How many milliliters of water must be mixed with 25 milliliters of a 30% sulfuric acid solution to obtain a 20% sulfuric acid solution?

Question

How many milliliters of water must be mixed with 25 milliliters of a 30% sulfuric acid solution to obtain a 20% sulfuric acid solution?       

 

Solve this please

Answer 1

25/2 = 12.5 ml of water mixed with 25 ml of 30% yield 37.5 ml of 20% solution

water is 0% solution. 0 is within 20 of 20 while 30 is within 10 of 20. 30 is twice as close, so you need twice as much, or half as much water. 25/2 = 12.5 ml of water

.30(25) + 0(X) = .20(25+X)

7.5 + 0 = 5 +.2X

.2X = 2.5

X = 2.5/.2 = 12.5 ml of water

mixed with 25 ml of 30% yields 37.5 ml of 20%