34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure.

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3

R = 0.083 bar dm3 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:

$p V=n \mathrm{R} T$

$\Rightarrow n=\frac{p V}{R T}$

$=\frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819}$

$=5.01 \times 10^{-5} \mathrm{~mol}$

Therefore, molar mass of phosphorus $=\frac{0.0625}{5.01 \times 10^{-5}}=1247.5 \mathrm{~g} \mathrm{~mol}^{-1}$

Hence, the molar mass of phosphorus is $1247.5 \mathrm{~g} \mathrm{~mol}^{-1}$.