{Zn}$

Quantity of charge passed (Q) =Current in amperes ×Time in seconds.
=$$\begin{gathered} (0.015\mathrm{A})\times (15\times 60)\mathrm{s} \\ =13.5\mathrm{C} \end{gathered}$$

Gram equivalent mass of zinc = $\frac{65.4}{2}=32.7$g
Now,  $96500$ C of charge deposit zinc =$32.7$ g
$13.5$ C charge will deposit zinc = $$\begin{gathered} \frac{(32.7\mathrm{g})}{\left(96500\mathrm{C}\right)}\times (13.5\mathrm{C}) \\ =4.57\times {10}^{-3} \mathrm{g} \end{gathered}$$

$=45.75\times {10}^{-4}\mathrm{gm}$