A solution of 1-propanol and 2-propanol having 3-4 by mass of 2-propanol has an equilibrium vapour pressure of 88-8 mm Hg- Another solution having 1-3 by mass of 2-propanol has an equilibrium vapour pressure of 68-3 mm Hg- Calculate vapour pressure of pure alcohols at 40oC assuming ideal solution mixture prepared at 40oC-

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A solution of 1-propanol and 2-propanol having $\frac{3}{4}$ by mass of 2-propanol has an equilibrium vapour pressure of 88.8 mm Hg. Another solution having $\frac{1}{3}$ by mass of 2.propanol has an equilibrium vapour pressure of 68.3 mm Hg. Calculate vapour pressure of pure alcohols at $40^{o} C$ assuming ideal solution mixture prepared at $40^{o} C$ .

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Solution

1-Propanol and 2-Propanol are structural isomers and thus have same molecular weight $(M . W = 60)$
Let mass of solution be $100 g$
First Case:
Vapour pressure of solution $= 88.8 ㎜ H g$
Let Vapour pressure of 1-Propanol be $P_{2}$
Let Vapour pressure of 2-Propanol be $P_{1}$
Raoult's Law-Partial Pressure (P.P)=mole fraction in solution x Vapor pressure
$P_{T} = {(P . P)}_{o f 1 - P r o p a n o l} + {(P . P)}_{o f 2 - P r o p a n o l} = 88.8$ (given)
Since mol. weights are same for both components, mass and mole fraction would be equal.
$∴ 88.8 = x_{1} P_{1} + x_{2} P_{2}$
$88.8 = 0.75 P_{1} + 0.25 P_{2} ⟶ 1 (x_{2} = 1 - x_{1})$
Second case:
$68.3 = 0.33 P_{1} + 0.66 P_{2} ⟶ 2$
Solving equation 1 & 2 for $P_{1}$ & $P_{2}$ , we get
$P_{1} = 75 ㎜ H g$
$P_{2} = 52 ㎜ H g$

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