SOLUTION: What quantity of pure acid must be added to 600 mL of a 50% acid solution to produce a 75% acid solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: What quantity of pure acid must be added to 600 mL of a 50% acid solution to produce a 75% acid solution?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 778238: What quantity of pure acid must be added to 600 mL of a 50% acid solution to produce a 75% acid solution?
Found 2 solutions by FrankM, josgarithmetic:
Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
You have 300ml acid, 300ml water. You want 75% acid, or 25% water.
300ml (water) = 25%
900ml (acid) = 75%
You must add another 600 ml acid

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Give a variable to how much pure acid.
Let y = volume in mL of pure acid to add.

Note that amount of acid in mixture or resulting solution will be 600%2A0.5%2By%2A1 and volume of resulting solution will be y%2B600 mL. We must assume you mean your concentration as volume/volume; otherwise, you will need the densities of your pure acid and of the 50% solution.

In your simplified description, your equation is highlight%28%28600%2A0.5%2By%29%2F%28y%2B600%29=0.75%29;
%28300%2By%29=0.75%28y%2B600%29
y%2B300=0.75y%2B%283%2F4%29600
y%2B300=0.75y%2B450
%281%2F4%29y=150
y=4%2A150
highlight%28y=600%29 mL of the pure acid.