Impure copper containing Fe ,Au ,Ag as impurities is elecrolytically refined A current of 140 A for 482.25 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g percentage of iron in impure copper is (Given molar mass of Fe =55.5 `mol^(-1)` molar mass of Cu =63.54 g `mol^(-1)`)

Options

A. 0.95
B. 0.85
C. 0.97
D. 0.9

Answer

D

Explanation

Amount pf impurity
=decreased mass of anode - increased mass of cathode
amount of pure Cu deposited
`W=Zit =(C )/(96500)xxit`
`=(63.54)/(2xx96500)xx140 xx482.5` =22.239 g
`therefore` amount of impurity (Fe) = 22.239 -22.011 = 0.228 g
Now from faraday second law of electrolysis
`("weight of fe deposited")/(0.228) =(27.55)/(31.77)`
`therefore` Weight of Fe `=(27.75 xx0.288)/(22.26) xx100=0.88 =0.09`