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A pharmacist has an 18% alcohol solution. How much of this solution and how much water must be mixed together to make 10 liters of a 12% alcohol solution?

Answer :

Answer:

[tex]6\frac{2}{3}[/tex] of 18%,  [tex]3\frac{1}{3}[/tex] water

Step-by-step explanation:

The 18% solution added to the water is equal to the 10 liters of 12% solution.

Let's say that the amount of 18% solution is x

18% of x = 0.18x

Let's say that the amount of water is 10-x

Because there is no alcohol in water, it will be 0(10-x), or just 0

0.18x + 0 = 10•0.12

0.18x = 1.2

x = [tex]6\frac{2}{3}[/tex]

10 - [tex]6\frac{2}{3}[/tex] = [tex]3\frac{1}{3}[/tex]

 The volume required to dilute the concentration to 12% is 6.6L

Data;

  • c1 = 18%
  • v1 = ?
  • c2 = 12%
  • v2 = 10L

Dilution Formula

To solve this problem, we have to use dilution formula which is given as

[tex]c_1v_1 = c_2v_2[/tex]

we can convert the percentages to decimal and solve

[tex]c_1 = 18\% = 0.18\\c_2 = 12\% = 0.12[/tex]

Let's substitute the values and solve

[tex]c_1v_1 = c_2v_2\\0.18v_1 = 0.12*10\\0.18v_1 = 1.2\\v_1 = \frac{1.2}{0.18} \\v_1= 6.6L[/tex]

The volume required to dilute the concentration to 12% is 6.6L

Learn more on dilution formula here;

https://brainly.com/question/7208939

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