COD calculation (1 Viewer)

[Steven12]

hi guys, i have question of BOD below,and i cant seem to match up the answer with the multiple choice given.(which is 5ml)

so here is the question.

A sample(sludge) contains 2.25 mg/litre organic material(CH2O), what volume of NaCr2O7 (0.01M) is consumed during a COD test in that 100 ml.

oxidation
CH2O+h2o= co2+4H+4e
reduction
cr2o7+14h+6e=2cr + 7 H20

overall

2Cr2O7 + 3 CH2O + 16h = 4Cr +11H20+ 3 Co2

so here is what i did

2.25mg/litre = 0.00225 g /litre=0.000225g/(100ml)
0.000225/30(MW)= 0.0000075(mol) of CH2O

0.0000075 * 2/3 = 0.000005 (mol) Cr2O7

0.000005* (1000ml/ 0.01mol) = 0.5 ml

but the only multiple choice given close to this answer is 5 ml.

 

[yodajones2000]

Try asking some one else in your coarse. I Cant give much help. I am 1st currently in my chemistry class but I am in year 11.