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Question

A mixture of chlorobenzene and wate (immiscible) boils at 90.3oC at an external pressure of 740.2 mm. The vapour pressure of pure water at 90.3oC is 530.1 mm. Calculate the % composition of distillate :

A
H2O= 35%
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B
H2O= 22%
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C
H2O= 29%
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D
H2O= 71%
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Solution

The correct option is C H2O= 29%
At boiling point PM=740.2mm
PH2O=530.1mm
Pchlorobenzene=740.2530.1
=210.1mm
Also, P=PM× mole fraction
530.1=740.2× mole fraction of H2O
mole fraction of H2O=0.716
mole fraction of C6H5Cl=0.284
Let a g H2O and b g chlorobenzene be present in distillate
or wH2O/18wCB112.5+WH2O18=0.716 ....(i)
and wH2O/112.5wCB112.5+WH2O18=0.284 ....(ii)
By eq. (i)/(ii)
or a18×112.5b=0.7160.284
or ab=12.8931.5=0.403 .....(iii)
Let total distillate be 100 g
a+b=100 .....(iv)
By eqs. (iii) and (iv)
b= 71.27%
a= 28.73%

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