How many liters of a 20% solution of acid should be added to 10 liters of a 30% solution of acid to obtain a 25% solution?
Equation:
Answer:
looks very similar to the question just posted by Lily.
let the amount of 30% solution we want to add be x l
now let's reason it through....
the volume of acid we have before adding is .3(10) l or 3 l
we are adding x l of a 20% solution so we are adding .2x l
but the final solution containing 10+x l is to contain 25% acid or .25(10+x) l of acid
so isn't our equation...
3 + .2x = .25(10+x) ?
solve for x and tell me what you got.
How many liters of a 8% solution of acid should be added to 20 liters of a 60% solution of acid to obtain a 40% solution?
To answer this question, we need to use the equation for mixture problems. The equation we will use is:
(Amount of acid in solution A) + (Amount of acid in solution B) = (Amount of acid in the final solution)
Let's break down the given information:
Solution A:
- Concentration: 30%
- Volume: 10 liters
Solution B:
- Concentration: 20%
- Volume: Unknown (let's call it x liters)
Final Solution:
- Concentration: 25%
- Volume: 10 liters + x liters (since we are adding solution B to solution A)
Now, let's use the equation to set up the problem:
(0.30)(10) + (0.20)(x) = (0.25)(10 + x)
We can simplify the equation to:
3 + 0.2x = 2.5 + 0.25x
Next, let's solve for x:
0.25x - 0.2x = 2.5 - 3
0.05x = -0.5
x = -0.5 / 0.05
x = -10
Uh-oh! It appears that we have a negative value for x. This means that it is not possible to obtain a 25% solution by adding a 20% solution and a 30% solution together.
In this case, you may want to double-check the information given or consider other potential strategies to achieve your desired solution.