A mixture contains 1 mole of volatile liquid A P-A-100 mm Hg and 3 moles of volatile liquid BP-B-80 mm Hg- If solution behaves ideally- the total vapour pressure of the distillate is -

The correct option is A 85 mm HgMole fraction of A is X_A=11+3=0.25 The mole fraction of B is X_B=1 0.25=0.75 The expression for the total pres ...

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Byju's Answer

Standard XII

Chemistry

Raoult's Law

A mixture con...

Question

A mixture contains 1 mole of volatile liquid $A (P_{A}^{\circ} = 100 m m H g)$ and 3 moles of volatile liquid $B (P_{B}^{\circ} = 80 m m H g)$ . If solution behaves ideally, the total vapour pressure of the distillate is :

85

mm Hg

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85.88

mm Hg

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90

mm Hg

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92

mm Hg

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Solution

The correct option is A $85$ mm Hg
Mole fraction of A is $X_{A} = \frac{1}{1 + 3} = 0.25$

The mole fraction of B is $X_{B} = 1 - 0.25 = 0.75$

The expression for the total pressure of the solution is

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

Substitute values in the above expression

$P = 100 \times 0.25 + 80 \times 0.75 = 85 m m H g$

Hence, the correct option is $A$

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Similar questions

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A mixture contains 1 mole of volatile liquid A(P∘A=100 mm Hg) and 3 moles of volatile liquid B(P∘B=80 mm Hg). If solution behaves ideally, the total vapour pressure of the distillate is :

The correct option is A 85 mm HgMole fraction of A is XA=11+3=0.25The mole fraction of B is XB=1−0.25=0.75The expression for the total pressure of the solution is P=PoAXA+PoBXBSubstitute values in the above expressionP=100×0.25+80×0.75=85 mm HgHence, the correct option is A

A mixture contains 1 mole of volatile liquid $A (P_{A}^{\circ} = 100 m m H g)$ and 3 moles of volatile liquid $B (P_{B}^{\circ} = 80 m m H g)$ . If solution behaves ideally, the total vapour pressure of the distillate is :