A waste stream contains 45 mg L^-1 of glucose as its only organic matter. Take C = 12, H = 1, O = 16 and complete oxidation C6H12O6 + 6O2 -> 6CO2 + 6H2O. If bacteria exert only 68% of this theoretical demand in 5 days, the BOD_5 is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3ev146

A waste stream contains $45 mg L^{- 1}$ of glucose as its only organic matter. Take C = 12, H = 1, O = 16 and complete oxidation $C X_{6} H X_{12} O X_{6} + 6 O X_{2} 6 CO X_{2} + 6 H X_{2} O$ . If bacteria exert only 68% of this theoretical demand in 5 days, the BOD $_{5}$ is:

Glucose is 180 g/mol, so 45 mg/L needs

(45/180) \times 6 \times 32 = 48.0

mg/L of oxygen (theoretical demand). The 5-day value is

0.68 \times 48.0 = 32.64 \approx 32.6

mg/L. The value 48.0 ignores the 68% factor, and 70.6 divides instead of multiplying.