A) 0.2 gm
B) 0.4 gm
C) 0.6 gm
D) 0.8 gm
Correct Answer: D
Solution :
\[C{{a}^{++}}+2{{e}^{-}}\to Ca\] \[{{E}_{Ca}}=\frac{40}{2}=20\] \[{{W}_{Ca}}={{E}_{Ca}}\]× No. of faradays \[=20\times 0.04=0.8\,gm\].You need to login to perform this action.
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