Alab assistant wants to make five liters of 23.6% acid solution. if the solutions of 30% and 14% are in stock, how many liters of each must be mixed to prepare the solution.

3 L of 30% and 2 L of 14% solution are required

Step-by-step explanation:

I like to work mixture problems using a device I learned from a nursing school student. It is an X diagram with the solution strengths on the left, the mixture strength in the middle, and numbers on the right that are the differences along the diagonals.

Here, the upper difference is 9.6 = 23.6 - 14; the lower difference is 6.4 = 30 - 23.6. These differences, 9.6 and 6.4 can be divided by their common factor (3.2) to see that they are in the ratio 3:2. That is,

... 9.6 : 6.4 = 3 : 2

The upper difference and lower difference represent the ratio of the upper and lower constituent solutions (30% and 14%). Our result means that the ratio of 30% solution to 14% solution is 3 : 2.

As it happens, these ratio units, 3 and 2 sum to 5, which is precisely the number of liters of solution we want. Hence we want ...

... 3 liters of 30% solution and 2 liters of 14% solution.

_____

Comment on the tool

This tool can be used with virtually any mixture problem—even problems involving dividing investment amounts between accounts that have different interest rates. (Figure the mix that gives the average rate of return on the total investment value.)

_____

Alternate solution method

If you want to write an equation, you write it for the total amount of acid in the solution. Let x represent the number of liters of 30% solution (the strongest contributor). Then that equation is ...

... 0.30x + 0.14(5-x) = 0.236·5

... 0.16x + 0.7 = 1.18 . . . . eliminate parenthses, simplify

... x = 0.48/0.16 = 3 . . . . subtract 0.7, divide by the coefficient of x

3 liters of 30% solution are required; 2 liters of 14% solution.

ab: 2 n83b8eb7ebi. 8e yen in

2500

step-by-step explanation: