Total vapour pressure of mixture of 1 mol of A P0A-150 torr and 2 mol of B P0B-240 torr is 200 torr- In this case -

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Thermal Expansion

Total vapour ...

Question

Total vapour pressure of mixture of 1 mol of A $(P_{A}^{0} = 150 t o r r)$ and 2 mol of B $(P_{B}^{0} = 240 t o r r)$ is 200 torr. 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 for calculating the deviation

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Solution

The correct option is A there is negative deviation from Raoult's law
Considering the mixture to be ideal,
$P_{t o t a l} = P_{A}^{\circ} χ_{A} + P_{B}^{\circ} χ_{B} = 50 + 160 = 210 t o r r$
But the experimental pressure is 200 torr. So, the mixture shows a negative deviation from Raoult's law.

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Total vapour pressure of mixture of 1 mol of A (P0A=150 torr) and 2 mol of B (P0B=240 torr) is 200 torr. In this case :

The correct option is A there is negative deviation from Raoult's lawConsidering the mixture to be ideal,Ptotal=P∘AχA+P∘BχB=50+160=210torrBut the experimental pressure is 200 torr. So, the mixture shows a negative deviation from Raoult's law.

Total vapour pressure of mixture of 1 mol of A $(P_{A}^{0} = 150 t o r r)$ and 2 mol of B $(P_{B}^{0} = 240 t o r r)$ is 200 torr. In this case :

The correct option is A there is negative deviation from Raoult's law
Considering the mixture to be ideal,
$P_{t o t a l} = P_{A}^{\circ} χ_{A} + P_{B}^{\circ} χ_{B} = 50 + 160 = 210 t o r r$
But the experimental pressure is 200 torr. So, the mixture shows a negative deviation from Raoult's law.