Chapter 11 - Liquids and Intermolecular Forces - Exercises - Page 473: 11.13c

$0.0351\ L/mol\ Cl_2$ $0.0203\ L/mol\ NH_3$

$\frac{1\ cm^3}{2.02\ g\ Cl_2}\times\frac{70.096\ g\ Cl_2}{1\ mol\ Cl_2}=35.1\ cm^3/mol\ Cl_2=0.0351\ L/mol\ Cl_2$ $\frac{1\ cm^3}{0.84\ g\ NH_3}\times\frac{17.031\ g\ NH_3}{1\ mol\ NH_3}=20.3\ cm^3/mol\ NH_3=0.0203\ L/mol\ NH_3$