Question 1033626
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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.