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

103  grams  water  reuired  29.4×10−3  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