Answer
See the answer below.
Work Step by Step
The number of moles of DMG: $53.0\ g\div 116.12\ g/mol=0.456\ mol$
The number of moles of ethanol: $525\ g\div 46.07\ g/mol= 11.40\ mol$
a) Mole fraction: $0.456/(0.456+11.40)=0.038$
b) Molality: $0.456\ mol\div (525/1000)kg=0.87\ m$
c) At the normal boiling point, the vapor pressure of pure ethanol is 1 atm
$\Delta P=-X_{solute}P^0=-0.038\times 1\ atm=-0.038\ atm$
Vapor pressure: $1-0.038=0.962\ atm$
d) $\Delta T_b=K_bm=1.22^{\circ}C/m\times 0.87\ m=1.06^{\circ}C$
Boiling point: $78.4+1.06=79.46^{\circ}C$
