CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper - ...

CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

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Question : 19 of 35

Marks: +1, -0

SECTION - B

The vapour pressure of pure liquid

X

and pure liquid

{Y}

at

25°{C}

are

120\ {mm} \ {Hg}

and

160 \ { mm} \ {Hg}

respectively. If equal moles of

X

and

Y

are mixed to form an ideal solution, calculate the vapour pressure of the solution.

Solution:

{X}=120 \ { mm} \ {Hg}, {P}_{{A}}={P}_{{A}}^0 \.{x}_{{A}}

{Y}=160 \ { mm} \ {Hg}, {P}_{{B}}={P}^0_{{B}} \.{x}_{{B}}

\ ∵\

When equal amount of

X

and

Y

are mixed

{x}_{{A}}+{x}_{{B}}=1

Acc. to Raoutt's law

{P}_{\text "total "}={P}_{{A}}+{P}_{{B}}

{P}_{{A}}^0, {x}_{{A}}+{P}_{{ }_{{B}} }^0, {x}_{{B}}

=(1-{x}_{{B}}) {P}_{{A}}^0+{x}_{{B}} {P}_{{B}}^0

{P}_{\text "Total "}={P}_{{A}}^0+({P}_{{B}}^0-{P}_{{A}}^0) {x}_{{B}}

=120+(160-120) \×1

{P}_{\text "Total "}=120+40

=160 \ {mm\ Hg}

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