A chemist is asked to make 10 liters of a a35% solution. She has only a 10% solution and a 50% solution to work with. how much of each solution must she mix to ge 10 liters of the 35% solution?

Question

A chemist is asked to make 10 liters of a a35% solution. She has only a 10% solution and a 50% solution to work with. how much of each solution must she mix to ge 10 liters of the 35% solution?

Answer 1

Using a double equation system;Let x be the 10% soln,so y is the 50% soln

Vol.Eqn1: x+y = 10L, then
Volx[conc]Eqn2:0.1x+0.5y= 0.35(10)
Eqn2-(Eqn1x0.5) gives xL
Therefore: 0.1x + 0.5y = 3.5
-0.5x + 0.5y = 5.0
-0.4x =-1.5
x = 1.5/0.4L= 3.75L
Therefore y is 10-3.75=6.25L of 50%soln

Answer 2

To find out how much of each solution she needs to mix, we can set up an equation using the concept of mixtures.

Let's assume she needs to mix x liters of the 10% solution and (10 - x) liters of the 50% solution to get a total of 10 liters of a 35% solution.

Now, let's calculate the amount of pure substance (in this case, the amount of acid) in each solution that she would use:

Amount of pure substance in the 10% solution = 0.10x
Amount of pure substance in the 50% solution = 0.50(10 - x)

Since we want to make a 35% solution, the total amount of pure substance in the mixture should be equal to 35% of the total volume of the mixture, which is 10 liters:

0.10x + 0.50(10 - x) = 0.35(10)

Now we can solve this equation for x:

0.10x + 5 - 0.50x = 3.5
-0.40x + 5 = 3.5
-0.40x = -1.5
x = -1.5 / -0.40
x = 3.75

Therefore, she needs to mix 3.75 liters of the 10% solution and (10 - 3.75), which is 6.25 liters, of the 50% solution to get 10 liters of the 35% solution.