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

34.05 mL of phosphorus vapour weights 0.0625 g at 546^(@)C and 0.1 bar

1.	34.05 mL of phosphorus vapour weights 0.0625 g at 546^(@)C and 0.1 bar pressure. What is the molar mass of phosphorus ?
Answer» Solution :According to ideal GAS law, `pV=nRT=(w)/(M)RT` `therefore M=(wRT)/(pV)` where, Molecular MASS of phosphorus, (w) = 0.0625 g Gas Constant `(R )=8.314xx10^(-2)` BAR `L mol^(-1)K^(-1)` Pressure of phosphorus vapour (p) = 1.0 bar Absolute TEMPERATURE `(T)=(546+273)K=819 K` Volume of Phosphorus vapour (V) = 34.05 mL = 0.03405 L `therefore M =((0.0625 g)(8.314xx10^(-2)"bar L mol"^(-1)K^(-1))(819K))/(("1 bar")(0.03405 L))` `= 124.98 = 125 g mol^(-1)` Note : If 0.1 bar is corrct for according to text book then answer will be 1249.8. It is not real because, `P_(4)=31xx4=124`