The total vapour pressure of a mixture of 1 mol A P A 0-130- torr and 2 mol B P B 0-240- torr is 200 mm- In this case-A- There is no deviation from Raoult-s LawB- There is positive deviation from Raoult-s LawC- Molecular masses of A and B are also requiredD- There is negative deviation from Raoult-s Law

The correct option is D There is negative deviation from Raoult's LawP_total=P^0_A× A+P^0_B× B =203.33mm gt; 200mm ∴Vapour pressure decreases. Hence, there ...

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Standard XII

Chemistry

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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The total vapour pressure of a mixture of 1 mol A (P0A=130 torr) and 2 mol B(P0B=240 torr) is 200 mm. In this case:

The correct option is B There is negative deviation from Raoult's LawPtotal=P0A×A+P0B×B =203.33mm>200mm ∴Vapour pressure decreases. Hence, there is an observed negative deviation from Raoult's Law.

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:

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.