An ideal solution made by mixing two volatile components A and B. IF the vapour pressure of solution is 170mm Hg, then mole fraction of A in vapour phase is

$(P_{A}^{o} = 200 m m H g, P_{B}^{o} = 100 m m H g)$

Option 1 -

0.72

Option 2 -

0.82

Option 3 -

0.68

Option 4 -

0.50

2 Views | Posted 4 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 weeks ago

Correct Option - 2

Detailed Solution:

$P_{T} = P_{A}^{o} X_{A} + P_{B}^{o} X_{B}$

$170 = 200 X_{A} + 100 (1 - X_{A})$

$170 = 200 X_{A} + 100 - 100 X_{A}$

X_A = 0.7 X_B = 0.3

$Y_{A} = \frac{P_{A}^{o} X_{A}}{P_{T}} = \frac{0.7 \times 200}{170} = 0.82$ f

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An ideal solution made by mixing two volatile components A and B. IF the vapour pressure of solution is 170mm Hg, then mole fraction of A in vapour phase is

$(P_{A}^{o} = 200 m m H g, P_{B}^{o} = 100 m m H g)$

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An ideal solution made by mixing two volatile components A and B. IF the vapour pressure of solution is 170mm Hg, then mole fraction of A in vapour phase is ( P A o = 2 0 0 m m H g , P B o = 1 0 0 m m H g )

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An ideal solution made by mixing two volatile components A and B. IF the vapour pressure of solution is 170mm Hg, then mole fraction of A in vapour phase is

$(P_{A}^{o} = 200 m m H g, P_{B}^{o} = 100 m m H g)$