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If the percent free SO(3) in an oleum is...

If the percent free `SO_(3)` in an oleum is 20% then label the sample of oleum in terms of percent `H_(2) SO_(4)`,

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`%` labelling of oleum sample `= (100 + x) %`
`{:(SO_(3),+,H_(2)Orarr,H_(2)SO_(4),),(20gm,,,,),(1//4"mole",,1//4"mole",,):}`
`4.5gm`
`:. %` labelling of oleum sample `= (100 + 4.5) % = 104.5%` .
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