Question

Vapour density of ${PCl}_5$ is $104.25$ at $T^{\circ }C.$ Then degree of dissociation of ${PCl}_5$ is: $(M_w=208.5)$

• A. $20$%
• B. $0$%
• C. $30$%
• D. $15$%

Answer 1

The correct option is B $0$%

Solution:

$PC{l}_5\leftrightharpoons PC{l}_3+C{l}_2$

$\alpha =\frac{(D-d)}{(n-1)d}$

Here, $\alpha =$ degree of dissociation,

D = vapour density before dissociation $=M_{theo}/2$

d = vapour density after dissociation $=M_{obs}/2$

n = number of moles of product obtained by one mole of reactant

Now, here $D=208.5/2=104.25$

And $d=104.25$ (given)

And $n=2$

On putting the values we get $\alpha =0$.

Hence, option "B" is the correct answer.