How much `PCl_(5)` must be added to a one little vessel at `250^(@)C` in order to obtain a concentration of 0.1 mole of `Cl_(2)` at equilibrium. `K_(c)` for `PCl(g)hArrPCl_(3)(g)+Cl_(2)(g)` is `0.0414M`

Explanation

`underset(a-x)underset(a)(PCl_(5))(g)hArrunderset(x)underset(0)(PCl_(3))(g)+underset(x "equii con.")underset(0 "initial conc.")(Cl_(2)(g))`
Given `K_(c)=0.0414`
`0.0414=([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`0.0414=((0.1)^(2))/(a-0.1)`
`a=0.3415` mol
So 0.3415 mol of `PCl_(5)` should be added to the 1 litre flask to ger 0.1 mol of `Cl_(2)` at equilibrium.