1.

Calculate the current ( in ma ) required to deposit 0.195 gr of platinum metal in 5.0 hours from a solution of PtCl_6^(-2) ( Atomic weight : pt 195)

Answer»

310
31
`21.44`
`5.36`

SOLUTION :`1F to 1 ` EQ. WT of Pt . `((Pt)/(+4) Cl_6)^(-2) , 96500 C to ((195)/(4))gm`
Q to 0.195 gm , `Q = (0.195 xx 96500)/(((195)/(4))^(1000))=C xx t `
` 96.5 xx 4 = C xx 5 xx 60 xx 60 , C = 0.021145 amp = 21.45 ` MILLI amp


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