Answer:
Volume in milliliters of the liquid should be used is 49 mL.
Explanation:
Mass of the ethylene glycol = 54.6 g = 0.0546 kg
Density = [tex]1114 kg/m^3[/tex]
Volume of ethylene gylcol = V
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]1114 kg/m^3=\frac{0.0546 kg}{V}[/tex]
[tex]V=\frac{0.0546 kg}{1114 kg/m^3}=4.90\times 10^{-5} m^3[/tex]
[tex]1 m^3=1000 L[/tex]
[tex]4.90\times 10^{-5} m^3=4.90\times 10^{-5}\times 1000=4.90\times 10^{-2} L=49 ml[/tex]
1 L = 1000 mL
Volume in milliliters of the liquid should be used is 49 mL.
