A chemist had 100 mL of a 40% acid solution. She wants to reduce the strength of the solution to 25% by adding distilled water. How much water should she add?
2006-11-23 14:56:01 · 5 answers · asked by usha_kaler 1 in Science & Mathematics ➔ Mathematics
C1V1=C2V2
40%x100mL=25%xV2
4000=25%xV2
4000/25=V2 ((25%xV2)/25=V2)
160=V2
V2-V1=60
she should add 60 mL of water.
2006-11-23 15:14:36 · answer #1 · answered by konala 3 · 1⤊ 0⤋
40% of 100mL is 40mL which is the acid you already have. We are not adding or taking away acid so it will always by 40 mL. So the equation is
40mL = 25% of 100mL plus the distilled water to be added
40 = .25 (100 + x)
40 = 25 +.25X
15 = .25 X
x= 60 mL
2006-11-23 15:06:43 · answer #2 · answered by MrWiz 4 · 0⤊ 0⤋
okay....40% of 100ml is 40 ml. If we want our solution to be 25% then we need 3 parts water to one part acid. (160 ml total) So, she needs to add 60 ml
Unless you have to write an eqation for it, I woulnd't recommend doing so. The best I can come up with is
x=40/.25 with x being the total amount in the solution.
Then x-100 is the amount to be added.
2006-11-23 15:02:10 · answer #3 · answered by pzratnog 3 · 0⤊ 1⤋
I'm a little rusty on chemistry but i think she should add 60 ml of water.
40/25 = 1.6
1.6*100 = 160
160-100 = 60
2006-11-23 15:01:26 · answer #4 · answered by guysexy420 1 · 0⤊ 0⤋
you should use the formula m1v1=m2v2 where v=volume in liters and m=concentration
substitue the numbers in:
.4x.1=.25x(v) solve for v
v=.16 L or 160 ml
2006-11-23 15:21:49 · answer #5 · answered by thulfiqar 2 · 0⤊ 0⤋