Question

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.

Answer 1

The required volume of hexane is 0.66245 Liters.

Step-by-step explanation:

Volume of octane = v=1.0 L=
`1000 cm^3`

Density of octane= d =
`0.703 g/cm^3`

Mass of octane ,m=
`d* v=0.703 g/cm^3* 1000 cm^3=703 g`

Moles of octane =
`(m)/(114 g/mol)=(703 g)/(114 g/mol)=6.166 mol`

Mole percentage of Hexane = 45%

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

`55\%=\frac{6.166 mol}{\text{Total moles}}* 100`

Total moles = 11.212 mol

Moles of hexane :

`45%=\frac{\text{moles of hexane }}{\text{Total moles}}* 100`

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 =
`0.655 g/cm^3`

Volume of hexane = V

`V=(M)/(D)=(433.9044 g)/(0.655 g/cm^3)=662.4494 cm^3\approx 0.66245 L`

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.