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Question

# The vapour pressure of pure liquids 1 and 2 are 450 mm Hg and 700 mm Hg respectively at 350 K. When total vapour pressureof the solution is 600 mm Hg, the mole fraction of component 1 in liquid phase will be:

A
0.4
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B
0.6
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C
0.3
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D
0.7
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Solution

## The correct option is A 0.4By Raoult's Law : PT=p∘1χ1+p∘2χ2 where PT is the total pressure of the solution. p∘1 is the vapour pressure of pure component 1 p∘2 is the vapour pressure of pure component 2 χ1 is the mole fraction of component 1 in liquid phase. χ2 is the mole fraction of component 2 in liquid phase. Putting the values: 600=450χ1+700χ2600=450χ1+700(1−χ1)600=700−250χ1χ1=0.4 χ2=1−0.4=0.6 Hence option (a) is correct

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