How many liters of pure water should be mixed with a 7 L solution of 90% acid to produce a mixture that is 20% water

Question

Please help solve it 

Answer 1

The way I like to do questions like this is to reduce them to their essence.  What do you really have?  7L of 90% acid means that you have some acid and some water already.  How much of each?  Of your 7L, 90% is acid and the rest is water.  90% of 7 L is 6.3 L, and then you have .7 L of water.

 

Now that you know what you have, look at what you want; a solution that is 20% water, and 80% acid.  So, 6.3L must represent 80% of the total amount of solution; we want to find out, first how much solution we will have, and then we can determine how much of that solution is water we need to add to the initial solution to get our "answer" solution.

 

Now the math is easy; just write down in symbols the procedure that has been outlined.  S is the amount of solution we have after adding water.  80%*S = 6.3L, or .8S=6.3L.  Solve for S: S = 6.3L/.8 = 7.875L.  Of that solution, 7L came from the solution we started with, leaving .875L that must be added water.