Answer
$151.25\ g/mol$
Work Step by Step
Boiling point elevation: $61.82-61.70=0.12^{\circ}C$
Molality: $\Delta T_b=K_bm\rightarrow m=0.12^{\circ}C\div 3.63^{\circ}C/m=0.033\ mol/kg$
Number of moles: $0.033\ mol/kg\times (25/1000)kg=8.26\times10^{-4}\ mol$
Molar mass: $0.125\ g\div 8.26\times10^{-4}\ mol=151.25\ g/mol$
