A chemist has 800ml of a 15% acidic solution. How much pure acid is to be added to make the solution 32% acidic?
2006-11-10 01:54:07 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
800*0.15+x=(x+800)*0.32
120+x=0.32x+256
x-0.32x=256-120
0.68x=136
x=136/0.68
=200
he has to add 200 ml
2006-11-10 02:04:02 · answer #1 · answered by raj 7 · 1⤊ 0⤋
15/100 x 800 = 120
let the amount of pure acid be X
so, (120+x) / (800+x) = 32/100
12000 + 100x = 25600 + 32x
68x = 13600
x = 200ml
Hope this helps=)
2006-11-10 02:26:15 · answer #2 · answered by luv_phy 3 · 0⤊ 0⤋
set up equation where chemical is what each term is
let x = amt pure acid to add
.15*800 acid in original
.35*(800+x) acid in mix
.15*800 + x = .35(800+x)
.65x = .2*800
x = 246.15 mg
2006-11-10 02:16:04 · answer #3 · answered by Anonymous · 1⤊ 0⤋
i think u hv posted ur question under the wrong section. this one comes under chemistry
2006-11-10 02:06:52 · answer #4 · answered by Srikanth 2 · 0⤊ 0⤋