Answer
$178.71\ g/mol$
Work Step by Step
Boiling point elevation:
$62.22-61.70=0.52^{\circ}C$
Molality:
$\Delta T_b=K_bm\rightarrow 0.52^{\circ} C=3.63^{\circ}C/m\times m\rightarrow m=0.14\ mol/kg$
Number of moles of solute: $0.14\ mol/kg_{solute}=n\div (25.0/1000)kg\rightarrow n=0.0036\ mol$
Molar mass: $0.640\ g\div 0.0036\ mol=178.71\ g/mol$
