SOLUTION: You need a 65% alcohol solution. On hand, you have a 15 mL of a 25% alcohol mixture. You also have 70% alcohol mixture. How much of the 70% mixture will you need to add to obtain t
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-> SOLUTION: You need a 65% alcohol solution. On hand, you have a 15 mL of a 25% alcohol mixture. You also have 70% alcohol mixture. How much of the 70% mixture will you need to add to obtain t
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Question 1144544: You need a 65% alcohol solution. On hand, you have a 15 mL of a 25% alcohol mixture. You also have 70% alcohol mixture. How much of the 70% mixture will you need to add to obtain the desired solution? Answer by ikleyn(52802) (Show Source):
Let x = the amount of the 70% alcohol needed (in milliliters.
Then the amount of the resulting 65% alcohol solution will be is (x+15) milliliters.
The amount of the "pure" alcohol in 15 mL of the 25% alcohol solution is 0.25*15 mL.
The amount of the "pure" alcohol in x mL of the 70% alcohol solution is 0.70*x mL.
The total amount of the "pure" alcohol in ingredients is the sum 0.25*15 + 0.70*x mL.
It should be equal to the amount of the "pure" alcohol in the mixture, which is 0.65*(x+15) ml.
So, your "pure alcohol" equation is
0.70*x + 0.25*15 = 0.65*(x+15).
Simplify and solve it for x:
0.70x + 0.25*15 = 0.65x + 0.65*15,
0.70x - 0.65x = 0.65*15 - 0.25*15,
0.05x = 6,
x = = 120.
Answer. 120 mL of 70% alcohol solution are needed.
Check. 0.07*120 + 0.25*15 = 87.5 mL of the pure alcohol in ingredients, and
0.65*(120+15) = 87.75 mL of the pure alcohol in the final mixture. ! Precisely correct !
