BOD is expressed in terms of ppm of oxygen. One kg of a water sample required 29.4 mg of acidified K2Cr2O7 to oxidize all the micro organisms. BOD of water sample is

BOD is expressed in terms of ppm of oxygen. One kg of a water sample required 29.4 mg of acidified K2Cr2O7 to oxidize all the micro organisms. BOD of water sample is

  1. A

    4.8

  2. B

    29.4

  3. C

    8.4

  4. D

    24.8

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    Solution:

    103  grams  water  reuired  29.4×103  grams  of  K2Cr2O7/H+106  grams  water  reuired  'X'  grams  of  K2Cr2O7/H+X=29.4One  GEW  of  K2Cr2O7/H+49  grams=one  GEW  of  oxygen8g29.4grams  of   K2Cr2O7/H+'Y'  grams  of  oxygenY=29.4×849g;Y=4.8  g;BOD  of  water  sample  is  4.8  ppm

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