Question

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

Answer 1

Correct Answer - `1249.8gmol^(-1)`
Given, p=0.1 bar
`V=34.05 mL = 34.05xx10^(-3) L =34.05xx10^(-3)dm^(3)`
R=0.083 bar `dm^(3)K^(-1)mol^(-1)`
T= `546^(@)C` = (546+273) K = 819 K
The number of moles (n) can be calculated using the ideal gas equation as:
pV=nRT
`rArrn=(pV)/(RT)`
`=(0.1xx34.05xx10^(-3))/(0.083xx819)`
`= 5.01xx10^(-5)` mol
Therefore, molar mass of phosphorus `=(0.0625)/(5.01xx10^(-5))=1247.5 g mol^(-1)`
Hence, the molar mass of phosphorus is 1247.5 g `mol^(-1).`