Two hundred kg of liquid contains 40% pentane, and rest hexane. Pentane MW is 72.15 g/...

Two hundred kg of liquid contains 40% pentane, and rest hexane. Pentane MW is 72.15 g/ gmol and SG (20C/4C) is 0.17. Hexane MW is 86.17 g/ gmol and SG (20C/4C) is 0.659. Determine a) the total number of moles, b) the mole fraction composition of the liquid, c) the average molecular weight and d) the average density of the solution.

Solutions

Expert Solution

Mass of total liquid= 2000 kg, mass of pentane= 0.4*2000= 800 kg

Moles of pentane= mass/molecular weight= 800*1000/72.15 =11088.01 gmoles

Mass of hexane= 2000-800= 1200 kg, moles of hexane= 1200*1000/86.17gmoles= 13925.96 gmoles

Total number of moles= Moles of pentane+ moles of hexane= 11088.01+13925.96=25013.97 gmoles

Mole fractions : pentane= 11088.01/25013.97= 0.4432, hexane= 1-0.4432=0.5568

The average molecular weight = mole fraction of pentane* molecular weight of pentane + mole fraction of hexane* molecular weight of hexane= 0.4432*72.15+0.5568*86.17=79.95 g/mol

Volume of the solution= volume of pentane+ volume of hexane ( assuming the solution to be ideal)

Volume =mass/ density

Density= Specific gravity* density of water (1000 kg/m³)

Density : pentane= 0.17*1000= 170 kg/m3, hexane= 0.659*1000= 659 kg/m³

Volume : Pentane= 800/170 m3 = 4.71 m³ and hexane= 1200/659m3=1.821 m³

Volume of mixture= 4.71+1.821=6.531 m³

Average density= 2000 kg/6,531m3=306.23 kg/m3

queen_honey_blossom answered 1 year ago

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