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Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which is obtained by passing SO_(3) is solution of H_(2)SO_(4).In order to dissolve SO_(3) in oleum, dilution of oleum is done by water in which oleum is converted into pure H_(2)SO as shown below: H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4) (pure) When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum. For example: 109% H_(2)SO_(4) labelling of oleum sample means that 109 gm pure H_(2)SO_(4) is obtained on diluting 100 gm oleum with 9 gm H_(2)O which dissolves al free SO_(3) in oleum. If the number of moles of free SO_(3), H_(2)SO_(4), and H_(2)O be x, y and z respectively in 118% H_(2)SO_(4) labelled oleum, the value of (x+y+z) is |
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Answer» `2.2` 100G oleum + 18G water `underset("80 g")(SO_(3))+underset("18 g")(H_(2)O)rarrH_(2)SO_(4)` `18gH_(2)O` COMBINES with `80_(g)SO_(3)` So, weight of `SO_(3)` in 100g oleum = 80g `therefore` No. of moles of `SO_(3)=(80)/(80)=1` No. of moles of `H_(2)SO_(4)=(20)/(98)=0.2` No. of moles of `H_(2)O=(18)/(18)=1` Total no. of moles = 2.2 |
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