When 3.2 g of Sulphur is vaporized at 450oC and 723 mm pressure, the vapors occupy a volume of 780 ml. What is the molecular formula of sulphur vapours under these conditions? Calculate the vapor density also.

Given-
Mass of sulphur  = 3.2 g
Temperature = 450^oC or 723 K
Volume = 780 mL or 0.780 L
Pressure = 723 mm Hg or 0.9513 atm ( 1 mm Hg = 0.0013158 atm)
Calculation - 
Using ideal gas equation -
PV = nRT
​Where P = pressure
V= volume 
n = number of moles
R = gas constant (0.082 L atm/ mol. K)
​T = temperature
Putting the values in the given equation -
$0.9513 atm \times 0.780 L = n \times 0.082 L atm/ mol. K \times 723 K$

0.7420 = n $\times$ 59.286 
n = 0.01252 
Now, number of moles  = mass / Molar mass
0.01252 = 3.2 / Molar mass
Molar mass = 255.59 g mol^​-1
We know the Molar mass of S = 32 g
so dividing by 16 we will get the number of sulphur atoms
number of atoms of S = 255.59/32
number of atoms of S = 7.98 $~$ 8
Thus, Molecular formula will be S8 . 
Vapour density = molecular mass/ 2
vapour density = 127.49