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Standard XII

Chemistry

Raoult's Law

The vapour pr...

Question

The vapour pressure of ethanol (A) and methanol (B) are $44.5 m m H g$ and $88.7 m m H g$ 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 $(P_{T})$ and the mole fraction of methanol in the vapour phase $(Y_{B})$

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Solution

$Molar mass of ethyl alcohol = 46 g m o l^{- 1}$
$No. of moles of ethyl alcohol = \frac{60}{46} = 1.304 m o l$
$Molar mass of methyl alcohol = 32 g m o l^{- 1}$
$No. of moles of methyl alcohol = \frac{40}{32} = 1.25 m o l$
$Mole fraction of ethyl alcohol, XA=1.3041.304+1.25=0.5107$
$Mole fraction of methyl alcohol, XB=1.251.304+1.25=0.4893$
$Partial pressure of ethyl alcohol = X_{A} \times P_{A}^{\circ} = 0.5107 \times 44.5 = 22.73 m m H g$
$Partial pressure of methyl alcohol = X_{B} \times P_{B}^{\circ} = 0.4893 \times 88.7 = 43.40 m m H g$
$Total vapour pressure of solution = 22.73 + 43.40 = 66.13 m m H g$
$Mole fraction of methyl alcohol in the vapour,YB=Partial pressure of CH3OHTotal vapour pressure=43.4066.13=0.6563$

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Raoult's Law

Standard XII Chemistry

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