A 5% alcohol solution is mixed with 15% alcohol solution to make a 25mlof a 9% solution. How much of each alcohol solution is used?
Let x be the number of ml of the 5% solution and y be the number of moles of the 9% solution.
0.09 * 25 = 0.05*x + 0.15y (alcohol content)
x + y = 25 (total volume)
Substitute 25-x for y in the first equation and solve for x.
2.25 = 0.05 x + 0.15(25-x) = 3.75 -0.10x
etc.
To solve this problem, we can use the concept of the concentration of a solution and the formula:
Concentration = (Amount of solute / Amount of solution) * 100
Let's assign variables to the two solutions:
Let x be the amount (in ml) of the 5% alcohol solution used.
Then, 25 - x will be the amount (in ml) of the 15% alcohol solution used.
Now, let's set up the equation based on the given information:
0.05x + 0.15(25 - x) = 0.09 * 25
The left side of the equation represents the amount of alcohol in the mixed solution, while the right side represents the amount of alcohol in a 9% solution of 25 ml.
Now, let's simplify and solve the equation:
0.05x + 3.75 - 0.15x = 2.25
Combine like terms:
-0.1x + 3.75 = 2.25
Subtract 3.75 from both sides:
-0.1x = 2.25 - 3.75
-0.1x = -1.5
Divide both sides by -0.1 to solve for x:
x = -1.5 / -0.1
x = 15
Therefore, 15 ml of the 5% alcohol solution is used.
To find the amount of the 15% alcohol solution, we can subtract the amount of the 5% solution used from the total solution amount of 25 ml:
25 - 15 = 10
Therefore, 10 ml of the 15% alcohol solution is used.
So, the solution involves using 15 ml of the 5% alcohol solution and 10 ml of the 15% alcohol solution.