34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus ?

Hello Student
Given,
p = 0.1 bar
V = 34.05 mL = 34.05 × 10^–3 L = 34.05 × 10^–3 dm³
R = 0.083 bar dm³ K^–1 mol^–1
T = 546°C = (546 + 273) K = 819 K

We can calculate the number of moles (n) by using the ideal gas equation : pV = nRT.
Replacing n by m/M, where, m = mass = 0.0625g
M = Molecular mass, which we need to find.

Therefore, pV = mRT/M,
So, M = mRT/PV = (0.0625 X 0.083 X 819) / ( 0.1 X 34.05 × 10–3 )
= 1247.5 g mol–1