A plating shop never asks for a mass. It asks for a setting: how many amperes, for how many minutes, to land the thickness the drawing calls for. That question is this chapter's last two formulas welded end to end — $Q = I\,t$ feeds $m = \dfrac{QM}{nF}$ — and the weld holds in both directions.

Forwards, from the setting: seconds times amperes give coulombs, and coulombs give grams. Backwards, from the drawing: $Q = \dfrac{mnF}{M}$ prices the wanted mass in coulombs, and then $I = \dfrac{Q}{t}$ or $t = \dfrac{Q}{I}$ splits that bill between the dial and the clock. The cast is unchanged — $m$ in grams, $Q$ in coulombs, $I$ in amperes, $t$ in seconds, $M$ in g/mol, and $n$ a bare count of electrons per ion — and whichever one the question leaves blank is the one you solve for.

Push the units through every rearrangement you write. $\dfrac{\mathrm{C} \cdot \mathrm{g/mol}}{\mathrm{C/mol}}$ cancels to grams, which is precisely why Faraday's law is shaped the way it is. If your units refuse to cancel into the answer's units, the rearrangement is wrong — no appeal. Units that DO work out never prove you right; the check is a one-way street.

One wink before you start. The current $9.65\ \mathrm{A}$ turns up in these problems the way 3-4-5 turns up in triangles: held for 1000 s it delivers 9650 C, which is exactly a tenth of a mole of electrons. Examiners are not being random there. They are being kind.