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1.	what is the vapour density of PCL5 at 250° C when it is dissociated to the extent of 80%
Answer» 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 Cl_² 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.

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

Answer»

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 Cl_² 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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