A chemist has 800ml of acidic solution. How much pure acid is to be added to make the solution 32% acidic?
2006-11-09 22:25:36 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
need more information
2006-11-09 22:36:11 · answer #1 · answered by deansubasinghe 3 · 1⤊ 0⤋
Are you sure you had all of the facts in the problem?
Missing: Either: Definite volume of acidic solution:
(800x)+(n*100)=(800+n)*32%
x is the percentage of acid in the first solution
n is the volume of the pure acid
or Amount of pure acid added.
(800n)+(x*100)=(800+x)*32%
n is the percentage of acid in the first solution
x is the volume of the pure acid
find x using equality
Hope it helps.
2006-11-10 07:03:05 · answer #2 · answered by Ryan 3 · 0⤊ 0⤋
I suppose there's some missing knowledge. dont you have to know the percentage of acid in 800ml solution?is it 50% or 75%? or sth else?
2006-11-10 06:31:51 · answer #3 · answered by denizxx 1 · 0⤊ 0⤋
how much is the solution already?
have to know the starting reference
800ml * ____ + Xml @ 100% = 800ml+x times 32%
2006-11-10 06:31:47 · answer #4 · answered by tom4bucs 7 · 0⤊ 0⤋
unless you know what is the % acid in the 800 ml of acidic solution you cannot get a firm answer
2006-11-10 06:50:38 · answer #5 · answered by grandpa 4 · 0⤊ 0⤋
question not clear!
he has 800 ml of acidic solution of what strength?
2006-11-10 06:48:43 · answer #6 · answered by raj 7 · 0⤊ 0⤋
something missing in this question..sure you got the rite question...didnt miss a thing ?
2006-11-10 06:33:54 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Question is not complete.
2006-11-10 06:37:57 · answer #8 · answered by rajesh bhowmick 2 · 0⤊ 0⤋