Problema Solution
how much pure water should be mixed with a 10L solution of 60% acid to produce a mixture that is 60% water?
Answer provided by our tutors
let x = L of pure water
we have a 10 L solution of 60% acid and 40% water
1.00*x + 0.40*10 = 0.60(x + 10)
by solving we find:
x = 5 L
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5 liters of pure water should be used for the mixture.