Answer:
[tex]V=5545.2mL[/tex]
Explanation:
Hello,
In this case, by considering the mass fractions and the solution's mass, one computes both ethylene glycol's and water's masses:
[tex]m_{etg}=12.9lbm*0.593=7.64lbm\\m_{H_2O}=12.9lbm-7.64lbm=5.26lbm[/tex]
Now, the densities are considered to find the volume of the solution as shown below:
[tex]V=V_{etg}+V_{H_2O}\\V=7.64lbm*\frac{1ft^3}{68.5lbm} +5.26lbm*\frac{1ft^3}{62.4lbm} \\V=0.196ft^3*\frac{28316.8mL}{1ft^3} \\V=5545.2mL[/tex]
Best regards.
