Question

1. The absolute density of phosphorus vapour at 310°C and 775 torr is 2.64 g dm
What is the molecular formula of phosphorus?
The absolute density of phosphorus vapour at 310-C and 775 tor is 2.64 g dm​

Answer 1

Here we have ρ = 2.64 gdm-3 = 2.64 gL-1

Pressure P = 775 mm = 775 / 760 atm = 1.02 atm

T = 3100 C = 310 + 273 = 583 K

R = 0.0821 litre atm-1K-1mol-1 

now PV = nRT

 = (mass / molar mass) X RT

Thus Molar mass of phosphorus molecule = [mass / (P X volume )] X RT

 = (ρ RT) / P

 = (2.64 X 0.0821 X 583 ) / 1.02

 = 123.88 g

Molar mass of 1 phosphorus atom = 31 g

Therefore 123.88 / 31 = 4

This means that 4 atoms of phosphorus constitute a molecule of phosphorus. Hence the molecular formula of phosphorus is P4

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Answer 2

Explanation:

Here we have ρ = 2.64 gdm-3 = 2.64 gL-1

Pressure P = 775 mm = 775 / 760 atm = 1.02 atm

T = 3100 C = 310 + 273 = 583 K

R = 0.0821 litre atm-1K-1mol-1 

now PV = nRT

 = (mass / molar mass) X RT

Thus Molar mass of phosphorus molecule = [mass / (P X volume )] X RT

 = (ρ RT) / P

 = (2.64 X 0.0821 X 583 ) / 1.02

 = 123.88 g

Molar mass of 1 phosphorus atom = 31 g

Therefore 123.88 / 31 = 4

This means that 4 atoms of phosphorus constitute a molecule of phosphorus. Hence the molecular formula of phosphorus is P4

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