Answer:
Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃
Explanation:
To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).
The mixture has a density of 2.07 g/mL, so <u>20 mL of the mixture would weigh</u>:
20 mL * 2.07 g/mL = 41.4 g
Let X be the volume of CHCl₃ and Y the volume of CHBr₃:
X + Y = 20 mL
The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:
X * 1.492 + Y * 2.890 = 41.4 g
So now we have a system of two equations and two unknowns, we use algebra to solve it:
1. Express Y in terms of X:
X + Y = 20
2. Replace Y in the second equation:
X * 1.492 + Y * 2.890 = 41.4
- 1.492*X + 2.890*(20-X) = 41.4
3. Solve for X:
- 1.492*X + 57.8 - 2.890*X = 41.4
4. Using the now known value of X, solve for Y:
X + Y = 20
So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of CHCl₃.
