`0.2 "mole"` of `NH_(4)Cl` are introduced into an empty container of `10 litre` and heated to `327^(@)C` to attain equilibrium as : `NH_(4)Cl_((s))hArrNH_(3(g))+HCl_((g))`, `(K_(p)=0.36 atm^(2))`. The quantity of solid `NH_(4)Cl` left is :A. `0.078 "mole"`B. `0.02 "mole"`C. `0.095 "mole"`D. `0.035 "mole"`

`0.2 "mole"` of `NH_(4)Cl` are introduced into an empty container of `

1.	`0.2 "mole"` of `NH_(4)Cl` are introduced into an empty container of `10 litre` and heated to `327^(@)C` to attain equilibrium as : `NH_(4)Cl_((s))hArrNH_(3(g))+HCl_((g))`, `(K_(p)=0.36 atm^(2))`. The quantity of solid `NH_(4)Cl` left is :A. `0.078 "mole"`B. `0.02 "mole"`C. `0.095 "mole"`D. `0.035 "mole"`
Answer» `{:(,NH_(4)Cl,hArr,NH_(3),+,HCl),("Moles at t=0",0.20,,0,,0),("Mole at eqm.",(0.20-a),,a,,a):}` Also, `K_(p)=P_(NH_(3))xxP_(HCl)=P^(2)` `P=sqrt(K_(p))=sqrt(0.36)=0.6 atm` Now, `NH_(3)` formed `n=(PV)/(RT)=(0.6xx10)/(0.0821xx600)` `=0.122 "mole" ="moles of" NH_(4)Cl` decomposed `:. NH_(4)Cl` left `=0.2-0.122=0.078 "mole"`

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