You will need more data about all densities (in kg/Litre)
and you must be sure of using mass% = (g solute)/(100 g solution)
Solve two equations for both X and Y:
In which:
dm = density of the 'm'% salt solution in kg/Litre)
X and Y = volume of the 20% and 70% salt solutions respectively
To create a 50% chlorinated solution from the 60% and 40% solutions, the chemist will need to mix the two in equal amounts. Therefore, 50 L of the 60% solution and 50 L of the 40% solution are needed to make a 100 L solution that is 50% chlorinated.
A 2% salt solution is hypotonic compared to a 4% salt solution because it has a lower concentration of salt. In osmosis, water flows from hypotonic to hypertonic solutions, so in this case, water would flow from the 2% solution to the 4% solution to try to equalize the concentrations.
Let x be the volume of the 53% solution and y be the volume of the 97% solution. The total volume will be x + y = 110 ml. The equation for the concentration is 0.53x + 0.97y = 0.65(110). Solving these two equations simultaneously gives x = 50 ml of the 53% solution and y = 60 ml of the 97% solution.
It would flow toward the weaker solution. The intent of osmosis is to gain equilibrium, so the 15 percent solution would gain sugar content until, if you allowed the osmosis to go to completion, the two solutions had the same amount of sugar in them. "Going to completion" doesn't necessarily mean 20 percent concentration on both sides. If you were to make a gallon bag out of dialysis membrane, fill it with 15 percent solution and put a stirrer in it, then drop it into a 25,000-gallon reaction vessel full of 25 percent solution with a stirrer in it, you might wind up with 24.9999999999 percent sugar solution in both bags.
Chemists typically use percent by weight or molarity to prepare and describe solutions because these measures are more accurate and reflective of the actual concentration of solute in the solution. Percent by volume can fluctuate with temperature changes and can be affected by differences in the volumes of the solute and solvent, making it less precise for analytical purposes.
To create a 50% chlorinated solution from the 60% and 40% solutions, the chemist will need to mix the two in equal amounts. Therefore, 50 L of the 60% solution and 50 L of the 40% solution are needed to make a 100 L solution that is 50% chlorinated.
some liquid volumes are not additive, leading to potentially confusing final solution volumes.
144liters
50liters
50
50 Liters of the 60% solution.
A 3 percent solution is 1.5 times as strong as a 2 percent solution.
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.
yes it is isotonic solution.
6 litres of 50% + 4 litres of 25%
Let v be the capacity in milliliters of the 75% solution required then (90 - v) is the required capacity of the 84% solution needed. 75/100v + 84/100(90 - v) = 77/100 x 90 75v + 84(90 - v) = 77 x 90 75v + 7560 - 84v = 6930 9v = 7560 - 6930 = 630 v = 70 : therefore (90 - v) = 20 The mixed solution requires 70ml of the 75% solution and 20ml of the 84% solution to create 90ml of 77% solution.
To create a 400 L solution that is 62% acid, you would need 200 L of the 80% acid solution and 200 L of the 30% acid solution. This would result in a final solution with the desired concentration.