Question

The density of an ideal gas is 1.25 × 10⁻³ g cm⁻³ at STP. Calculate the molecular weight of the gas.

Use R=8.31J K^-1 mol^-1

Answer 1

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas,\[\rho\]= 1.25 × 10⁻³ gcm⁻³ =1.25 kgm⁻³
Pressure, P = 1.01325\[\times\]10⁵ Pa   (At STP)
Temperature, T = 273 K    (At STP)
Using the ideal gas equation, we get

\[PV   =   nRT                                         .  .  . (1)\] 

\[n   =   \frac{m}{M}                                                 .  .  . (2)\] 

\[ \therefore   PV   =   \frac{m}{M}RT\] 

\[ \Rightarrow M = \frac{m}{V}\frac{RT}{P}\] 

\[ \Rightarrow M = \rho\frac{RT}{P}\] 

\[ \Rightarrow M = 1 . 25 \times \frac{8 . 31 \times 273}{{10}^5}\] 

\[ \Rightarrow M = 2 . 83 \times  {10}^{- 2}   \] 

\[                   = 28 . 3  g -  {\text { mol }}^{- 1}   \]