At a certain temperature, the vapour pressure of pure either is $640 m m$ and that of pure acetone is $280 m m$ . Calculate the mole fraction of each component in the vapour state if the mole fraction of ether in the solution is $0.50$ .

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Solution

Let $x_{A}$ & $x_{B}$ denote mole fractions of Ether & Acetone respectively in the solution.
$P_{s o l u t i o n} = P_{s} = P_{A}^{\circ} \cdot x_{A} + P_{B}^{\circ} \cdot x_{B}$
$= 640 \times 0.5 + 280 \times 0.5$
$= 460 m m$
Mole fraction of ether in vapour state $= \frac{x_{A} \cdot P_{A}^{\circ}}{P_{s}}$
$= \frac{0.5 \times 640}{460} = 0.695$
Mole fraction of acetone in vapour state $= \frac{x_{B} \cdot P_{B}^{\circ}}{P_{s}}$
$= \frac{0.5 \times 280}{460} = 0.305$

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At a certain temperature, the vapour pressure of pure either is 640 mm and that of pure acetone is 280 mm. Calculate the mole fraction of each component in the vapour state if the mole fraction of ether in the solution is 0.50.

Let xA & xB denote mole fractions of Ether & Acetone respectively in the solution.Psolution=Ps=P∘A⋅xA+P∘B⋅xB=640×0.5+280×0.5=460mmMole fraction of ether in vapour state=xA⋅P∘APs=0.5×640460=0.695Mole fraction of acetone in vapour state=xB⋅P∘BPs=0.5×280460=0.305

At a certain temperature, the vapour pressure of pure either is $640 m m$ and that of pure acetone is $280 m m$ . Calculate the mole fraction of each component in the vapour state if the mole fraction of ether in the solution is $0.50$ .