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Find the mass of Free `SO_(3)` present in 100gm, 109% oleum sample |
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Answer» 109% means, 9gm of `H_(2)O` is required `{:(SO_(3),+,H_(2)O,rarr,H_(2)SO_(4)),(,,9gm,,),(1//2 "mole",,1//2 "mole",,),(40gm,,,,):}` `:.` Mass of free `SO_(3) = 40 gm`, Mass of `H_(2)SO_(4) = 60 gm` Note: Work out, what are the maximum and minimum value of the % labelling |
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