A Chemist has one solution that is 30% pure salt and another is 60% pure salt.

Question

How many ounces of each solution must he use to produce 60 ounces of a solution must he use to produce 60 ounces of a solution that is 50% pure salt.

Answer 1

This problem is a dilution problem where some volume of the 30% solution is used to dilute some volume of the 60% to a total of 60oz of a 50% solution.

 

The formula for this process is C1·V1 + C2·V2 = C3·V3, where C = concentration and V = volume

 

let x = the volume of the 30% solution; so C1·V1 = 30%x

then the volume of the 60% solution will be 60oz (the final volume) - x

so C2·V2 = 60%(60oz - x)

 

thus, our equation becomes 30%x + 60%(60oz - x) = 60oz·50%

 

when we expand this equation, we get

30%x + 3600%oz - 60%x = 3000oz%

 

we combine like terms to get -30%x = -600oz%

when we divide both sides by -30%, we get x = 20oz

 

This number is the Volume of the 30% solution which will be added to 40oz of the 60% solution to yield 60oz of 50% salt.

(The volume of the 60% solution was 60oz - x = 40oz)

 

Proof: 40oz · 60%/100% = 24; 20oz · 30%/100% = 6; 24+6=30, which is 50% of 60oz