Question

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)`.

Answer 1

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`