a pharmacist mixed some 10% saline solution with some 15% saline solution to obtain 100ml of a 12%saline solution.How much of 10% saline solution did the pharmist use in the mixture.dO NOT UNDERSTAND TO MUCH INFO.
14 years ago
14 years ago
how did you get that
14 years ago
let the amount of 10% solution be x ml
then the amount of 15% solution must be 100-x ml
.10x + .15(100-x) = .12(100)
10x + 15(100-x) = 12(100)
10x + 1500 - 15x = 1200
-5x = -300
x = 60
8 years ago
31.25 liters
11 months ago
To solve this problem, we can use the concept of the concentration of a solution.
Let's assume that the pharmacist used x milliliters of the 10% saline solution.
Since the final volume of the mixture is 100 ml, the remaining volume would be (100 - x) ml.
Now, let's calculate the amount of salt (saline) in the 10% saline solution.
In a 10% solution, 10% of the volume is salt. Therefore, the amount of saline in x ml of the 10% solution is 0.10x ml.
Similarly, the amount of salt in the (100 - x) ml of the 15% saline solution is 0.15(100 - x) ml.
According to the problem, the mixture should have a 12% concentration, which means the amount of salt in the mixture is 0.12 * 100 ml.
Since the amount of salt in the total mixture should be equal to the sum of the amounts of salt in the individual solutions,
0.10x + 0.15(100 - x) = 0.12 * 100
Now you can solve this equation to find the value of x, which represents the amount of the 10% saline solution used by the pharmacist.