Doc has available to have a 20% alcohol solution and a 70% alcohol solution how many liters of each solution should be mixed to make 30 L of a 50% alcohol solution?

12 L of the 20% solution should be used with 18 L of the 70% alcohol solution

Step-by-step explanation:

assuming that the concentration given is in terms of mass/volume , then doing an mass balance

C₁*V₁ + C₂*V₂ =C₃*V₃

where C= concentration , V= volume , 1,2,3 denotes first , second and final solution respectively.

and since we are mixing the same substance, the total volume  V₃ will be the sum of the solutions mixed

V₁ + V₂ = V₃ → V₂=V₃-V₁

thus

C₁*V₁ + C₂*(V₃-V₁) =C₃*V₃

(C₁-C₂)*V₁ + C₂*V₃ = C₃*V₃

(C₁-C₂)*V₁  =  (C₃- C₂) *V₃

V₁ = V₃ *  (C₃- C₂)/(C₁-C₂)

replacing values

V₁ = V₃ *  (C₃- C₂)/(C₁-C₂) = 30 L * (50%-70%)/(20%- 70%) = 12 L

then

V₂=V₃-V₁=30 L - 12 L = 18 L

therefore 12 L of the 20% solution should be used with 18 L of the 70% alcohol solution

The solution is ,

Step-by-step explanation:

Equation:

-Combine like terms:

Add both sides by 2:

-Divide both sides by 26:

So, the answer is .