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Question

When the mixture of two immiscible liquids (water and nitrobenzene) boils at 372 K and the vapour pressures at this temperature are 97.7 kPa(H2O) and 3.6 kPa(C6H5NO2), calculate the weight percentage of nitrobenzene in the vapour.

A
10 %
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B
28.5 %
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C
72 %
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D
20.1 %
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Solution

The correct option is D 20.1 %
Since the two liquids given are immiscible, the mole fraction for both in liquid mixture will be taken as 1
total vapour pressure of liquid mixture =97.7+3.6=101.3kPa
Now mole fraction of nitrobenzene in vapour phase =3.6101.3=0.0355
Now weight percentage of NO2 in vapour phase =0.0355×123(10.0355)×18+0.0355×123×100
=20.1%

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