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Question

For her chemistry experiment, Marisol needs 2 liters of a 50% alcohol solution. The lab has two containers, one with 40% alcohol solution and the other with 80% alcohol solution. 

Answer 1

This is a combination-mixture problem. If the resulting mixture needs to be a 50% solution, then the pure portion needs to remain consistent throughout. Thus, the pure portion of the 40% solution plus the pure portion of the 80% solution should be equal to the pure  portion of the 50% solution discussed earlier. Thus, the equation for this is as follows with the constraint being the 2 liters being the total volume of solution: 

 

.40x + .80(2-x) = .50(2)

.40x + 1.60 - .80x = 1

-.40x = - .60

 

x = 3/2 or 1.50 of the 40% solution is required.

 

Thus, .50 Liters of the 80% solution is needed to be mix to make a 50% solution.

 

Hope his helps! 

 

Answer 2

I'm not entirely certain what the question is.

 

The following is based on a question along the lines of "How can Marisol create her 2 liters of 50% alcohol solution?"

 

If that isn't the question please post the original.

 

You can solve this using simultaneous equations. The first equation will be for the total amount of fluid and the second will be for the alcohol content.

 

If this is enough of a hint then by all means go to it!

 

Remember that if α is the total amount of the first solution, than .4α would be the amount of alcohol in that solution.

 

If you need more help setting things up let me know, set up is the real trick for simultaneous equations.