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Oleum in water is treated with 0.5lof 2.75 M Ca(OH)2solution. The resulting solution required 15.7 gm of H3PO3(Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3in 100 gms of oleuma)7.111gmb)0.7111gmc)71.11 gmd)None of theseCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Oleum in water is treated with 0.5lof 2.75 M Ca(OH)2solution. The resulting solution required 15.7 gm of H3PO3(Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3in 100 gms of oleuma)7.111gmb)0.7111gmc)71.11 gmd)None of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Oleum in water is treated with 0.5lof 2.75 M Ca(OH)2solution. The resulting solution required 15.7 gm of H3PO3(Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3in 100 gms of oleuma)7.111gmb)0.7111gmc)71.11 gmd)None of theseCorrect answer is option 'C'. Can you explain this answer?.

Solutions for Oleum in water is treated with 0.5lof 2.75 M Ca(OH)2solution. The resulting solution required 15.7 gm of H3PO3(Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3in 100 gms of oleuma)7.111gmb)0.7111gmc)71.11 gmd)None of theseCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

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