Vapour pressure

Asked by  | 17th Jan, 2010, 09:54: PM

Expert Answer:

Molar mass of methanol = 32 g mol-1

Molar mass of ethanol = 46 g mol-1

nmethanol = 40 / 32 = 1.25 moles

nethanol = 60 / 46 = 1.30 moles

n total = 1.25 + 1.30 = 2.55 moles

X methanol = 1.25 / 2.55 = 0.49

X ethanol = 1.30 / 2.55 = 0.51

ptotal = poethanol + (pomethanol  - poethanol)  x  Xmethanol

           = 44.5 + (88.7 – 44.5) x 0.49

           = 66.16 mm Hg

Mole fraction in vapour phase,

pmethanol = 0.49 x 66.16 = 32.64 mm Hg

pethanol = 0.51 x 66.158 = 33.74 mm Hg

ymethanol = 32.64 / 66.16 = 0.49

yethanol = 33.74 / 66.16 = 0.51

 

       

 

 

 

Answered by  | 18th Jan, 2010, 10:48: AM

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