In what proportion should 95% alcohol be mixed with 30% alcohol to make 70% alcohol?
To find the proportion in which the two alcohols should be mixed to obtain a desired concentration, we can use a basic mixture equation. Let's assume we have x liters of 95% alcohol and y liters of 30% alcohol.
The total volume of the mixture will be x + y liters.
Now, let's consider the concentration of alcohol in the mixture. We want to have a 70% concentration of alcohol.
To calculate the concentration, we need to consider the amount of pure alcohol in each component of the mixture. In the 95% alcohol, x liters contain 0.95x liters of pure alcohol, while in the 30% alcohol, y liters contain 0.3y liters of pure alcohol.
Now, we can set up an equation based on the amount of pure alcohol:
0.95x + 0.3y = 0.7(x + y)
Simplifying the equation:
0.95x + 0.3y = 0.7x + 0.7y
0.95x - 0.7x = 0.7y - 0.3y
0.25x = 0.4y
Simplifying further:
x/y = 0.4/0.25
Dividing both sides by y:
x/y = 1.6
So, the proportion in which the 95% alcohol should be mixed with the 30% alcohol to obtain a 70% alcohol mixture is 1.6:1.
To calculate the actual volumes, you can choose any value for y and find the corresponding x using the equation above. For example, if you set y = 1 liter, then x = 1.6 liters, and the mixture would be 1.6 liters of 95% alcohol mixed with 1 liter of 30% alcohol.