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Standard XII

Chemistry

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At a given te...

Question

At a given temperature, the vapour pressure in mm of Hg of a solution of two volatile liquids A & B is given by the equation:

$P = 120 - 80 X_{B}$ (Where $X_{B}$ is molefraction of B)

Vapour pressures of pure A & B at the same temperature are respectively:

120, 80

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120, 200

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120, 40

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80, 40

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Solution

The correct option is C 120, 40
When $x_{B} = 0$ we have only pure A

$∴ P_{A}^{0} = 120 - (80 \times 0) = 120 m m$

The given expression $P = 120 - 80 X_{B}$ can also be written as

$P = 120 - 80 (1 - X_{A}) [∵ X_{A} + X_{B} = 1]$

If $X_{A} = 0$ , we have only pure B

$\Rightarrow P_{B}^{0} = 120 - 80 (1 - 0) = 40 m m$

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At a given temperature, the vapour pressure in mm of Hg of a solution of two volatile liquids A & B is given by the equation: P=120–80 XB (Where XB is molefraction of B) Vapour pressures of pure A & B at the same temperature are respectively:

The correct option is C 120, 40When xB=0 we have only pure A ∴P0A=120−(80×0)=120mm The given expression P=120−80 XB can also be written as P=120−80(1−XA)[∵XA+XB=1] If XA=0, we have only pure B ⇒P0B=120−80(1−0)=40mm

At a given temperature, the vapour pressure in mm of Hg of a solution of two volatile liquids A & B is given by the equation:

$P = 120 - 80 X_{B}$ (Where $X_{B}$ is molefraction of B)

Vapour pressures of pure A & B at the same temperature are respectively: