Answer:
To prepare 20,0 mL of the liquid mixture you should mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃
Explanation:
Here you have two variables: The volume of both CHCl₃ (X) and CHBr₃ (Y). To find these two variables you must have, at least, two equations.
You know total volume is 20,0 mL. Thus:
X + Y = 20,0 mL (1)
The other equation is:
× 1,492 g/mL +
× 2,890 g/mL = 1,82 g/mL (2)
If you replace (1) in (2):
× 1,492 g/mL +
× 2,890 g/mL = 1,82 g/mL
Solving:
X = 15,3 mL
Thus, using (1):
20,0 mL - 15,3 mL = Y = 4,7 mL
Thus, to prepare 20,0 mL of the liquid mixture you must mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃.
I hope it helps!