A) one mole of oxygen
B) one gram atom of oxygen
C) 8 g oxygen
D) 22.4 lit. of oxygen
Correct Answer: C
Solution :
[c] According to the definition 1 F or 96500 C is the charge carried by 1 mol of electrons when water is electrolysed \[2{{H}_{2}}O\xrightarrow{{}}4{{H}^{+}}+{{O}_{2}}+4{{e}^{-}}\] So, 4 Faraday of electricity liberate = 32 g of\[{{O}_{2}}\]. Thus 1 Faraday of electricity liberate \[=\frac{32}{4}g\,of\,{{O}_{2}}=8g\,of\,{{O}_{2}}\]
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