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At 250°c;the vapour density of pcl5 is Y(at equilibrium)and molar mass is Q(intially) Its degree of dissociation is then equal to- At 250°c;the vapour density of pcl5 is Y(at equilibrium)and molar mass is Q(intially) Its degree of dissociation is then equal to-
The Reaction for the PCl5 is:PCl5 ====> PCl3 + Cl2n = 1 + 1 = 2Now the degree of dissociation related o the theoretical mass and the observed mass is:a = (Mth – Mobs.)/Mth.(n -1)a = (208.24 – 200)/(208.4*(2-1)) => 0.039 or 3.9% disscocitaion.
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