400g of 20%,240g of 43%and350g of 66% sulphuric acid are mixed .how much watershould be added to make a 10% solution of sulphuric acid?
If pure water is added, the % of sulphuric acid = 0%
So....we have the following equation :
400(.20) + 240(.43) + 350(.66) + 0x = .10(400 + 240 + 350 + x)
80 + 103.2 + 231 = .10x + .10( 400 + 240 + 350)
414.2 = .10x + .10(990)
414.2 = .10x + 99 subtract 99 from both sides
315.2 = .10x divide both sides by .10
3152 = x
So.....3152g of pure water should be added