Answer
See the answer below.
Work Step by Step
Boiling point elevation: $104.4-100.00=4.4^{\circ} C$
Molality: $\Delta T_b=K_bm\rightarrow 4.4^{\circ}C=0.5121^{\circ}C/m\times m\rightarrow m=8.59\ m$
Number of moles of glycerol: $8.59\ mol/kg\times0.735\ kg=6.32\ mol$
Mass of glycerol added: $6.32\ mol\times92.09\ g/mol=581.56\ g$
Number of moles of water: $735\ g\div18.015\ g/mol=40.80\ mol$
Mole fraction of solute: $6.32/(40.80+6.32)=0.134$
