Final answer:
To make a 60% sulfuric acid solution from 5 liters of 20% sulfuric acid solution, 7 liters of pure sulfuric acid must be added.
Step-by-step explanation:
To determine how many liters of pure sulfuric acid must be added to 5 liters of a 20% sulfuric acid solution to make a solution that is 60% sulfuric acid, we can set up an equation based on the concentration of sulfuric acid before and after the addition. Let x represent the volume of pure sulfuric acid added.
Before mixing, the mass of sulfuric acid in the 20% solution is 20% of 5 L. Since the density of the solution is not provided, we will assume it is 1.0 g/mL, which is the density of water. This is a simplistic assumption, as the actual density of the solution would differ, but it allows us to focus on the relationship between volumes and percentages. Thus, we have 1000 g/L * 5 L * 20% = 1000 g of sulfuric acid.
After adding x liters of pure sulfuric acid (100%), the total volume becomes (5 + x) liters, and the new concentration is supposed to be 60%. We set up the equation:
(5 L * 20% + x L * 100%) = (5 + x) L * 60%
Solving this equation yields:
1 L * 20% + x * 100% = (5 + x) * 60%
0.2 L + x = 3 L + 0.6x
0.4x = 2.8 L
x = 7 L
Therefore, 7 liters of pure sulfuric acid must be added to the 5 liters of 20% sulfuric acid solution to achieve a 60% sulfuric acid solution.