Answer to Question #230431 in General Chemistry for Vee

Question

mix 6.80 mL of CHCl3 (d=1.492 g/mL) and 6.90 mL of CHBr3 (d= 2.890 g/mL) giving 13.7 mL

Answer 1

Question:

Consider the following: you mix 6.80mL of CHCl3 (d = 1.492 g/mL) and 6.90mL of CHBr3 (d = 2.890 g/mL), giving 13.7mL of solution. What is the density of this mixture?

Explanation:

The density of mixture will be calculated by taking ratio of total mass of the mixture obtained by mixing the two liquids and the total volume of the mixture

The total volume = volume of $CHCl_3$ + volume of $CHBr_3$ = $6.80+6.90 =13.70mL$

The mass of CHCl3-l; = density of CHCI₃ × volume of ICI₃

= $1.492 × 6.80= 10.15 \ g$

the mass of CHBr3-I; = density of CHBr₃ × volume of CHBrl₃ $= 2.890 × 6.90= 19.94\ g$

Thus total mass $= 10.15\ g + 19.94\ g = 30.09\ g$

Density of mixture = total mass / total volume $=30.09/13.70 = 2.20\ g/mL$