Grade upto college level Physical Chemistry

The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Shane Macguire

12 Years agoGrade upto college level

3 Answers

Deepak Patra

12 Years ago

Hello Student,

Please find the answer to your question

P_total = p_A + p_B

Molecular weight of CH₃OH = 12 + 3 + 16 + 1 = 32

Molecular weight of C₂H₅OH = 24 + 5 + 16 + 1 = 46

According to Raoult’s law

P_total = p₁ + p₂

Where P_total = Total vapour pressure of the solution

P₁ = Partial vapour pressure of one component

P₂ = Partial vapour pressure of other componenet

Again, p₁ = Vapour pressure (p^o₁) * mole fraction

Similarly, p₂ = Vapour pressure (p^o₂) * mole fraction

Mole fraction of CH₃OH = 40/32 /40/32+60/46 = 0.49

Mole fraction of ethanol = 60/46 /60/46+40/32 = 0.51

NOTE THIS STEP : Thus now let us first calculate the partial vapour pressures, i.e., p₁ and p₂ of the two component.

Partial vapour pressure of CH₃OH(p₁)

= 44.5 * 0.51 = 22.69 mm

∴ Total vapour pressure of the solution

= 43.48 + 22.69 mm = 66.17 mm

Mole fraction of CH₃OH in vapour = 43.48/66.17 = 0.65

Thanks
Deepak patra
askIITians Faculty

Sandeep

8 Years ago

P=P1+P2MOLECULAR WEIGHT OF CH3OH=32(p1)MOLECULAR WEIGHT OF C2H5OH=46(p2)NOW WE HAVEP1=vapour pressure(P`1)×mole fractionP2=vapour pressure (P`2)×mole fraction ThenMole fraction of CH3OH=0.49Mole fraction of C2H5OH=0.51P1=44.5×0.49=22.69P2=88.7×0.51=44.8P=P1+P2=66.9mm=total vapour pressure of the mixture 66.9mm Mole fraction of CH3OH in vapour is 44.8/66.17 = 0.65