Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg c. the densities of hexane and octane are 0.655 g/cm3 and 0.703 g/cm3, respectively. assume you have 1.0 l of octane. calculate the required volume of hexane. report your answer in liters.

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L=

Density of octane= d =

Mass of octane ,m=

Moles of octane =

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

Total moles = 11.212 mol

Moles of hexane :

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 =

Volume of hexane = V

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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