Calculate the mass of silver deposited when 0.4A of electricity is passed through a silver chloride solution for 2 hours , given that the chemical equivalent of silver is 0.000098g/c

Answer 1

I believe you have made a typo or the problem has the wrong information. According to my information the chemical equivalency of Ag is 1.118 mg Ag/C. But forget that for the moment and solve the problem using coulomgs like this.
coulombs = amperes x seconds
C = 0.4 x 2 hrs x (60 min/hr) x (60 sec/min) = 2880
96,485 coulombs will deposit 107.9 g Ag so
107.9 x (2,880/96,485) = 3.22 g Ag
Or from my number above of 1.118 mg Ag/C it will be
1.118 mg Ag/C x 2,880 C = 3219.8 which rounds to 3220 mg or 3.22 g so the answer is 3.22 g Ag will be deposited. I don't know where the 0.000098 g/c comes from.