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How many gallons of a 40 percent antifreeze solution should be mixed with a 10 percent solution to make 50 gallons of a 30 percent solution?
Let the amount of pure alcohol be x, so the amount of 25% solution would be x + 5. So we have:
x + 0.15(5) = 0.25(x + 5)
x + 0.75 = 0.25x + 1.25
x - 0.25x = 1.25 - 0.75
0.75x = 0.5
x = 0.5/0.75 = 0.666... = 6/9 = 1/3 of the liter
Added note
to deal with the so called 'dilution contraction' of total volume
If it were % by MASS ( %m/m), it's quite easy to do (based on the 'Mass Conservation Law). You calculate with mass (kg) and mass-% (%m/m) i.s.o. volume (L) and vol% (%v/v).
However if the meaning was: % by Volume ( %v/v) then calculation appears to become quite complicated, but not impossible if you know at least the density values of all solutions (original 100%v/v or 15%v/v and final 25%v/v).
DO NOT use: (orig. volume) + (added volume) = final volume, as done above, if exact figures are necessary.
It's only a rule of thump, an approximation. This is because fluids can contract on mixing at dilution. There is no rule such as: conservation of volume.
Your case: 0.33 L + 5 L (is not equal but) < 5.33 L final solution.
15 L of 60% & 25 L of 20%. Here's how to solve:
Let A = volume of 60% solution, and B = volume of 20% solution, then:
- Volume of solute in A is 0.6*A
- Volume of solute in B is 0.2*B
- Volume of final solute is 0.35*(final total volume), so:
- Total vol is 40: A + B = 40
- Solute is 0.35*40 = 14: 0.6*A + 0.2*B = 14.
Two equations and two unknowns: A = 15, B = 25
Let x be the gallons of 40% antifreeze solution needed. Then, (0.40)x + (0.10)(50 - x) = (0.30)(50). Solving for x, we get x = 20 gallons of the 40% solution.
Let x be the gallons of 40% antifreeze solution needed. Then, (0.40)x + (0.10)(50 - x) = (0.30)(50). Solving for x, we get x = 20 gallons of the 40% solution.
You would use 100/3 = 33.33 gallons of the 40 percent antifreeze solution and 50/3 = 16.67 gallons of the 10 percent antifreeze solution.
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How much of water should be add to obtain a solution that is twenty percent antifreeze?
To obtain a solution that is twenty percent antifreeze, you would add 4 parts water to 1 part antifreeze. This means that for every 1 part of antifreeze, you would add 4 parts of water. This would result in a total of 5 parts of solution, with 1 part being antifreeze and 4 parts being water, achieving a solution that is twenty percent antifreeze.
How much water should be mixed with 2 gallons of a 50 percent acid solution in order to get an 2 percent acid solution?
To get a 2% acid solution, you need to dilute the 50% acid solution with water. Since the final volume is 2 gallons, you will need to mix 2 gallons of water with the 2 gallons of 50% acid solution to get a 2% acid solution.
A solution of 30 percent acid and a solution of 60 percent acid are to be mixed to produce 50 liters of 57 percent acid How many liters of each solution should be used?
Let x be the liters of the 30% acid solution and y be the liters of the 60% acid solution. We can set up a system of equations: x + y = 50 (total liters) and 0.3x + 0.6y = 0.57*50 (acid content). Solving this system of equations, we find that x = 20 liters of the 30% acid solution and y = 30 liters of the 60% acid solution.
How much uan 32 percent fertilizer should you mix with a gallon of water?
To mix a 32% UAN fertilizer with water, you would typically dilute it at a rate of 1 part fertilizer to 2 parts water. Therefore, you would mix approximately 0.33 gallons of 32% UAN fertilizer with 0.67 gallons of water to make a solution.
How many milliliter of a 20 percent and a 65 percent solutions should you mix together to make 500 milliliter of a 45 percent solution?
Let x be the amount of the 20% solution and y be the amount of the 65% solution. The total volume equation is: x + y = 500. The total amount of the solute equation is: 0.20x + 0.65y = 0.45(500). Solve these two equations to find the amounts of each solution needed.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How many gallons of 70 percent antifreeze solution should be mixed with 60 gallons of 10 percent antifreeze to get a mixture that is 60 percent antifreeze?
