A) 0.66 mol
B) 0.33 mol
C) 0.66 g
D) 0.33 g
Correct Answer: C
Solution :
Key Idea: The problem will be solved by Faraday's law of electrolysis. \[Q=it\] Given,\[i=1\text{ }A,\text{ }t=30\times 60=1800\text{ }s\]. \[\therefore \] \[Q=1\times 1800\] \[=1800C.\] \[\underset{\begin{smallmatrix} 1\,mol \\ 71\,g. \end{smallmatrix}}{\mathop{2C{{l}^{-}}}}\,\xrightarrow{{}}\underset{2\times 96500\,C}{\mathop{C{{l}_{2}}\text{ }+\text{ }2{{e}^{-}}}}\,\] (at anode) \[\therefore \]the amount of\[C{{l}_{2}}\]gas liberated by passing 1800 C of electric charge \[=\frac{1\times 1800\times 71}{2\times 96500}=0.66g\]
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