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Azeotropic Mixtures

The vapour pr...

Question

The vapour pressure of a solution of the liquids A $(P^{0} = 80 mm of Hg and X_{A} = 0.4)$ and B $(P^{0} = 120 mm of Hg and X_{B} = 0.6)$ is observed to be $100 mm of Hg$ . It shows that the solution exhibits:

Positive deviation from ideal behaviour

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Negative deviation from ideal behaviour

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Ideal behaviour

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Positive deviation at low concentrations and negative at higher concentrations

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Solution

The correct option is B Negative deviation from ideal behaviour
From Raoults law,
$P_{total} = p_{A}^{0} \times X_{A} + p_{B}^{0} \times X_{B}$
$= 80.0 \times 0.4 + 120.0 \times 0.6$
$= 104 mm of Hg$
The observed $P_{total}$ is $100 mm of Hg$ which is less than $104 mm of Hg$ . Hence the solution shows negative deviation.

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Azeotropic Mixtures

Standard XII Chemistry

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The vapour pressure of a solution of the liquids A (P0=80 mm of Hg and XA=0.4) and B (P0=120 mm of Hg and XB=0.6) is observed to be 100 mm of Hg. It shows that the solution exhibits:

The correct option is B Negative deviation from ideal behaviourFrom Raoults law, Ptotal=p0A×XA+p0B×XB =80.0×0.4+120.0×0.6 =104 mm of Hg The observed Ptotal is 100 mm of Hg which is less than 104 mm of Hg. Hence the solution shows negative deviation.

The vapour pressure of a solution of the liquids A $(P^{0} = 80 mm of Hg and X_{A} = 0.4)$ and B $(P^{0} = 120 mm of Hg and X_{B} = 0.6)$ is observed to be $100 mm of Hg$ . It shows that the solution exhibits: