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Grade: 12

                        

Tge vapour pressure of methyl alcohol and ethyl alcohol at 25°c are 90 and 40 mm of Hg respectively. If 12 g of CH3OH and 48 g of C2H5OH are mixed, calculte tge vapour pressure exerted by each component, the total vapour pressure of mixture and the composition of mixture.

5 months ago

Answers : (1)

Arun
24742 Points
							
Hello Student,
Please find the answer to your question
Ptotal = pA + pB
Molecular weight of CH3OH = 12 + 3 + 16 + 1 = 32
Molecular weight of C­2H5OH = 24 + 5 + 16 + 1 = 46
According to Raoult’s law
Ptotal = p1 + p2
Where Ptotal = Total vapour pressure of the solution
P1 = Partial vapour pressure of one component
P2 = Partial vapour pressure of other componenet
Again, p1 = Vapour pressure (po1) * mole fraction
Similarly, p2 = Vapour pressure (po2) * 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 and regards
5 months ago
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