SOLUTION: a 70% acid solution is to be mixed with water to produce a 5 liter mixture containing 28% acid. How many milliliters of water and how many milliliters of the 70% acid solution are

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Question 1130950: a 70% acid solution is to be mixed with water to produce a 5 liter mixture containing 28% acid. How many milliliters of water and how many milliliters of the 70% acid solution are needed to create the 28% acid solution?

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x liters of the 70% acid
70x%2F5=28
highlight%28x=2%29
water=3

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = the amount of water (in liters), and y = the amount of the 70% acide solution.



First equation is for the total volume of the solution

x + y = 5     liters.     (1)



Second equation is for the final mixture concentration

%280.7y%29%2F5 = 0.28     (2)  (concentration is the ratio of the pure acid volume to the total liquid volume)



From equation (2)  you get   y = %280.28%2A5%29%2F0.7 = 2.


Then from equation (1) the rest,  5-2 = 3 liters is the water volume.



Answer.  2 liters of water and 3 liters of the 70% solution to be mixed.


Check.   %280.7%2A2%29%2F5 = 0.28 = 28%.    ! Correct !

Solved.