Answer:
Volume of CHCl3 required, V(CHCl3) = 8.02 mL
Volume of CHBr3 required, V(CHBr3) = 11.98 mL
Explanation:
Note: The complete question is incomplete. The complete question is given below:
Suppose now that you wanted to determine the density of a small yellow crystal to confirm that it is silicon. From the literature, you know that silicon has a density of 2.33 g/cm3. How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl3 (d = 1.492 g/mL) and CHBr3 (d = 2.890 g/mL)? (Note: 1 mL = 1 cm3.)
V(CHCl3) = mL
V(CHBr3) = mL
Density of required mixture of CHCl3 and CHBr3 equals the density of silicon = 2.33 g/mL
Density of CHCl3 = 1.492 g/mL
Density of CHBr3 = 2.890 g/mL
Let the fraction of CHCl3 required be y
Fraction of CHBr3 required will be 1 - y
Density of mixture = ( fraction of CHCl3 * density of CHCl3) + (fraction of CHBr3 * density of CHBr3)
2.33 = (y * 1.492) + (1 - y) * 2.890
2.33 = 1.492y + 2.890 - 2.890y
-0.56 = -1.398y
y = 0.401
Therefore, volume of CHCl3 required = 0.401 * 20 mL = 8.02 mL
Volume of CHBr3 required = (20 - 8.02) = 11.98 mL
