The volume percentage of `Cl_2` at equilibrium in the dissociation of `PCl_5` under a total pressure of 1.5atm is (Kp = 0.202) ,

Explanation

`overset(1)(PCl_(5)) Leftrightarrow overset(0)(PCl_(3))+overset(0)(Cl_(2))" Initial moles"`
`(1-x)" x x Moles at equilibrium"`
Total moles at equilibrium =1-x+x+x= (1+x)
`K_(p)=(p_(PCl_(3)).p_(Cl_(2)))/(p_(PCl_(5)))=((x)/(1+x).p) ((x)/(1+x).p)/((1-x)/(1+x).p)`
`{"partial pressure of a species" =("moles of species")/("total moles") xx " total pressure"}`
`K_p=(x^2)/(1-x^2).p`
Substituting the values of `K_p and p`
`0.202=x^2/(1-x^2) xx 1.5, x =0.343`
moles of `Cl_2` at equilibrium =0.343
and total moles at equilibrium =1+x
=1+0.343
=1.343
volume percentage of chlorine =mole percent of chlorine
`=("moles of "Cl_2)/("total moles") xx 100`
`=(0.343)/(1.343) xx 100`
=25.5%