What is the boilong point of this solution?

Question

What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 23.0 L of water to produce an antifreeze solution with a freezing point of -29.0 degrees C? (The density of ethylene glycol is 1.11g/cm^3, and the density of water is 1.00 g/cm^3.)
What is the boilong point of this solution?

Answer 1

ΔTf= Kf*m

m=molality = mole solute /kg of solvent

So if G is the mass of solvent in grams,
m= 1000*mole/G
but mole= g/MW

where g is the grams of solute and MW the molecular weight of the solute

so ΔTf= Kf*1000*g/ (MW*G)

also
d1=g/Vx => g=d1*Vx and
d2=G/Vsolvent => G=d2*Vwater

so ΔTf= Kf*1000* d1*Vx/(MW*d2*Vwater)

Vx= MW*d2*Vwater*ΔTf/ (1000*Kf*d1) =>
Vx= 62*1*23000*29 / (1000*1.858*1.11) =20052 mL= 20.0 L

ΔΤb= Kb*1000*d1*Vx / (MW*d2*Vwater)=
= 0.512*1000*1.11*20052 / (62*1*23000)
=7.99= 8.0 C

For the solution
Tb= Tb0+ΔΤb =100+8.0=108.0 C at P=1 atm.