Hellmann, zelles, palojarvi, and bai published a paper in 1997 about extracting lipids from organic material. they reported using a mixture that was 85% n-hexane and 15 % dichloromethane (vol/vol). what is the mole fraction of each compound in the mixture? what is the mass fraction of each compound in the mixture? skills: unit conversions, choosing a temporary basis

Mass fraction: 73,6% n-hexane; 26,4% dichloromethane

Mole fraction: 73,0% n-hexane; 27,0% dichloromethane

Explanation:

With a basis of 100 mL:

Mass of n-hexane:

85 mL × = 55,7 g

Mass of dichloromethane

15 mL × = 20,0 g

Total mass = 20,0 g + 55,7 g = 75,7 g

Mass fraction of n-hexane:

=73,6%

Mass fraction of dichloromethane:

= 26,4%

Moles of n-hexane:

55,7 g × = 0,65 moles

Mass of dichloromethane

20,0g × = 0,24 moles

Total moles: 0,65 moles + 0,24 moles = 0,89 moles

Molar fraction of n-hexane:

=73,0%

Molar fraction of dichloromethane:

= 27,0%

I hope it helps!

the correct answer would be b) volume versus temperature, because they are also directly proportional.