SOLUTION: A hospital needs to dilute a 60% boric acid solution to a 10% solution. If it needs 40 liters of the 10% solution, how much of the 40% solution and how much water should it use? F
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Question 155010: A hospital needs to dilute a 60% boric acid solution to a 10% solution. If it needs 40 liters of the 10% solution, how much of the 40% solution and how much water should it use? For some reason i always come up with a negative 24 and that isn't right. Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! I THINK YOU NEED TO RECHECK YOUR PROBLEM. I THINK THE "40%" SHOULD BE "60%". ANYWAY, I WILL MAKE THAT ASSUMPTION.
Let x=amount of 60% solution needed
Then 40-x=amount of water needed
Now we know that the amount of pure boric acid in the water(0) plus the amount of pure boric acid in the 60% solution (0.60x) has to equal the amount of pure boric acid in the final mixture(0.10*40). So our equatuion to solve is:
0.60x=0.10*40
0.60x=4 divide each side by 0.60
x=6.67 liters --------------of the 60% solution
40-x=40-6.67=33.33 liters-------------of water
Ck
6.67*0.60=40*0.10
4=4
