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Raoult's Law

The total vap...

Question

The total vapour pressure of a mixture of 1 mol A $(P_{A}^{0} = 130 t o r r)$ and 2 mol B $(P_{B}^{0} = 240 t o r r)$ is 200 mm. In this case:

There is positive deviation from Raoult's Law

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There is negative deviation from Raoult's Law

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There is no deviation from Raoult's Law

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Molecular masses of A and B are also required

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Solution

The correct option is B There is negative deviation from Raoult's Law
$P_{t o t a l} = P_{A}^{0} \times A + P_{B}^{0} \times B$

$= 203.33 m m > 200 m m$
$∴$ Vapour pressure decreases.
Hence, there is an observed negative deviation from Raoult's Law.

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Similar questions

(P_{A}^{o} = 150 t o r r)

(P_{B}^{o} = 240 t o r r)

(P_{A}^{0} = 150 t o r r)

(P_{B}^{0} = 240 t o r r)

1

P^{o}_{X} = 150 t o r r)

Y

P^{0}_{Y} = 300 t o r r

240

(P_{A}^{0} = 130 t o r r)

(P_{B}^{0} = 240 t o r r)

25^{o} C

25^{o} C

P_{A}^{0} - P_{B}^{0} = 40

P_{A}^{0}, P_{B}^{0}

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