Respuesta :
The question is incomplete, here is the complete question:
In the lab, keith has two solutions that contain alcohol and is mixing them with each other. Solution A is 20 % alcohol and Solution B is 6 % alcohol. He uses 400 milliliters of Solution A. How many milliliters of Solution B does he use, if the resulting mixture is a 12% alcohol solution?
Answer: The volume of solution B used is 533.3 mL
Explanation:
We are given:
Volume percent of alcohol in solution A = 20 %
Volume percent of alcohol in solution B = 6 %
Let the volume of solution B used be 'x' mL
Volume of solution A used in the mixture = 400 mL
Total volume of the mixture = (400 + x) mL
Calculating the volume of solution B used:
[tex]\Rightarrow (20\% \text{ of }400)+(6\% \text{ of }x)=12\% \text{ of }(400+x)[/tex]
[tex]\Rightarrow (\frac{20}{100}\times 400)+(\frac{6}{100}\times x)=\frac{12}{100}\times (400+x)\\\\\Rightarrow 80+0.06x=0.12(400+x)\\\\\Rightarrow 0.06x=32\\\\\Rightarrow x=\frac{32}{0.06}=533.3mL[/tex]
Hence, the volume of solution B used is 533.3 mL
