What volume of `98%` sulphuric acid should be mixed with water to obtain `200mL` of `15%` solution of sulphuric acid by weight ? Given density of `H_(2)O=1.00 g cm^(-3)`, sulphuric acid `(98%)=1.88g cm^(-3)` and sulphuric acied `(15%)=1.12g cm^(-3)`.

Answer

`V_(1)=18.2 mL`

Explanation

It is a case of dilution, simplest way is to determine normally of the `98%` and `15% H_(2)SO_(4)`.
(i) For `98% H_(2)SO_(4) : 100g H_(2)SO_(4)` solution `=(100)/(1.88)cm^(3)` solution has `H_(2)SO_(4)=98g`
`N_(1)(98%)=(1000w_(1))/(E_(1)V)=(100xx98)/(49xx(100)/(1.88))=37.6N`
where `w_(1)` is the weight of solute of equivalent weight `E,` in `V mL` solution
(ii) For `15% H_(2)SO_(4) : 100g H_(2)SO_(4)` solution `=(100)/(1.12)cm^(3)` has `H_(2)SO_(4)=15g`
`N_(2)(15%)=(1000w_(1))/(E_(1)V)=(1000xx15)/(49xx(100)/(1.12))=3.43`
Using `N_(1)V_(1)=N_(2)V_(2)`
`37.6xxV_(1)=3.43xx200`
`V_(1)=18.2 mL`