Chapter 17 - Phases and Phase Changes - Problems and Conceptual Exercises - Page 604: 16

$M=15.64g/mol$

We can find the required molecular mass as follows: $PV=nRT$ $\implies n=\frac{PV}{RT}$ We plug in the known values to obtain: $n=\frac{(153\times 10^3)(515\times 10^{-6})}{(8.314)(322)}$ $n=2.94\times 10^{-2}$ Now the molecular mass is given as $M=\frac{0.460}{2.94\times 10^{-2}}$ $M=15.64g/mol$