Oleum!
0.5 grams of Oleum is diluted with . The resultant solution required 26.7 ml of 0.4 N NaOH solution for complete neutralisation. Find the percentage of free in the sample of Oleum and also find Label for oleum
Enter your answer as sum of the values (up to 2 decimal places).
Note : Label is actually the final weight of after complete conversion of 100gm oleum mixture( & ) to sulfuric acid after addition of water.
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This is a part of my set: Aniket's Chemistry Challenges . (BEST OF JEE - ADVANCED)
The answer is 125.39.
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1 solution
1st calculate the weight of sulphuric acid neutralised = 0.5233200 ..then let the oleum sample contain x gm of SO3 and rest (0.5-x) gm of H2SO4. ....Find x by 0.52332=(0.5-x) + 98x/80.......then find percentage of SO3 (200x)
...then label = 100+ 200x * 18/80...In the last step o calculate the final weight of H2SO4 I have added the initial weight 100 to the extra added weight of water