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  4. At 300° C, the vapour density of PCl5 is 60. At this temperature, its dissociation percentage will be

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Q. At $ 300{}^\circ C, $ the vapour density of $ PC{{l}_{5}} $ is 60. At this temperature, its dissociation percentage will be

Rajasthan PETRajasthan PET 2003

Solution:

$ PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}} $ Normal molecular weight of $ PC{{l}_{5}} $ $ =2\times $ vapour density $ =2\times 60=120 $ Observed molecular weight of $ PC{{l}_{5}}=30+5\times 35.5 $ $ =207.5 $ $ i=\frac{Observed\text{ }molecular\text{ }weight}{Normal\text{ }molecular\text{ }weight}=\frac{207.5}{120}=1.73 $ Degree of dissociation $ =\frac{i-1}{n-1}=\frac{1.73-1}{2-1}=0.73 $ % dissociation $ =0.73\times 100=73%. $