Let us say the chemist mix V1 volume from 9% acid solution and V2 from 4% acid solution.

Moles in 9% acid solution `n1 = (9V1)/100`

Moles in 4% acid solution `n2 = (4V2)/100`

It is given that in the final solution the acid concentration is 7%.

We have 3 L of this solution.

Moles in 7% acid solution` n = (7*3)/100`

But we know;

`n1+n2 = n`

`V1+V2 = 3`

`n1+n2 = n`

`(9V1)/100+(4V2)/100 = (3*7)/100`

`9V1+4V2 = 21 -----(1)`

`V1+V2 = 3 -----(2)`

`(1)*9-(2)`

`9V2-4V2 = 27-21`

`V2 = 1.2L`

From (2)

`V1 = 3-1.2 = 1.8L`

*So the chemist needs to mix 1.8L from the 9% acid solution and 1.2L from the 4% acid solution to get the final acid solution.*

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