How many liters of pure water should be mixed with a 15​-L solution of 80​% acid to produce a mixture that is 80 % ​water?

Question

 How many liters of pure water should be mixed with a 15-L solution of 80% acid to produce a mixture that is 80 % water?

Answer 1

You can tell that the person who made this question is probably not a chemist! So the total volume of your final mixture is going to be 15 liters + the liters of pure water you are going to add. 15+x. 

 

You want the % of pure water by volume in the end to be 80% (0.8). So you need to set up a fraction

0.8= (volume of pure water in final mixture)/total volume

 

What makes this tricky is that the 80% acid mixture is presumably 20% pure water. So the volume of pure water in the final mixture is (15*0.2)+x instead of just x.  

 

0.8=(3+x)/(15+x)     --> x=45 L