# Liquids A and B form an ideal solution in the entire composition range . At 350 K , the vapor pressures of pure A and pure B are$7\times 10^{3}Pa\: \: and\: \: 12\times 10^{3}Pa$ , respectively . The composition of the vapor in equilibrium with a solution containing 40 mol percent of A at this temperature is :Option 1)$x_{A}=0.37; \: x_{B}=0.63$Option 2)$x_{A}=0.28; \: x_{B}=0.72$Option 3)$x_{A}=0.4; \: x_{B}=0.6$Option 4)$x_{A}=0.76; \: x_{B}=0.24$

Mole Fraction -

$\dpi{100} Mole\: Fraction= \frac{Moles\: o\! f\: solute}{Moles\: o\! f solute+Moles\: o\! f solvent}$

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Vapour Pressure -

It is defined as pressure exerted by vapours on liquid surface at equilibrium and condensation.

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Negative deviation from Raoult's Law behaviour -

When the pressure exerted by vapours of mixture is less than that in case of ideal behaviour.

- wherein

$P_{T}< P_{A}^{0}x_{A}+ P_{B}^{0}x_{B}$

Factors affecting vapour pressure -

$V.P.\: \alpha \: T\! emp$

- wherein

It depends only on temperature and nature of liquid.

as we have learned in partial pressure

we know that

$y_{a} = \frac{P_{A}}{P_{total}} = \frac{P_{A}^{0}\times A}{P_{A}^{0}\times A\times P_{B}^{0}\times B}$

$y_{A} = \frac{7\times 10^{3}\times 0.4}{7\times 10^{3}\times 0.4 + 12\times 10^{3}\times 0.6}= \frac{2.8}{10} = 0.28$

$y_{B} = 0.72$

Option 1)

$x_{A}=0.37; \: x_{B}=0.63$

Option 2)

$x_{A}=0.28; \: x_{B}=0.72$

Option 3)

$x_{A}=0.4; \: x_{B}=0.6$

Option 4)

$x_{A}=0.76; \: x_{B}=0.24$

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