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Standard XII

Chemistry

Dalton's Law of Partial Pressures

The total pre...

Question

The total pressure for volatile components $A$ and $B$ is 0.02 bar at equilbibrium. If the mole fractions of component $A$ is $0.2$ , then what will be partial pressure of compount $B$ ?

0.02 b a r

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0.04 b a r

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0.016 b a r

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0.2 b a r

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Solution

The correct option is B $0.016 b a r$
Partial pressure of A $= P_{A} = x_{A} \times P_{t o t a l} = 0.2 \times 0.02 = 0.004$

Partial pressure of B $= P_{B}$

$P_{t o t a l} = P_{A} + P_{B}$

$P_{B} = 0.020 - 0.004 = 0.016 b a r$

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Similar questions

Q. The total vapour pressure,

P_{total}

(in torr) for a mixture of two volatile components,

A

and

B

is given by

P_{total}

= 220 - 110

X_{B}

Where,

X_{B}

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B

in mixture.

Hence, pressure (in torr) of pure components,

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and

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are respectively:

Q. Total vapour pressure of mixture of

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and

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90 m m H g

. For such case

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?

P_{A}^{0}

and

P_{B}^{0}

are the vapour pressure of pure liquid components,

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, respectively of an ideal binary solution. If

X_{A}

represents the mole fraction of component

A

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Q. Mole fraction of component

A

in the vapour phase is

X_{1}

and mole fraction of component

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The total pressure for volatile components A and B is 0.02 bar at equilbibrium. If the mole fractions of component A is 0.2, then what will be partial pressure of compount B?

The correct option is B 0.016 barPartial pressure of A=PA=xA×Ptotal=0.2×0.02=0.004Partial pressure of B=PBPtotal=PA+PBPB=0.020−0.004=0.016bar

The total pressure for volatile components $A$ and $B$ is 0.02 bar at equilbibrium. If the mole fractions of component $A$ is $0.2$ , then what will be partial pressure of compount $B$ ?

The correct option is B $0.016 b a r$
Partial pressure of A $= P_{A} = x_{A} \times P_{t o t a l} = 0.2 \times 0.02 = 0.004$

Partial pressure of B $= P_{B}$

$P_{t o t a l} = P_{A} + P_{B}$

$P_{B} = 0.020 - 0.004 = 0.016 b a r$