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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?
Wiki User
∙ 13y ago
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.
Wiki User
∙ 11y ago
Wiki User
∙ 12y ago15 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
Wiki User
∙ 13y agoYou 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.
Wiki User
∙ 14y ago144
Wiki User
∙ 14y ago69
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How much of water should be add to obtain a solution that is twenty percent antifreeze?
80% water
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?
t = number of liters of 30% acid solution s = number of liters of 60% acid solution t+s=57 .30t+.60s=.50*57 t=57-s .30(57-s)+.60s=.50*57 .30*57 -.30s +.60s = .50*57 .30s = .50*57 - .30*57 = .20*57 s = .20*57/.30 = 38 liters of 30% solution t = 57 - s = 57-38 = 19 liters of 60% solution
How much water should be mixed with 2 gallons of a 50 percent acid solution in order to get an 2 percent acid solution?
tv=tv, so 0,25x50=0,1v; v=12,5/0,1; v=125ml of water. So the solution must have 125ml of water for the title to be 10%, then we must add 125-50ml(75ml) of water to it.
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?
222.223 ml @ 20% solution = 44.444 ml 277.777 ml @ 65% solution = 180.555 ml total = 225 ml out of 500 ml = 45%
How would I answer this question How many grams to 0.1 g of ethanol should be dissolved into 135 g of water to produce a 23 percent by mass ethanol solution?
The answer is 31,05 g ethanol.
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 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 much of water should be add to obtain a solution that is twenty percent antifreeze?
80% water
How many gallons of a 10 percent ammonia solution should be mixed with 50 gallons of a 30 percent ammonia solution to make a 15 percent ammonia solution?
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
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.
