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Standard XII

Chemistry

Percentage Composition

At 80∘C the...

Question

At $80^{\circ} C$ the vapour pressure of pure liquid A is 250 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at $80^{\circ} C$ and 1 atm pressure, the amount of A in the mixture is (1 atm = 760 mm Hg)

50 mole percent

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52 mole percent

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34 mole percent

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48 mole percent

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Solution

The correct option is C 34 mole percent
$P$ = $P °_{a}$ * $X_{a}$ + $P °_{b}$ ( $1$ - $X_{a}$ )
$760$ = $250$ $X_{a}$ + $1000$ *( $1$ - $X_{a}$ )
$X_{a}$ = $34$ mole percent

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Similar questions

Q. At

80^{o}

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A

250

mm of

H g

and that of pure liquid

B

1000

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and

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(

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Q. At

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^{'} A^{'}

520 mm Hg

and that of pure liquid

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1000 mm Hg

. If a mixture of

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and

^{'} B^{'}

boils at

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and

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Q. At

80^{\circ} C

, the vapour pressure of a pure liquid

A

520 mm Hg

and that of a pure liquid of

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and

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Q. At

80^{\circ}

C the vapour pressure of pure liquid ‘A’ is 520 mm of Hg and that of pure liq ‘B’ is 1000 mm of Hg. If the mixture of both liquids from an ideal solution and boils at

80^{\circ}

C and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

At $80^{\circ}$ C the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liq ‘B’ is 1000 mm Hg. If a mixture of both the liquids form an ideal solution which boils at $80^{\circ} C$ and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

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At 80∘C the vapour pressure of pure liquid A is 250 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at 80∘C and 1 atm pressure, the amount of A in the mixture is (1 atm = 760 mm Hg)

The correct option is C 34 mole percentP= P°a*Xa+ P°b*(1 - Xa)760 = 250*Xa + 1000*(1 - Xa)Xa = 34 mole percent

At $80^{\circ} C$ the vapour pressure of pure liquid A is 250 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at $80^{\circ} C$ and 1 atm pressure, the amount of A in the mixture is (1 atm = 760 mm Hg)

The correct option is C 34 mole percent
$P$ = $P °_{a}$ * $X_{a}$ + $P °_{b}$ ( $1$ - $X_{a}$ )
$760$ = $250$ $X_{a}$ + $1000$ *( $1$ - $X_{a}$ )
$X_{a}$ = $34$ mole percent