How much of an 18% solution of sulfuric acid should be added to 360 ml of a 10% solution to obtain a 15% solution?
.18 x + 36 ml = .15(x+360) solve for x
To find out how much of an 18% solution of sulfuric acid should be added to 360 ml of a 10% solution in order to obtain a 15% solution, we can use the method of mixtures.
Let's start by assigning variables to the unknowns:
Let x be the amount (in ml) of the 18% sulfuric acid solution to be added.
Now, let's set up the equation based on the amount of acid in the solution:
The amount of acid in the 10% solution is equal to 10% of 360 ml, which is (10/100) * 360 = 36 ml.
The amount of acid in the 18% solution is equal to 18% of x ml, which is (18/100) * x = 0.18x ml.
The amount of acid in the final 15% solution is equal to 15% of (360 + x) ml, which is (15/100) * (360 + x) = 0.15(360 + x) ml.
Since acid is conserved in the process of mixing, we can equate the amount of acid before and after the mixture:
36 ml + 0.18x ml = 0.15(360 + x) ml.
Let's simplify the equation:
36 + 0.18x = 0.15(360 + x).
Next, let's solve the equation for x:
36 + 0.18x = 0.15 * 360 + 0.15x
36 + 0.18x = 54 + 0.15x
0.18x - 0.15x = 54 - 36
0.03x = 18
x = 18 / 0.03
x = 600.
Therefore, you would need to add 600 ml of the 18% solution of sulfuric acid to the 360 ml of the 10% solution in order to obtain a 15% solution.