Question

A hospital needs to dilute a 50% boric acid solution to a 10% solution. If it needs 25 liters of the 10% solution, how much of the 50% solution and how much water should be used?

Answer 1

You need one equation of this problem to balance the amount of a boric acid.

The amount of boric acid in 50% solution is expressed as 0.50*x

Next, the amount of water is taken from 25 liters and x (based on 50% solution), that is 25 - x. However, there is NO amount of BORIC ACID in water, and that is 0.

For the output solution, the amount of boric acid will be 0.10*2.5 = 2.5

Then, the balance equation is
0.50*x + 0*(25 - x) = 2.5
Solving for x,
0.50*x + 0 = 2.5
x = 2.5/0.5 = 5

Therefore the amount of liters in 50% solution is
5*0.50 = 2.5 liters
The amount of water is 25 - 5 = 20 liters

Answer 2

Answer:

20 liters of water and 5 liters of 50% solution

Step-by-step explanation:

Balance the amount of acid:

amount of 50% solution: x liters  

amount of acid in 50% solution: .5x liters (concentration * volume)

amount of water: 25 - x liters (since 25 altogether, and x is 50% solution)

amount of acid in water: 0

final solution: 25 liters

amount of acid in final 10% solution: .10 * 25 = 2.5

balance the "goes into" and the "comes outta"

.50x + 0 = 2.5

x = 2.5 / .5 = 5

therefore 5 liters of 50% solution (for 2.5 liters of acid)

mixed with 25 - 5 = 20 liters of water

will yield 25 liters of a 10% solution (2.5 liters of acid)