At $80^{\circ}$ C the vapour pressure of pure liquid ‘A’ is 520 mm of Hg and that of pure liq ‘B’ is 1000 mm of Hg. If the mixture of both liquids from an ideal solution and boils at $80^{\circ}$ C and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

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Solution

$P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B} = P_{m i x} (520) X_{A} + (1000) (1 - X_{A}) = 760 520 X_{A} - 1000 X_{A} = 760 - 1000 - 480 X_{A} = - 240 ∴ X_{A} = \frac{1}{2} = 0.5$

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At $80^{\circ}$ C the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liq ‘B’ is 1000 mm Hg. If a mixture of both the liquids form an ideal solution which boils at $80^{\circ} C$ and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

At $80^{\circ} C$ the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liq 'B' is 1000 mm Hg. If mixture of both liquids from ideal solution and boils at $80^{\circ} C$ and 1 atm pressure, the mole fraction of 'A' in the liquid mixture is:

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At 80∘C the vapour pressure of pure liquid ‘A’ is 520 mm of Hg and that of pure liq ‘B’ is 1000 mm of Hg. If the mixture of both liquids from an ideal solution and boils at 80∘C and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

P∘AXA+P∘BXB=Pmix(520)XA+(1000)(1−XA)=760520XA−1000XA=760−1000−480XA=−240∴XA=12=0.5

At $80^{\circ}$ C the vapour pressure of pure liquid ‘A’ is 520 mm of Hg and that of pure liq ‘B’ is 1000 mm of Hg. If the mixture of both liquids from an ideal solution and boils at $80^{\circ}$ C and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

$P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B} = P_{m i x} (520) X_{A} + (1000) (1 - X_{A}) = 760 520 X_{A} - 1000 X_{A} = 760 - 1000 - 480 X_{A} = - 240 ∴ X_{A} = \frac{1}{2} = 0.5$