`34.05 mL` of phosphorus vapours weighs `0.0625 g` at `546^(@)C` and `0.1` bar pressure. What is the molar mass of phossphorus ?

Explanation

According to ideal gas equation :
`PV = nRT , PV = (WRT)/(M)` or `M = (WRT)/(PV)`
According to available data ,
Mass of phosphrous vapours `(W) = 0.0625 g`
Volume of vapours `(V) = 34.05 mL = 34.05 xx 10^(-3) L`
Pressure of vapours `(P) = 1.0 "bar"`
Gas constant `(R) = 0.083 "bar" L K^(-1) mol^(-1)`
Temperature `(T) = 546+273 = 819 K`
`:. M = ((0.0625 g) xx (0.083 "bar" LK^(-1) mol^(-1)) xx (819 K))/((1.0 "bar") xx (34.05 xx 10^(-3) L)) = 125 g mol^(-1)`.