A) 0.66 mol
B) 0.33 mol
C) 0.66 g
D) 0.33 g
Correct Answer: C
Solution :
\[2C{{l}^{-}}\to \underset{\text{1}\,\,\text{mole}}{\mathop{C{{l}_{2}}}}\,+\underset{\text{2}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{96500}\,\,\text{coulomb}}{\mathop{2{{e}^{-}}(At\,\,anode)}}\,\] \[Q=it=1\times 30\times 60\] = 1800 coulomb The amount of chlorine liberated by passing 1800 coulomb of electric charge \[=\frac{1\times 1800\times 71}{2\times 96500}=0.66\,g\]
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