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
50 liters
x+y=110 .53x+.97y=.65*110 x=110-y .53(110-y)+.97y=.65*110 .53*110-.53y+.97y=.65*110 .97y-.53y=(.65-.53)*110 .44y=.12*110 y=.12*110/.44 = 30 ml x=110-30 = 80 ml So, you need 80 ml of the 53% solution and 30 ml of the 97% 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%
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.
50
some liquid volumes are not additive, leading to potentially confusing final solution volumes.
144liters
128 liters
50liters
256 liters
50 liters
50
50 Liters of the 60% solution.
A 3 percent solution is 1.5 times as strong as a 2 percent solution.
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.