The total vapour pressure of a mixture of 1 mol A $(P_{A}^{o} = 150 t o r r)$ and 2 mol B $(P_{B}^{o} = 240 t o r r)$ 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 A there is negative deviation from Raoult's law.
$P_{t o t a l} = P_{A}^{o} + P_{B}^{o}$

$= 210 m m > 200 m m$

Vapour pressure is decreased. So, there is a negative deviation from Raoult's law.

So, the correct option is $B$

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The total vapour pressure of a mixture of 1 mol A (PoA=150 torr) and 2 mol B (PoB=240 torr) is 200 mm. In this case:

The correct option is A there is negative deviation from Raoult's law.Ptotal=PoA+PoB =210mm>200mmVapour pressure is decreased. So, there is a negative deviation from Raoult's law.So, the correct option is B

The total vapour pressure of a mixture of 1 mol A $(P_{A}^{o} = 150 t o r r)$ and 2 mol B $(P_{B}^{o} = 240 t o r r)$ is 200 mm. In this case:

The correct option is A there is negative deviation from Raoult's law.
$P_{t o t a l} = P_{A}^{o} + P_{B}^{o}$

$= 210 m m > 200 m m$

Vapour pressure is decreased. So, there is a negative deviation from Raoult's law.

So, the correct option is $B$