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Standard XII

Chemistry

Vapour Pressure

At 80∘C the v...

Question

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:

0.4

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0.5

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0.8

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Solution

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

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At 80∘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∘C and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

The correct option is C 0.5P∘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 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:

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