the density of phosphorous vapour at 310°c and 775 torr is 2.64 g/cc

1.	the density of phosphorous vapour at 310°c and 775 torr is 2.64 g/cc what is the molecular formula of phosphorous
Answer» 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 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

the density of phosphorous vapour at 310°c and 775 torr is 2.64 g/cc what is the molecular formula of phosphorous

Answer»

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