500 ml sample of H2O2 was labelled as "11.2V"H2O2 whereas on complete decomposition of the given sample it liberates 5 lit of O2 at 1 atm & 273K Calculate percentage error in labelling.

  1. A

    12%

  2. B

    15%

  3. C

    10.7%

  4. D

    9.7%

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    Solution:

    2H2O22H2O+O2                  0.4464            0.2232 M=0.4464×2 VS=11.2×M VS=11.2×2×0.4464=10 VSgiven=11.2 %error=1.210×100=12%

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