Question

At $300{}^{\circ }C,$ the vapour density of $PC{l}_5$ is 60. At this temperature, its dissociation percentage will be

• A. 73
• B. 77
• C. 83
• D. 87

Answer 1

Answer: (A)

$PC{l}_{5}PC{l}_3+C{l}_2$ Normal molecular weight of $PC{l}_5$ $=2\times$ vapour density $=2\times 60=120$ Observed molecular weight of $PC{l}_5=30+5\times 35.5$ $=207.5$ $i=\frac{Observedmolecularweight}{Normalmolecularweight}=\frac{207.5}{120}=1.73$ Degree of dissociation $=\frac{i-1}{n-1}=\frac{1.73-1}{2-1}=0.73$ % dissociation $=0.73\times 100=73$