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

\frac{P_{A}^{0} x_{2}}{x_{1}}

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\frac{P_{A}^{0} x_{1}}{x_{2}}

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\frac{P_{B}^{0} x_{1}}{x_{2}}

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\frac{P_{B}^{0} x_{2}}{x_{1}}

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Solution

The correct option is A $\frac{P_{A}^{0} x_{2}}{x_{1}}$

$P_{t o t a l} x_{1} = P_{A}^{0} x_{2}$

$∴ P_{t o t a l} = \frac{P_{A}^{0} x_{A}}{x_{1}}$

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Similar questions

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. 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}^{o} =

vapour pressure of pure

A; P_{B}^{o} =

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

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

x_{1}

and the mole fraction of component A in liquid mixture is

x_{2}

. The total vapour pressure of liquid mixture is.
Here,

p_{A}^{\circ} =

vapour pressure of pure A

p_{B}^{\circ} =

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

The correct option is A P0Ax2x1 Ptotalx1=P0Ax2 ∴Ptotal=P0AxAx1

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 correct option is A $\frac{P_{A}^{0} x_{2}}{x_{1}}$

$P_{t o t a l} x_{1} = P_{A}^{0} x_{2}$

$∴ P_{t o t a l} = \frac{P_{A}^{0} x_{A}}{x_{1}}$