Answer : The correct option is, (B) 0.11 M
Solution :
First we have to calculate the concentration
and
.




The given equilibrium reaction is,

Initially 0.70 0.70 0
At equilibrium (0.70-x) (0.70-x) x
The expression of
will be,
![K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BPCl_3%5D%5BCl_2%5D%7D)

Now put all the given values in the above expression, we get:

By solving the term x, we get

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.
Thus, the concentration of
at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of
at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of
at equilibrium = x = 0.59 M
Therefore, the concentration of
at equilibrium is 0.11 M