Ethanol (CH3 CH2OH) and methanol (CH3OH) form nearly ideal solutions (Raoult’s L

Question

Ethanol (CH3 CH2OH) and methanol (CH3OH) form nearly ideal solutions (Raoult’s Law is obeyed). The vapor pressure of pure ethanol is 5.93 kPa and that of pure methanol is 11.83 kPa at 20oC . (a) A solution is formed by mixing 100g of methanol and 100g of ethanol. What are the mole fractions of ethanol and methanol? (b) What is the total vapor pressure of the solution (i.e. the pressure of the gas phase in equilibrium with the liquid solution)? (c) What are the mole fractions of ethanol and methanol in the vapor phase?

Explanation / Answer

Molecular weights

Ethanol (C2H5OH)= 2*12+5+16+1= 46

Methanol (CH3OH)= 12+3+16+1=32

Moles

Methanol= 100/32= 3.125 Ethanol= 100/46= 2.173

Total moles= 3.125+2.173 = 5.298

Mole fraction= moles/ Total moles

Methanol (x1)=3.125/ 5.298 =0.59 ethanol (x2)=1-0.59=0.41

From y1P= x1p1sat and y2P= x2P12sat where x1 and x2 are mole fractinons of methanol and ethanol respectively and P1sat and P2sat are saturation pressures of methanol and ethanol respectively y1and y2 are mole fraction of methanol and ethanol in the vapor phase.

P= x1p1sat+x2p2sat = 0.59*11.83+ 0.41*5.93 =9.411 Kpa

From y1P= x1p1sat y1= 0.59*11.83/9.411 =0.742 and y2= 1-0.742= 0.258

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