Question

q8. how many liters of a 25% sulfuric acid solution should be mixed with 10 l of a 10% sulfuric acid solution to obtain a 15% solution of sulfuric acid?

Explanation:

Step1: Set up the equation

Let $x$ be the volume of the 25% sulfuric - acid solution. The amount of pure sulfuric acid in the 10% solution is $0.1\times10$ liters, the amount of pure sulfuric acid in the 25% solution is $0.25x$ liters, and the total volume of the final 15% solution is $(x + 10)$ liters. So, the equation based on the conservation of the amount of pure sulfuric acid is $0.1\times10+0.25x=0.15(x + 10)$.

Step2: Expand the right - hand side

Expand $0.15(x + 10)$ to get $0.15x+1.5$. The equation becomes $1 + 0.25x=0.15x+1.5$.

Step3: Move the $x$ terms to one side

Subtract $0.15x$ from both sides: $0.25x-0.15x=1.5 - 1$.

Step4: Simplify the equation

$0.1x=0.5$.

Step5: Solve for $x$

Divide both sides by 0.1: $x=\frac{0.5}{0.1}=5$.

Answer:

5