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How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to get obtain 18 liters of 30 percent alcohol solution?
Wiki User
∙ 13y ago
The total volume of solution equals 18 L. So if we let X be the 50% solution & Y be the 20% solution we can solve a set of two equations with 2 unknowns:
X + Y = 18 (total volume of solution is 18 L)
0.5X + 0.2Y = 18*(0.30) (the total amount of alcohol in the new 18L solution)
Solving simultaneously, multiply the 1st equation by 0.5 then substract to solve for Y.
X + Y = 18 {*0.5} => 0.5X + 0.5Y = 9
0.5X + 0.2Y = 5.4 => 0.5X + 0.2Y = 5.4
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0.3Y = 3.6
Y = 12
Then substitute for Y back into the 1st equation & solve to get X=6.
So you need 12 L of the 20% solution & 6 L of the 50% solutions to mix together to get 18 L of a 30% alcohol solution.
Wiki User
∙ 13y ago
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A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
How many liters of water must be added to 7 liters of a 20 percent acid solution to obtain a 10 percent acid solution?
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
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How many liters of a 7 percent solution of salt should be added to a 19 percent solution in order to obtain 456 liters of a 15 percent solution?
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How many liters of pure antifreeze must be added to 30L of 60 percent antifreeze solution to obtain a 75 percent solution?
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10 liters.
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A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
Pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution How much of the 20 percent solution did the pharmacist use in the mixture?
10
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How many liters of a 20 percent solution of saline should a nurse mix with 10 liters of a 50 percent solution to obtain a mixture that is 40 percent saline?
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A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to obtain 3 liters of 40 percent alcohol solution?
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What fraction of 10 percent alcohol should be mixed with 70 percent alcohol to obtain 50 percent alcohol?
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600ml.
How many liters of water must be added to 7 liters of a 20 percent acid solution to obtain a 10 percent acid solution?
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
