Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=[tex]1000 cm^3[/tex]
Density of octane= d = [tex]0.703 g/cm^3[/tex]
Mass of octane ,m= [tex]d\times v=0.703 g/cm^3\times 1000 cm^3=703 g[/tex]
Moles of octane =[tex]\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol[/tex]
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%
[tex]55\%=\frac{6.166 mol}{\text{Total moles}}\times 100[/tex]
Total moles = 11.212 mol
Moles of hexane :
[tex]45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100[/tex]
Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D = [tex]0.655 g/cm^3[/tex]
Volume of hexane = V
[tex]V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L[/tex]
(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.
