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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.

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5 months ago

```							Hello Student,Please find the answer to your questionPtotal = pA + pBMolecular weight of CH3OH = 12 + 3 + 16 + 1 = 32Molecular weight of C­2H5OH = 24 + 5 + 16 + 1 = 46According to Raoult’s lawPtotal = p1 + p2Where Ptotal = Total vapour pressure of the solutionP1 = Partial vapour pressure of one componentP2 = Partial vapour pressure of other componenetAgain, p1 = Vapour pressure (po1) * mole fractionSimilarly, p2 = Vapour pressure (po2) * mole fractionMole fraction of CH3OH = 40/32 /40/32+60/46 = 0.49Mole fraction of ethanol = 60/46 /60/46+40/32 = 0.51NOTE 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 mmMole fraction of CH3OH in vapour = 43.48/66.17 = 0.65 Thanks and regards
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5 months ago
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