A) \[2.24\,d{{m}^{3}}\]
B) \[1.12\,d{{m}^{3}}\]
C) \[4.48\,d{{m}^{3}}\]
D) \[0.56\,d{{m}^{3}}\]
Correct Answer: C
Solution :
During electrolysis volumes of\[{{O}_{2}}\]and\[{{H}_{2}}\] liberated in the ratio of 1:2. Hence, volume of\[{{H}_{2}}\]liberated will be\[4.48\text{ }d{{m}^{3}}\].
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