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At 80 oC, the vapour pressure of pure liquid ′A′ is 520 mm Hg and that of pure liquid ′B′ is 1000 mm Hg. If a mixture of ′A′ and ′B′ boils at 80 oC and 1 atm pressure, the amount of ′A′ (in mole percent) in the mixture is:(1 atm=760 mm Hg) |
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Answer» At 80 oC, the vapour pressure of pure liquid ′A′ is 520 mm Hg and that of pure liquid ′B′ is 1000 mm Hg. If a mixture of ′A′ and ′B′ boils at 80 oC and 1 atm pressure, the amount of ′A′ (in mole percent) in the mixture is: (1 atm=760 mm Hg) |
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