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in Equilibrium by (45 points)

what is the vapour density of PCL5 at 250° C when it is dissociated to the extent of 80%

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PCl5(g) → PCl3(g) + Cl2(g)  

80% 20% 20%

Assuming the pressure is 1 atm.  

The density of the mixture will have an 80% contribution from PCl5  

20% contribution from PCl3 and  20% contribution from Cl2

Density of PCl5 at 250C = (208 g/mol /22.4L) x 273K /523K  x 0.80  = 3.87 g/L 

Density of PCl3 at 250C = (137 g/mol /22.4L) x 273K /523K  x 0.20  = 0.63 g/L 

Density of Cl2 at 250C = (70.9 g/mol /22.4 L) x 273K /523K  x 0.20  = 0.33 g/L  

Density of mixture = 3.87 + 0.63 + 0.33 = 4.83 g/L.

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