SOLUTION: How many millilitres of water must be added to 240 milliliters of a 25% sulphuric acid solution to reduce the percentage of sulphuric acid in the new solution to 10%?

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Question 1033626: How many millilitres of water must be added to 240 milliliters of a 25% sulphuric acid solution to reduce the percentage of sulphuric acid in the new solution to 10%?
Found 3 solutions by mananth, addingup, ikleyn:
Answer by mananth(16946)   (Show Source):
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-------------- percent ---------------- Amount
Solution I 25 ---------------- 240 ml
Water 0 ---------------- x ml
Mixture 10 ---------------- 240 + x ml

25 * 240 + 0 x = 10 ( 240 + x )
6000 + 0 x = 2400 + 10 x
0 x -10 x = -6000 + 2400
-10 x = -3600
/ -10
x = 360 ml Water

m.ananth@hotmail.ca

Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website!
240*0.25 = x+240*0.10
60 = x+24
x = 60-24 = 36

Answer by ikleyn(52802)   (Show Source):
You can put this solution on YOUR website!
.
How many milliliters of water must be added to 240 milliliters of a 25% sulphuric acid solution to reduce the percentage of sulphuric acid in the new solution to 10%?
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240*0.25 = (x+240)*0.10
60 = 0.1x + 24
0.1x = 60 - 24 = 36
x = 360 milliliters of water must be added.