# Chapter 13 Solutions and Their Behavior - Study Questions - Page 505b: 45

$167.4\ g/mol$

#### 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: $0.180\ g\div 1.08\times10^{-3}\ mol=167.4\ g/mol$

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