The vapour-pressure of ethanol and methanol are 44.5 and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by the 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.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Total vapour-pressure of solution :
Pm = Pºeth × M.F.eth + Pºmeth × M.F.meth
Given that :
Pºeth = 44.5 mm Hg, Pºmeth = 88.7 mm Hg
wt. of ethanol = 60 g, wt. of methanol = 40 g
∴ neth = 
= 1.304 ( mol. wt. of ethanol=40)
nmeth= 
= 1.25 ( mol. wt. of methanol=32)
xeth = 
= 
= 
= 0.51
xmeth = 
= 
= 0.49
∴ Pm = 44.5 × 0.51 + 88.7 × 0.49
= 22.67 + 43.46
= 66.13 mm Hg
mole-fraction of methanol in vapour
= 
= 
= 0.66 mm Hg
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