Question

How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?



11 L
15 L
25 L

Answer 1

please help ill be more than greatful

Answer 2

Ok, I have the answer. I was working it out first before explaining it. I can help.

Answer 3

thank ou sooo much sorry if i was bugging u

Answer 4

It is important to understand this conceptually as it will allow you to solve similar problems and help in further question in math that will not necessarily be directly related. And you are not bugging, you're fine. You start with % concentration to get "amount". You have a relationship of x/10 = 0.25 which means you have an "amount" of "x" acid in 10 liters to give 25% solution. Take a second to go over that fraction because it is the whole key to the problem. Take many seconds if you like. It is important.

Answer 5

thank you sooo much

Answer 6

ok, but we're not done yet.

Answer 7

Once you get or accept that fraction, we have "x" is 2.5. We set up another relationship that is in the same format, but this one is now. 2.5/(y+10) = 0.10 Notice that this is the same format. The 2.5 is the amount of acid as that will not change. "y + 10" means that we have the original 10 liters and we will add y more to get 0.10 or 10%. You can easily solve for y and get 15, which means 15 MORE liters of solution which is what you are looking for.

Answer 8

So, that's it! Do you want to go over any part of the solution or problem at this point?