Question

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?

Answer 1

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × ${10}^{-3}$ L = 34.05 × ${10}^{-3}d{m}^3$

R = 0.083 bar $d{m}^3{K}^{-1}mo{l}^{-1}$

T = 546°C = (546 + 273) K = 819 K

From the gas equation PV = w. RT / M, we get

M = w. RT/ Pv ……….

(1)The molar mass of phosphorus, $M=\frac{wRT}{PV}=\frac{0.0625\times 0.08314\times 819}{1\times 34.05\times {10}^{-3}}=124.46g/mol$.