0.2 mole of NH4Cl are introduced into an empty container of 10 litre and heated to 327ºC to attain equilibrium as:
NH4Cl(s) ⇋ NH3(g) HCl(g) ; KP = 0.36 atm2 . The quantity of solid NH4Cl left is
(A) 0.078 mole
(B) 0.02 mole
(C)0.095 mole
(D) 0.035 mole
