Question

34.05 mL of phosphorus vapour weighs 0.0625 g at ${546}^{\circ }C$ and 1 bar pressure. What is the molar mass of phosphorus?

Answer 1

Given,$P=1.0barV=34.05mL=34.05\times 10–3L=34.05\times {10}^{-3}d{m}^{3}R=0.083bard{m}^3{K}^{-1}mo{l}^{-1}T={546}^{\circ }C=(546+273)K=819K$
The number of moles (n) can be calculated using the ideal gas equation as:
$PV=nRT\Rightarrow n=\frac{pV}{RT}=\frac{1.0\times 34.05\times {10}^{-3}}{0.083\times 819}=5.01\times {10}^{-4}mol$
Therefore, molar mass of phosphorus $=\frac{0.0625}{5.01\times {10}^{-4}}124.75gmo{l}^{–1}$
Hence, the molar mass of Phosphorus, $P_4$ is $124.75gmo{l}^{-1}$.