How could you prepare a 3.08 m aqueous solution of Glycerine,\({\bf{C}}_3{\bf{H}}_8{\bf{O}}_3\)? What is the freezing point of this solution?

At standard atmospheric pressure, the freezing point may be defined as the temperature at which a liquid is converted into solid.

\({\bf{\Delta Tf = i \times Kf \times m}}\)

m = molality of solution

i = Van’t Hoff factor

\({\bf{Kf}}\) = Cryoscopic constant

\({\bf{\Delta Tf}}\)= Freezing point depression

First, find the molar mass of glycerine.

The molar mass of Glycerine is 92.095 g mole^-1

A 3.08m aqueous glycerine solution require

\({\rm{3}}{\rm{.08 mole \times 92}}{\rm{.095 g \;mol}}{{\rm{e}}^{{\rm{ - 1}}}}{\rm{ = 284 g}}\)

Dissolve 284g of glycerine in 1.00 kg of water to prepare the solution.

Glycerine is a non-electrolyte, so in this case, the value of i = 1

\(\begin{aligned}{l}{\rm{\Delta }}{{\rm{T}}_{\rm{f}}} &= {\rm{1}}{\rm{.86 ^\circ C}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{ \times 3}}{\rm{.08 m }}\\{\rm{\Delta }}{{\rm{T}}_{\rm{f}}} &= {\rm{5}}{\rm{.73^\circ C}}\end{aligned}\)

The freezing point is 5.73^oC.