SATHEE: Solutions Question 311

Solutions Question 311

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

Options:

A) positive deviation from ideal behaviour

B) negative deviation from ideal behaviour

C) ideal behaviour

D) positive deviation for lower cone. and negative for higher cone.

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Answer:

Correct Answer: B

Solution:

$P_{t o t a l} = P_{A}^{0} \times X_{A} + P_{B}^{0} \times X_{B}$

$= 80.0 \times 0.4 + 120.0 \times 0.6 = 104 m m$ Hg The observed $P_{t o t a l}$ is 100 mm Hg which is less than 104 mm Hg. Hence the solution shows negative deviation.

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Solutions Question 311

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

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Solutions Question 311

Question: The vapour pressure of a solution of the liquids A ( P∘=80 mm Hg and xA=0.4 ) and B ( P∘=120 mm Hg and XB=0.6 ) is found to be 100mm Hg. It shows that the solution exhibits

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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