Best Answer

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.

More answers

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

Q: 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?

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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.

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.

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.

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.

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.

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How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?

2 gallons.

0.25 gallons of water (or 1 quart)

.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300

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.

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.

Approx 1.86 gallons.

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.

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.

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

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.

x=45