Question

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

• A. $124.77gmo{l}^{-1}$
• B. $1247.74gmo{l}^{-1}$
• C. $12.47gmo{l}^{-1}$
• D. $30gmo{l}^{-1}$

Answer 1

The correct option is B $1247.74gmo{l}^{-1}$

Given, $V=34.05mL$, $m=0.0625g$, $T={546}^{\circ }C=819K$

and $P=0.1bar$

Thus, molar mass of the phosphorus can be given as;

$M=\frac{mRT}{PV}=\frac{0.0625\times 0.083\times 819}{0.1\times \left(34.05/1000\right)}$
$=1247/74gmo{l}^{-1}$.