$34.05 \, mL$ of phosphorus vapour weighs $0.0625 \, g$ at $546^{\circ} C$ and $0.1 \, bar$ pressure. What is the molar mass of phosphorus?

(N/A) Given: $p = 0.1 \, bar$,$V = 34.05 \, mL = 34.05 \times 10^{-3} \, L$,$T = 546 + 273 = 819 \, K$,$m = 0.0625 \, g$.
Using the ideal gas equation $pV = nRT$,where $n = \frac{m}{M}$:
$M = \frac{mRT}{pV}$
Substituting the values:
$M = \frac{0.0625 \times 0.08314 \times 819}{0.1 \times 34.05 \times 10^{-3}}$
$M = \frac{4.255}{0.003405} \approx 1249.6 \, g \, mol^{-1}$.
Rounding to significant figures,the molar mass is approximately $1250 \, g \, mol^{-1}$.