Question

What is the boiling point of a solution of 1.0 g of Glycerine, \({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{5}}}{\left( {{\bf{OH}}} \right)_{\bf{3}}}\), in 47.8 g of water? Assume an ideal solution.

Short answer

The boiling point of a solution \({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{\left( {{\rm{OH}}} \right)_{\rm{3}}}\)is 100.12^oC.

Step-by-step solution

Boiling Point

At standard atmospheric pressure, the boiling may be defined as the temperature at which the liquid is converted into vapours.

The boiling point is

\({\bf{\Delta }}{{\bf{T}}_{\bf{b}}}{\bf{ = }}{{\bf{K}}_{\bf{b}}}{\bf{m}}\)

where,

\({{\bf{K}}_{\bf{b}}}\)= Base Dissociation Constant

m = Molality of the solute

Explanation

Mass of Glycerine,\({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{\left( {{\rm{OH}}} \right)_{\rm{3}}}\)= 1.0 g

Molar Mass of Glycerine\({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{\left( {{\rm{OH}}} \right)_{\rm{3}}}\), = 92 g/mole

\({\rm{Number of Moles of\; = }}\frac{{\rm{1}}}{{{\rm{92}}}}{\rm{ = 0}}{\rm{.011}}\)

Mass of Solvent, Water = 47.8 g or 0.0478 kg

Molality = Number of Moles of Glycerine/ Mass of Solvent (kg)

\({\rm{Molality = }}\frac{{{\rm{0}}{\rm{.011}}}}{{{\rm{0}}{\rm{.0478}}}}{\rm{ = 0}}{\rm{.23m}}\)

Calculate the change in boiling point.

\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{K}}_{\rm{b}}}{\rm{m = 0}}{\rm{.52 \times 0}}{\rm{.23 = 0}}{\rm{.1}}{{\rm{2}}^{\rm{o}}}{\rm{C}}\)

\(\)

Therefore, the boiling point of the solution is given by

\({\rm{Boiling point temperature = 10}}{{\rm{0}}^{\rm{o}}}{\rm{C + 0}}{\rm{.1}}{{\rm{2}}^{\rm{o}}}{\rm{C = 100}}{\rm{.1}}{{\rm{2}}^{\rm{o}}}{\rm{C}}\)