AITS & Test SerieshardNUMERICAL

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Question

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Answer: 0.25

💡 Solution & Explanation

(for the Q. No. 33 and 34) In neutral or faintly alkaline medium 4 2 2 3 2MnO H O I 2MnO 2OH IO          Millimoles of 4 KMnO 500  Millimoles of I oxidized = 250 Moles of I oxidized = 0.25 In acidic medium 2 4 2 2 10I 2MnO 16H 2Mn 8H O 5I          Millimoles of 4 KMnO 50  Millimoles of I oxidized = 250 Moles of I oxidized = 0.25 For More Material Join: @JEEAdvanced_2024 AITS-FT-VIII (Paper-2)-PCM(Sol.)-JEE(Advanced)/2024 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11 Mathematics PART – III SECTION – A

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