34.05 mL of phosphorus vapour weights 0.0625 g at `546^(@)C` and 1 bar pressure. What is the molar mass of phosphorus?

Explanation

Step 1. Calculation of volume at `0^(@)C` and 1 bar pressure
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (1xx34.05)/(546+273)=(1xxV_(2))/(273)" or " V_(2)=11.35 mL`
11.35 mL of vapour at `0^(@)C` and 1 bar pressure weigh `=0.0625" g"`
Step 2. Calculation of mass of 22700 mL of `0^(@)C` and 1 bar pressure
`:. 22700" mL"` of vapour at `0^(@)C` and 1 bar pressure will weigh `=(0.0625)/(11.35)xx22700=125" g"`
`:. " " "Molar mass" =125" g "mol^(-1)`
Alternatively, using
`R=0.083" bar "dm^(3)K^(-1)mol^(-1)`
PV=nRT, i.e., `n=(PV)/(RT)=(1.0" bar"xx(34.05xx10^(-3)dm^(3))/(0.083" bar " dm^(3)K^(-1)xx819 K)=5xx10^(-4)mol`
`:.` Mass of 1 mole `"=(0.0625)/(5xx10^(-4))"g"=125 g`
`:. " "` Molar mass`=125 g mol^(-1)`