Answer
$167.4\ g/mol$
Work Step by Step
Freezing point depression: $-0.04^{\circ}C$
Molality: $\Delta T_f=K_fm\rightarrow m=-0.04^{\circ}C\div -1.86^{\circ}C/m=0.0215\ mol/kg$
Number of moles: $0.0215\ mol/kg\times (50.0/1000)kg=1.08\times10^{-3}\ mol$
Molar mass: $0.180\ g\div 1.08\times10^{-3}\ mol=167.4\ g/mol$