.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300
How many gallons of 80 percent solution must be mixed with 100 gallons of a 15 percent antifreeze solution to get a mixture that is 70 percent antifreeze?
For these types of problems, I recommend finding the initial amount of antifreeze. To find this value, multiply the percentage of antifreeze (15%) by the amount of solution (100 gallons). This gets:0.15*100=15 gallons of antifreeze*Now we need make an equation that represents this problem. We let x be the number of gallons of 80% solution.(.80x+15)/(x+100)=0.70The top of the equation represents the amount of antifreeze, while the bottom is the total volume of the solution. If we divide these values, we should get the final concentration %, .70.Now we just need to manipulate the equation algebraically:.80x+15=.70x+70.10x=55x=550 gallons of 80% solution*If the concentration of a solution is given in percentage, it is probably referring to mass, not volume. Because of this, the values for the volume of antifreeze are actually untrue. But because the density of both substances remains the same, the final answer should be correct.
How much water should be added to 30 gallons of a solution that is 70 percent antifreeze in order to get a mixture that is 60 percent antifreeze?
First off do not use tap water. Use only distilled water. You will need to add 3 gallons of distilled water to the solution to get 60% antifreeze and 40% distilled water or a 60/40 mix.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How much of water should be add to obtain a solution that is twenty percent antifreeze?
To obtain a solution that is twenty percent antifreeze, you would add 4 parts water to 1 part antifreeze. This means that for every 1 part of antifreeze, you would add 4 parts of water. This would result in a total of 5 parts of solution, with 1 part being antifreeze and 4 parts being water, achieving a solution that is twenty percent antifreeze.
How many gallons of a 50 percent antifreeze solution must be mixed with 90 gallons of 30 percent antifreeze to get a mixture that is 40 percent antifreeze?
Suppose x gallons of 50% antifreeze is added to 90 gallons of 30% antifreeze. Total active ingredient in mixture = 0.5*x + 0.3*90 = 0.5*x + 27 Total volume of mixture = x + 90 gallons Concentration of mixture = (0.5*x + 27)/(x + 90) = (50x + 2700)/(x + 90) percent This should be 40% So 40 = (50x + 2700)/(x + 90) That is, 40x + 3600 = 50x + 2700 so that 900 = 10x and x = 90 So, add 90 gallons of the 50% antifreeze. In fact, for this particular example, there is a short cut. You are adding antifreeze of two concentrations to get a result whose concentration is EXACTLY halfway. This requires the same amount of the two concentrations to be added together.
How many gallons of 30 percent antifreeze should be mixed with 10 percent gallons of 93 percent antifreeze to obtain a 65 percent antifreeze mixture?
Suppose there are G gallons of the 30% mix. Then G gallons of 30% contain 0.3*G gallons of the active ingredient. Also 10% gallons = 0.1 gallons of 93% contain 0.093 gallons of the active ingredient. Therefore, the total volume is G+0.1 gallons which contains 0.3*G + 0.093 gallons of the active ingredient. So its strength is (0.3*G + 0.393)/(G+0.1) which is 65% or 0.65 Thus 0.3*G + 0.093 = 0.65*G + 0.065 So that 0.028 = 0.35G Or G = 0.08 gallons
A garage owner wants to fill a 55-gallon drum with a 20 percent winter mixture of antifreeze for his customers How many gallons of 100 percent antifreeze should he mix with some 10 percent antifreeze?
Let a be the volume of 100% antifreeze then 55 - a is the volume of 10% antifreeze. 100a + 10(55 - a) = 20 x 55 = 1100 100a + 550 - 10a = 1100 90a = 550 a = 55/9 = 61/9 so 55 - a = 55 - 61/9 = 488/9. The mix is 61/9 gallons (6.111) 100% antifreeze and 488/9 gallons (48.889) 10% antifreeze.
How much pure acid should be mixed with 5 gallons of 70 percent acid solution in order to get a 90 percent acid solution?
x=45
