The ThOD required to oxidize 2175 mgL of glucose 1S0ymol to

The ThOD required to oxidize 217.5 mgL of glucose (1S0-ymol) to carbon dioxide and water is: C,H 120, t 602 6co, +6H,0 a 108,.75 mgL b. 116 mgLteLnone

Solution

d. 232 mg/L

Explanation: Molecular weight of glucose = 180 g

for 180g --> 192g of oxygen is required

for 1g --> (192/180) of oxygen is required

Thus, for 217.5mg/L of glucose --> 217.5*(192/180) =232 mg/L