Question

Calculate the ebullioscopic constant for water if the enthalpy of vaporization is equal to $ 40.685 $ $ {\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}} $
(A) $ {\text{0}}{\text{.512K}} $ $ {\text{kg mo}}{{\text{l}}^{{\text{ - 1}}}} $
(B) $ {\text{1}}{\text{.86 K}} $ $ {\text{kg mo}}{{\text{l}}^{{\text{ - 1}}}} $
(C) $ {\text{5}}{\text{.12K}} $ $ {\text{kg mo}}{{\text{l}}^{{\text{ - 1}}}} $
(D) $ {\text{3}}{\text{.56 K}} $ $ {\text{kg mo}}{{\text{l}}^{{\text{ - 1}}}} $

Answer 1

Hint: The ebullioscopic constant of water depends upon the molar mass of the solute in the solution. This constant relates the molality of a solution to the elevation in boiling point which is directly proportional to the molality of the solution. We can find the constant using the formula given below.
Formula Used: $ {{\text{K}}_{\text{b}}} = \dfrac{{{\text{R}}{{\text{T}}_{\text{b}}}^{\text{2}}{\text{M}}}}{{\Delta {{\text{H}}_{{\text{vap}}}} \times 1000}} $
Where, R is the universal gas constant, $ {{\text{T}}_{\text{b}}} $ is the boiling point of the solvent, M is the molecular weight of the solvent, and $ \Delta {{\text{H}}_{{\text{vap}}}} $ is the enthalpy of vaporization.

Complete step by step solution:
From the formula for the elevation in boiling point of a solution we know that,
 $ \Delta {{\text{T}}_{\text{b}}} = {\text{i}} \times {{\text{K}}_{\text{b}}} \times {\text{m}} $
Where “i” is the Van’t Hoff factor, $ {{\text{K}}_{\text{b}}} $ is the ebullioscopic constant, and m is the molality of the solution. When rearranged, this formula can be written as,
 $ {{\text{K}}_{\text{b}}} = \dfrac{{{\text{R}}{{\text{T}}_{\text{b}}}^{\text{2}}{\text{M}}}}{{\Delta {{\text{H}}_{{\text{vap}}}} \times 1000}} $
 $ \Delta {{\text{H}}_{{\text{vap}}}} $ is the enthalpy of vaporization and is equal to $ 40.685 $ $ {\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}} $ .
Now, the R = $ {\text{8}}{\text{.314 J/mol K}} $ ,
The molecular weight of water = $ \left[ {16 + \left( {2 \times 1} \right)} \right] = 18 $ $ {\text{gram/ mol}} $
 $ {{\text{T}}_{\text{b}}}{\text{ = 100 + 273 K = 373 K}} $
Therefore putting the values of these parameters in the equation for the ebullioscopic constant, we get,
  $ {{\text{K}}_{\text{b}}} = \dfrac{{{\text{8}}{\text{.314}} \times {{\left( {373} \right)}^{\text{2}}} \times {\text{18}}}}{{1000 \times 40.685}} = 0.512 $
Therefore, the ebullioscopic constant for water is $ {\text{0}}{\text{.512K}} $ $ {\text{kg mo}}{{\text{l}}^{{\text{ - 1}}}} $ .
The correct answer is option A.

Note:
The properties that are not dependent upon the nature of the solute present in the solution, but only on the number of the solute particles or the moles of the solute particles are called the “colligative properties”. There are four colligative properties overall and they are: the lowering of vapour pressure, the depression of freezing point, the elevation in the boiling point, and the osmotic pressure.
There are also some solutes that either associate or they dissociate when they are mixed with the solvents, for such liquids the Van’t Hoff factor “i” is added to the equations of these colligative properties.