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Grade: 12th pass

                        

At 250 degree c and 1 atm pressure , the vapour density of PCl5 is 57.9 . What will be the dissociation of PCl5 ?

3 years ago

Answers : (1)

Arun
24742 Points
							
 

PCl5(g) →  PCl3(g) + Cl2(g)

We are given the vapour density at equilibrium at 250oC.

The initial vapour density will be 

it would be  MPCl5 / 2. 

∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25

 

∴ Total moles at equilibrium / Total moles initial  = 1 + α = Vapour density initial / Vapour density at equilibrium

At 250oC,       1 + α = 104.25 / 57.9 = 1.8

 ∴  a = 0.8

3 years ago
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