Question

I'm kinda stumped here :/

Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, you know that phosphorus has a density of 1.82 g/m^3 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl^3 ( d= 1.492 g/mL) and CHBr^3( d= 2.890 g/mL)? (Note: 1 mL = 1 cm^3 .)

Answer 1

To prepare 20,0 mL of the liquid mixture you should mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃

Step-by-step 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:

`(X)/(20,0mL\\)` × 1,492 g/mL +
`(Y)/(20,0 mL)` × 2,890 g/mL = 1,82 g/mL (2)

If you replace (1) in (2):

`(X)/(20,0mL\\)` × 1,492 g/mL +
`(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!