Question

The vapour density of $PC{l}_5$ at $300{}^{\circ }C$ is 60. Its dissociation percentage will be at this temperature

• A. 73
• B. 77
• C. 83
• D. 87

Answer 1

Answer: (A)

Normal molecular weight of $PC{l}_5=2\times$ vapour density $=2\times 60$ $=120$ Observed molecular weight of $PC{l}_5=31+(35.5)\times 5$ $=31+177.5$ $=208.5$ van-t Hoff factor $(i)=\frac{208.5}{120}$ $=1.7375=1.73$ Degree of dissociation $=\frac{i-1}{n-1}$ $=\frac{1.73-1}{2-1}$ $=0.73$ % dissociation $=0.73\times 100=73$