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)`

Answer

`0.3415 "mole"`

Explanation

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Intial","a mole",,,,),(" At eqm.",(a-0*1),,0*1,,0*1 "mol " L^(-1)):}`
`K_(c) = ([PCl_(3)][CL_(2)])/([PCl_(5)]) , 1.e., 0*0414 = (0*1 xx 0*1)/(a-0*1)`
This gives a = `0*3415` mole