0.64g of an oxide of a sulphur occupies 0.224L at 2 bar and 273 degree celcius. Identify the compound. Also find out the mass of one molecule of gas.

Dear student,
P = 2 atm
V = 0.224L
Given, mass(m) = 0.64 g
T = 273⁰ C
   = (273 + 273) K  = 546 K
R = 0.082 L atm mol^-1K^{-1
PV = nRT
PV = mRT/PV
M = 0.64*0.082*546/ 2*0.224}
    = 63.96
   = 64
The oxide is SO₂ as its molar mass = 32 + 2 * 16  = 64 g/mol
Mass of 6.022* 10²³ molecule of oxide = 64 g
Mass of 1 molecule of oxide = 64/ 6.022* 10²³ g  = 1.063 * 10^-22g

Regards
Aparna