A current of 45A is passed through a solution of gold salt for 1 hour 45 minutes calculate (a) the mass of gold deposited (b) the numbers of moles of gold deposited (c) if the same current is used, find the time taken for 5.5g of gold to be deposited (Ah=197, 1F=96500C

6 answers

I assume this is AuCl3 or similar with Au in the +3 state. Also I assume you made a typo with that Ah when you meant to type Au for gold.
1 hr 45 minutes = 60 min + 45 min = 105 min
Coulombs = amperes x seconds = 45 x (105 min x 60 sec/min) = 283,500 C
96,500 C will deposit 197/3 = 65.7 g Au so
65.7 g Au x (283,500/96,500) = ? g Au deposited.
mols Au = grams Au/atomic mass Au = ?
(c) coulombs needed is
96,500 C will deposit 65.7 g Au. To deposit 4.5 g Au will require
96,500 C x 5.5 g/65.7 g = 8,078 C needed, then
C = amperes x seconds.
8078 C = 45 A x seconds. Solve for seconds.
Post your work if you get stuck.

0.334G

Not exactly what I got in my solvings

2700sec

127575

578.8 of Au