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JEE Advanced levelNumerical answerDifficulty 3 of 3eq147

Oleum is labelled by the mass of obtainable when water is added: '118%' oleum gives 118 g of per 100 g of oleum. How many mL of 1.00 M NaOH are needed to neutralise completely the acid obtained from 1.00 g of 118% oleum after adding sufficient water ( = 98 g/mol; give answer to one decimal place)?

Numerical answer type, answer in mL, tolerance 0.06.

Answer: 24.1 mL

1.00 g of 118% oleum yields 1.18 g = 1.18/98 = 12.04 mmol. is dibasic (n = 2), so equivalents = 24.08 meq. NaOH volume = 24.08/1.00 = 24.1 mL.
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