# How to solve this Q.72. Calculate the current (in mA) required to deposit 0.195 g of platinum metal in 5.0 hours from a solution of ${\left[PtC{l}_{6}\right]}^{2-}$: (atomic weight : Pt = 195) (1) 310 (2) 31 (3) 21.44 (4) 5.36

21.44
• -5
195g ----- 4×96500 0.195-----(4×96500)÷1000 = 4×96.5 Now we know , Q=it i = (4×96.5)÷(5×60×60) = 21.44
• -9