Question

I'm kinda stumped here :/

Answer 1

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:

$\frac{X}{20,0mL\\}$ × 1,492 g/mL + $\frac{Y}{20,0 mL}$ × 2,890 g/mL = 1,82 g/mL (2)

If you replace (1) in (2):

$\frac{X}{20,0mL\\}$ × 1,492 g/mL + $\frac{20,0 mL - X}{20,0 mL}$ × 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!