The mole fraction of component A in vapour phase is x1 and mole fraction of component A in liquid mixture is x2 P0A- vapour pressure of pure A- P0B - vapour pressure of pure B- What is the total pressure-

X_1 =P_A^∘x A/P_Total=P_A^∘X_2/P_Total ∴ P_Total=P_A^∘x_2/x_1 ...

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Byju's Answer

Standard XII

Chemistry

Raoult's Law

The mole frac...

Question

The mole fraction of component A in vapour phase is $x_{1}$ and mole fraction of component A in liquid mixture is $x_{2} (P_{A}^{0}$ = vapour pressure of pure A; $P_{B}^{0}$ = vapour pressure of pure B). What is the total pressure?

$\frac{(P_{A^{\circ}} x_{2})}{x_{1}}$

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$\frac{P_{A^{\circ}} x_{1}}{x_{2}}$

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$\frac{P_{A^{\circ}} x_{1}}{x_{2}}$

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$\frac{P_{B^{\circ}} x_{2}}{x_{1}}$

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Solution

The correct option is A
$\frac{(P_{A^{\circ}} x_{2})}{x_{1}}$

$x_{1} = \frac{P_{A^{\circ}} x A}{P_{T o t a l}} = \frac{P_{A^{\circ}} X_{2}}{P_{T o t a l}} ∴ P_{T o t a l} = \frac{P_{A^{\circ}} x_{2}}{x_{1}}$

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Similar questions

Q. Mole fraction of the component A in vapour phase is

x_{1}

and mole fraction of component A in liquid mixture is

x_{2}

, then (

p_{A}^{0} =

vapour pressure of pure A;

p_{B}^{0} =

vapour pressure of pure B), the total vapour pressure of liquid mixture is :

Q. The mole fraction of component A in vapour phase is

x_{1}

and mole fraction of component A in liquid mixture is

x_{2}

(P_{A}^{0} =

vapour pressure of pure A,

P_{B}^{0} =

vapour pressure of pure B). Then total vapour pressure of the liquid mixture is:-

The mole fraction of component A in the vapour phase is $x_{1}$ and mole fraction of component A in liquid mixture is $x_{2} (P_{A}^{0}$ = vapour pressure of pure A; $P_{B}^{0}$ = vapour pressure of pure B).

Q. The mole fraction of component A in the vapour phase is

χ_{1}

and mole fraction of component A in liquid mixture is

χ_{2}

(

P_{A}^{o}

= vapour pressure of pure A;

P_{B}^{o}

= vapour pressure of pure B). Then the total vapour pressure of the liquid mixture is:

Q. In a binary mixture of two ideal volatile liquids A and B, calculate the mole fraction of component A in vapour phase if:
Vapour pressure of A in pure state is

500 m m H g

Vapour pressure of B in pure state is

700 m m H g

Total pressure

(P_{T})

of the solution is

600 m m H g

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The mole fraction of component A in vapour phase is x1 and mole fraction of component A in liquid mixture is x2(P0A= vapour pressure of pure A; P0B = vapour pressure of pure B). What is the total pressure?

The correct option is A (PA∘x2)x1 x1=PA∘xAPTotal=PA∘X2PTotal∴PTotal=PA∘x2x1

The mole fraction of component A in vapour phase is $x_{1}$ and mole fraction of component A in liquid mixture is $x_{2} (P_{A}^{0}$ = vapour pressure of pure A; $P_{B}^{0}$ = vapour pressure of pure B). What is the total pressure?

The correct option is A
$\frac{(P_{A^{\circ}} x_{2})}{x_{1}}$

$x_{1} = \frac{P_{A^{\circ}} x A}{P_{T o t a l}} = \frac{P_{A^{\circ}} X_{2}}{P_{T o t a l}} ∴ P_{T o t a l} = \frac{P_{A^{\circ}} x_{2}}{x_{1}}$