4 mole of pure liquid A(PA0 = 80 mm Hg) and 6 mole of pure liquid B(PB0 = 80 mm of Hg) aremixed. The vapour pressure of the resulting solution is found to be 90 mm of Hg Hen

Ptotal as per Raoult’s law=410×80+610×100= 32 + 60 = 92 - Ve Deviation from Raoults Law

4 mole of pure liquid A(PA0 = 80 mm Hg) and 6 mole of pure liquid B(PB0 = 80 mm of Hg) aremixed. The vapour pressure of the resulting solution is found to be 90 mm of Hg Hen

Ptotal as per Raoult’s law=410×80+610×100= 32 + 60 = 92 - Ve Deviation from Raoults Law

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4 mole of pure liquid A(P_A⁰ = 80 mm Hg) and 6 mole of pure liquid B(P_B⁰ = 80 mm of Hg) are
mixed. The vapour pressure of the resulting solution is found to be 90 mm of Hg Hen

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It is an ideal solution

The solution shows negative deviation from Raoul’s law.

The solution is formed by the absorption of heat

The solution will boil at higher temperature than expected

answer is B, D.

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Detailed Solution

P_totalas per Raoult’s law
$= \frac{4}{10} \times 80 + \frac{6}{10} \times 100$
= 32 + 60 = 92 - Ve
Deviation from Raoults Law

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4 mole of pure liquid A(PA0 = 80 mm Hg) and 6 mole of pure liquid B(PB0 = 80 mm of Hg) aremixed. The vapour pressure of the resulting solution is found to be 90 mm of Hg Hen

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4 mole of pure liquid A(P_A⁰ = 80 mm Hg) and 6 mole of pure liquid B(P_B⁰ = 80 mm of Hg) are
mixed. The vapour pressure of the resulting solution is found to be 90 mm of Hg Hen