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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. |
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Answer» `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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