10
80 liters
50 Liters of the 60% solution.
There are 3 litres of alcohol in your starting mixture of 4 litres. If you add 6 litres of water you will have 3 litres of alcohol in a total of 10 litres. This is the required strength.
2 litersLet X represent the amount of pure alcohol to be added to the 7 liters of 10% alcohol solution to get a 30% alcohol solution. We want .3(7+x)=.1*7+x [because we need the alcohol in the final solution to equal the amount of alcohol in the original solution plus the amount of alcohol we are adding). Solving for x, we get:.3(7+x)=.1*7+x2.1+.3x=.7+x1.4=.7x2=xThus, 2 liters of pure alcohol needs to be added to the solution.Added noteto deal with the so called 'dilution contraction' of total volumeIf 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 10%v/v and final 30%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: 7 L + 2 L (is not equal but) < 9 L final solution.
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).
10
The chemist will use 100 liters of the 80% acid solution and 100 liters of the 30% acid solution to make a 200-liter solution that is 62% acid. The amount of acid in the 80% solution will be 0.8 * 100 = 80 liters, and in the 30% solution, it will be 0.3 * 100 = 30 liters.
10 liters
80 liters
50 Liters of the 60% solution.
4 litres
16 2/3 liters
70/19 or 3.684 liters. Here's how. 5 liters of 70% alcohol contains 3.5 liters of pure alcohol. So if X equals the answer to the question, then 0.95 * X = 3.5. That means X = 3.5/0.95 or "normalizing" the fraction gives 70/19 = 3.684L.
6 liters of a solution that is 10% ethanol contains 0.6 liters of ethanol.
Well, isn't that a lovely little problem to solve? To decrease the concentration from 25% to 20%, we need to dilute the solution. Since the concentration is decreasing by 5%, we can calculate that we need to add 60 liters of water to the 300 liters of solution to achieve the desired concentration of 20%. Just like painting, a little change can make a big difference in creating the perfect mixture.
10 liters of alcohol (above 22% alcohol) or 110 liters of bee