SOLUTION: a 60% sulphuric acid solution will be mixed with a 90% sulphuric acid solution to produce 8 L of a 75% solution. how much of each is required

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: a 60% sulphuric acid solution will be mixed with a 90% sulphuric acid solution to produce 8 L of a 75% solution. how much of each is required      Log On


   

Question 841871: a 60% sulphuric acid solution will be mixed with a 90% sulphuric acid solution to produce 8 L of a 75% solution. how much of each is required
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You will need the densities of sulfuric acid solutions. Check this:
http://www.chembuddy.com/?left=CASC&right=density_tables

60%_________1.5031 g/cc
90%_________1.8197
Unless you can find another data source (like maybe a handbook), you might try estimating the 75% concentration density, %281.615%2B1.7323%29%2F2=1.6736 g/cc, only an estimate or prediction.

You did not say if your concentrations are w/w or w/v. You probably expect to use percent w/v. But the data in the reference source is percents w/w.

Let x = volume, ml, or the 60%
Let y = volume, ml, of the 90%
And assuming volumes are simply additive:

Pure sulfuric acid: x%2A1.5031%2A%280.60%29%2By%2A1.8197%2A%280.90%29 grams of sulfuric acid.
Volume of 75% solution: %281.6736%29%28x%2By%29 grams of solution.

Compose the percentage equation to be all as percent by weight.


The volume sum equation is just highlight%28x%2By=8000%29, using milliliters.

Those two red-highlited equations are the system to solve, for x and y.