Amount of `PCl_(5)` (in moles) need to be added to one litre vessel at `250^(@)C` in order to obtain a concentration of `0.1 "mole"` of `Cl_(2)` for the given change is:
`PCl_(5)hArrPCl_(3)+Cl_(2)` , `K_(c)=0.0414 mol litre^(-1)`

Explanation

`{:(,PCl_5 hArr , PCl_3+ , Cl_2) ,("Initial mole", a, 0,0),("Mole at equilibrium", (alpha - 0.1),0.1,0.1):}`
`therefore K_C=(0.1/1xx0.1/1)/((a-0.1)/1)` ( `therefore` volume is 1 litre)
or `0.0414=0.01/((a-0.1))` a = 0.3415 mole