Chemistry Forum

31 g of pure K₃PO₄ are dissolved in 250 g of distilled water at a density of 1 g / mL. The resulting solution is then electrolyzed for 14 hours under constant electric current. After electrolysis, the mass fraction of salt in the solution increases by 3.4%, and the mass and appearance of the electrodes do not change during electrolysis. Calculate how much electric current in amperes passed through K₃PO₄ solution for 14 hours?

Mass Fraktion before is 31 g/281 g = 0,11032
Mass Fraktion after is 31g*1,034/281 g =0,11407
This value equal to the 31 g gives new total mass of 271,76
The difference is 9,237 g water. What gives about = 0,5131 mol H₂ equal 1,026 mol H what gives 1,018 g H
m = I*t* c
c= 0,037607 g/ Ah
t = 14 h
I = m/(t*c)
I = 1,018 g/(14h * 0,037607g/ Ah) = 1,93 A

Thanks, but the problem is that the solutions say the answer is 14 A. Is that possible?

With this Datas no. Show the calculation.

Even I calculated it, the result is the same as yours. This is an example from a chemical competition, and the official answer is 14 A. But now I'm not sure what to do, so I turned to this forum for help. So you don't believe it's possible to achieve 14 amperes in any case?

Then time cannot be 14 h. If the time is 1,93 h then it works.