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-

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Arun · Answer

The Reaction for the PCl5 is: PCl5 ====> PCl3 + Cl2 n = 1 + 1 = 2 Now 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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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-

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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-

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