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A sample of oleum is such that ratio of free `SO_(3)` by combined `SO_(3)` is equal to unity. Calculate its labelling in terms of percentage oleum. |
Answer» Correct Answer - `110.11%` Let free `SO_(3)rarr xg` `SO_(3)` in form of `H_(2)SO_(4)` `rarr(x)/(80)xx98=1.225x` so total `x+1.225x=100` `x=449.49` water required `=(44.94)/(80) xx18 = 10..11 g%` oleum `=100+10.11=110.11%` |
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