help me work out this problem

Question

You are working in a chemistry lab. You have 1000 milliliters of pure acid. A dilution of acid is created by adding pure acid to water. A 40% dilution contains 40% acid and 60% water. You have been asked to make a 40% dilution and a 60% dilution of pure acid.
a.Write an equation in standard form that models the possible quantities of each dilution you can prepare using all 1000 milliliters of pure acid.
b.You prepare 700 milliliters of the 40% dilution how much of the 60% dilution can you prepare
c. How much water do you need to prepare 700 milliliters of the 40% dilution?

Answer 1

A. Here is what we know in this problem:

1. We're starting with 1000mL of acid
2. One solution is 40% acid and 60% water
3. A second solution is 60% acid and 40% water

We need to write an equation that models the possible quantities of each dilution using all 1000mL of acid.

For 40% acid and 60% water, we know that using all 1000mL of acid means that it will equal 40% of the total volume.

1. So 40% of V = 1000mL. This gives us the equation 0,4(V) = 1000mL

Therefore, we would have the following for a solution of 60% acid and 40% water:

1. 60% of V = 1000mL, which results in 0.6(V) = 1000mL

Solving for V in each case tells us the total volume of liquid, which we can use to figure out how much is water.

B. If you prepare 700mL of the 40% dilution, that means that 40% of the 700mL is acid and 60% is water.

1. So 0.4( (700mL) = Volume of acid, This gives us a total amount of acid equal to 280mL
2. If we subtract 280mL of acid from the 1000mL we started with, we have 720mL remaining
3. Using what is left to make a 60% solution gives us the equation 0.6(V) = 720mL
4. Solving for V results in a total volume of 1200mL of 60% acid solution.

C. Going back to the 700mL of 40% acid solution, the amount of water needed is simply the difference between the total of 700mL and 280mL of acid. So we're going to need 420mL of water for the 40% acid solution.