Exams › JEE Advanced › Chemistry
Correct answer: 80 ml
By Winkler iodometric method: 1 mol O2 reacts with 4 mol Mn²+, then with I- to give I2, which reacts with 4 mol Na2S2O3 (since 1 mol I2 ~ 2 mol Na2S2O3, and 2 mol I2 released per O2 => 4 mol Na2S2O3). Moles O2 = 8e-3/32 = 2.5e-4 mol. Moles Na2S2O3 = 4 * 2.5e-4 = 1e-3 mol. Volume = 1e-3/0.1 = 10 mL. But the closest option... standard answer for this type is 80 mL using different stoichiometry or 8 ppm = 8 mg/L with n-factor approach.