A) 2.7 g
B) 2.7 mg
C) 0.27 g
D) 0.54 g
Correct Answer: C
Solution :
Given, Current = 241.25 columb 1 coulomb current will deposite \[=1.118\times {{10}^{-3}}gm\ Ag\]. \[\therefore \] 241.25 current will deposite =\[1.118\times {{10}^{-3}}\times 241.25\] \[=0.27\ gm\] silver.You need to login to perform this action.
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