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A bottle is labelled `122.5%` oleum. `22.7 mL` of `Ca(OH)_(2)` is unknown molarity are used to completely neutralise `1 g` oleum. Find the normality of `Ca(OH)_(2)`. |
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Answer» Correct Answer - 2 `(100+22.5)%` oleum means `222.5 g H_(2)SO_(4)` `H_(2)O+SO_(3)rarrH_(2)SO_(4)` `because 18 gH_(2)O "gives" =98g H_(2)SO_(4)` `therefore 22.5 g H_(2)O"gives" =(98xx22.5)/(18)=122.5gH_(2)SO_(4)` `therefore` Total `H_(2)SO_(4)=100+122.5=222.5g` 1 g oleum equiv `2.225gH_(2)SO_(4)=(2.225)/(49)xx1000` `45.41 meq.` `therefore "Meq.of" H_(2)SO_(4)=45.41` `therefore22.7xxN=45.41` `N=2` |
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