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During electrolysis of water the volume of O2 liberated is 2.24 dm3 . The volume of hydrogen liberated, under same conditions will be _______.(A) 0.56 dm3 (B) 1.12 dm3 (C) 2.24 dm3 (D) 4.48 dm3 |
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Answer» Correct option: (D) 4.48 dm3 H2O → H2 + \(\frac{1}{2}\) O2 H2 : O2 = 2 : 1 ∴ for 2.24 dm3 of O2, H2 liberated will be 4.48 dm3 |
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