What volume of 36 M and 1 M sulphuric acid must be mixed to get 1L of 6 M sulphuric acid ?

Explanation

Suppose the volume of 36 M `H_(2)SO_(4)` solution `=xL`
`:.` the volume of `1 M H_(2)SO_(4)` solution `= (1-x)L`
Both these acid solutions upon mixing are to form 1 L of `6 M H_(2)SO_(4)` solution
The value of x can be calculated as follows :
`M_(2)V_(1)+M_(2)V_(2)=M_(3)V_(3),36xxx+1xx(1-x)=6xx1`
or `36x+1-x=6or 35x=5`
`x=(5)/(35)=0.1428L=142.8mL`
`:.` Volume of `36 M H_(2)SO_(4)` solution `= 142.8 mL`
Volume of `1 M H_(2)SO_(4)` solution `= 1000 - 142.8 = 857.2 mL`.