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:
Two equations and two unknowns: A = 15, B = 25
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.
144
69
80% water
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
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.
222.223 ml @ 20% solution = 44.444 ml 277.777 ml @ 65% solution = 180.555 ml total = 225 ml out of 500 ml = 45%
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?
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.
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.
80% water
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
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.