+257 A chemist wishes to dilute a 90-cc solution that is 5% to that is 3% acid. How much water should the chemist add to the acid?
+128406 Let pure water = 0% acid
Let the amount of water to be added = W....so we have
(90 cc) * (5% acid solution) + W *( 0% acid) = ( Final amt) * (3% acid)
(90) (.05) + W (0) = ( 90 + W) (.03) simplify
4.5 = 2.7 + .03W subtract 2.7 from both sides
1.8 = .03W divide both sides by .03
1.8 / .03 = W = 60 cc of water
