A) \[0.95\]
B) \[0.85\]
C) \[0.97\]
D) \[0.90\]
Correct Answer: D
Solution :
: \[C{{u}^{2+}}+2{{e}^{-}}\to Cu,\] Equivalent wt. \[=\frac{63.54}{2}=31.77\] \[F{{e}^{2+}}+2{{e}^{-}}\to Fe\], equivalent wt. \[=\frac{55.5}{2}=27.75\] Mass increased at cathode is due to deposition of \[Cu\]. Hence, no. of gram equivalents of \[Cu\] deposited \[=\frac{22.011}{31.77}=0.6928\] Now, using Faradays first law, \[Q=it=140A\times 482.5s=67550C\] \[96,500C-31.77g\,Cu\] \[=67550\,\,\,C-?\] \[\frac{31.77}{96,500}\times \frac{67550}{1}\] \[=22.239\text{ }g\]of pure \[Cu\] deposited But, according to the question, the mass of cathode only increases \[=22.011g\] Hence, \[22.239-22.011=0.228g\] Now, \[31.77g\text{ }Cu\equiv 0.228\text{ }g\]deposited \[27.75g\,\,\,Fe\equiv \frac{0.228}{31.77}\times 27.75\] \[=0.199g\,\,Fe\] deposited % of \[Fe=\frac{Mass\,of\,Fe}{Mass\,of\,impurities}\times 100\] \[=\frac{0.199}{22.26}\times 100=0.894\approx 0.90%\]
You need to login to perform this action.
You will be redirected in
3 sec
