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Impure copper containing Fe ,Au ,Ag as...

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)`)

A

0.95

B

0.85

C

0.97

D

0.9

Text Solution

Verified by Experts

The correct Answer is:
D
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`
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Knowledge Check

  • Assume that impure copper contains only iron, silver and a gold as impurities. After passage of 140A , for 482.5 sec, of the mass of the anode decreased by 22.260g and the cathode increased in mass by 22.011g . Estimate the % iron and % copper originally present.

    A
    `Cu =98.88%, Fe = 0.85%`
    B
    `Cu =0.85%, Fe = 98.88%`
    C
    `Cu =90%, Fe = 10%`
    D
    `Cu =60%, Fe = 40%`

  • Number of atoms in 558.5 g Fe (molar mass Fe= 55.85 g mol^(-1) )

    A
    twice than in 60 g carbon
    B
    `6.023 xx 10^(22)`
    C
    half that of 8g He
    D
    `558.6 xx 6.023 xx 10^(23)`

  • Excess of NaOH (aq) was added to 100 mL of FeCI_(3) (aq) resulting into 2.14 g of Fe(OH)_(3). The molarity of FeCI_(3) (aq) is: (Given molar mass of Fe = 56g mol^(-1) and molar mass of CI = 35.5g mol^(-1) )

    A
    0.3M
    B
    0.2M
    C
    0.6M
    D
    1.8M
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