A) 0.44 moles
B) 0.24 moles
C) 0.66 g
D) 0.21 g
Correct Answer: C
Solution :
\[2C{{l}^{-}}\to C{{l}_{2}}+2{{e}^{-}}\](At anode) \[\text{1}\,\text{mole}\,2\times 96500\,\,\text{coulomb}\] \[Q=i.t=1\times 30\times 60\] = 1800 coulomb Hence, the amount of chloride liberated by passing 1800 coulomb of electric charge \[=\frac{1\times 1800\times 71}{2\times 96500}\] Hence, the correct option is [c].
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