Question

# Calculate the mole fraction of ethane (C2H6) in water at 10ºC when the ethane concentration in...

Calculate the mole fraction of ethane (C2H6) in water at 10ºC when the ethane concentration in air that is in equilibrium with the water is 10,000 ppm. The total pressure is 5.0 atm.

we will be using Antoine equation and Raoult's Law for solving this problem

T= 10 oC

Total P = 5 atm

conc. of ethane in air =10,000 ppm

1 ppm = 1 gas volume/106 air volume

mole % = Volume % = Pressure %

so,

mole fraction = volume of gas/( volume of gas+ volume of air)

(y) mole fraction of C2H6 in air = 10,000/(10,000+1,000,000) =0.09090909

Raoult's Law

yiP =xiPisat

For this first we need to know Pi sat of ethane at 10oC

for this use antoine equation for ethane

calculating vapour P

where T in K & P in kPa

ln P=13.88 -1582/(283.15-13.76)

lnP=8.00747318

P=3003.318672 kPa

{as    1 atm= 101.325 kPa}

P=29.640450747594 atm

now solving Raoult's Law

0.09090909*5=xi*29.640450747594

xi =0.01533

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