Question

At 500°C and 93.2 kPa, the mass density of sulfur vapour is 3.710 kg m-'. What is the molecular formula of sulfur under these conditions? 1.8(b) At 100°C and 1.60 kPa, the mass density of phosphorus vapour is 0.6388 kg m-'. What is the molecular formula of phosphorus under these conditions?

Answer 1

At 500°C and 93.2 kPa, the mass density of sulfur vapour is 3.710 kg m-'. What is the molecular formula of sulfur under these conditions?

Solution :- T = 500 C +273 = 773 K

Pressure = 93.2 kpa * 1 atm / 101.325 kpa = 0.9198 atm

Density of sulfur vapor = 3.710 kg/m3 = 3.710 g/L

Lets calculate the molar mass of the vapor

PM = dRT

M= dRT/P

   = 3.710 g/ L * 0.08206 L atm per mol K * 773 K / 0.9198 atm

= 256 g/ mol

Now lets calculate the number of sulfur atoms present in this

256 g /mol / 32.066 g per mol = 8 mol S

So the molecular form of the sufur vapor is S8

1.8(b) At 100°C and 1.60 kPa, the mass density of phosphorus vapour is 0.6388 kg m-'. What is the molecular formula of phosphorus under these conditions?

Solution :- T= 100 C +273 = 373 K

Pressure = 16.0 kpa * 1 atm / 101.325 kpa = 0.1579 atm

Density = 0.6388 kg/m3 = 0.6388 g/L

PM = dRT

M= dRT/P

   = 0.6388 g per L * 0.08206 L atm per mol K *373 K / 0.1579 atm

   = 124 g/ mol

Now lets find the moles of the P in the formula

124 g / 31.0 g per mol = 4 mol P

So the formula of the phosphorous vapor is P4