Question

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 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of ( = 1.492 g/mL) and ( = 2.890 g/mL)? (Note: 1 mL = 1 .)

Answer 1

Answer:

Volume of $CHCl_3$   = 15.31 mL

Volume of $CHBr_3$ = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of $CHCl_3$ = 1.492 g/mL

The density of $CHBr_3$ = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  $CHCl_3$ = P ml

and the volume of $CHBr_3$ = Q ml

the sum of their volumes should be equal to the total volume of the mixture

$P \ ml + Q \ ml = 20 ml$  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

$\mathtt{(Density \ of \ CHCl_3 \times Vol. \ of \ CHCl_3 ) + (Density \ of \ CHBr_3 \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density \ of \ mixture \times volume \ of \ the \ mixture)}$

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of $CHCl_3$   = 15.31 mL

Volume of $CHBr_3$ = 4.69 mL