InterviewSolution
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InterviewSolution
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The vapour pressure of ethanol and methanol are `44.0 mm` and `88.0 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60 g` of ethanol with `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. |
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Answer» Mol. Mass of ethyl alcolhol `(C_(2)H_(5)OH) = 46` Number of moles of ethylalcohol `= (60)/(46) = 1.304` Mol. Mass of methyl alcohol `(CH_(3)OH)=32` Number of moles of methyl alcohol `= (40)/(32) = 1.25` Mole fraction of ethyl alcohol , `chi_(A)=(1.304)/(1.304+1.25)` `=0.5107 mm Hg` Mole fraction of methyl alcohol ,`chi_(B)=(1.25)/(1.304+1.25)` `=0.4893 mm Hg` Partial pressure of ethyl alcohol `= chi_(A)P_(A)^(@)` `=22.47 mm Hg` Partial pressure of ethyl alcohol `chi_(B)P_(B)^(@)` `= 43.05 mm Hg` Total vapour pressure of solution `= 22.47 + 43.05` `65.52 mm Hg` Mole fraction of methyl alcohol in the vapour `=("Partial pressure of" CH_(3)OH)/("Total vapour pressure" ) =(43.05)/(65.52) =0.6570 ` |
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