Question

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.

benzene boiling point=80.1 Kb=2.53

carbon tetrachloride boiling point=76.8 Kb=5.03

Answer 1

Concepts and reason

The physical properties of a solution such as boiling point elevation, freezing point depression, vapor pressure lowering and osmotic pressure are together called as colligative properties which depend on the number of solute particles or ions and independent of identity of ions.

Difference in boiling point before an addition of solute and after the addition of solute is directly proportional to molality of a solute.

For the two solutions with the same boiling point, the molal concentration can be calculated by equating.

Fundamentals

Elevation of boiling point:

Normal boiling point is a temperature at which liquid state changes to gaseous state at one atmospheric pressure. Elevation of boiling point is one of the colligative properties that work upon the addition of solute molecules to pure solvent.

Mathematically,

$\begin{array}{l}\\\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{K}}_{\rm{b}}} \times {\rm{m}}\\\\{{\rm{T}}_{\rm{b}}}\left( {{\rm{solution}}} \right) - {{\rm{T}}_{\rm{b}}}\left( {{\rm{Solvent}}} \right){\rm{ = }}{{\rm{K}}_{\rm{b}}} \times {\rm{m}}\\\\ & {\rm{Where,}}\,\Delta {{\rm{T}}_{\rm{b}}}\;{\rm{is boiling point temperature}}\\\\ & & '{\rm{i'}}\;{\rm{is the van't Hoff factor }}\\\\ & & {\rm{'m' is molality}}\\\\ & & {\rm{'}}{{\rm{K}}_{\rm{b}}}'{\rm{ is molar boiling point constant}}\\\end{array}$

Given that,

The boiling point of benzene solution = ${80.1^{\rm{o}}}{\rm{C}}$

The boiling point of carbon tetrachloride = ${76.8^{\rm{o}}}{\rm{C}}$

The molal boiling point elevation constant of benzene $\left( {{{\rm{K}}_{\rm{b}}}} \right){\rm{ = 2}}{\rm{.5}}{{\rm{3}}^{\rm{o}}}{\rm{C/m}}$

The molal boiling point elevation constant of carbon tetrachloride $\left( {{{\rm{K}}_{\rm{b}}}} \right){\rm{ = 5}}{\rm{.0}}{{\rm{3}}^{\rm{o}}}{\rm{C/m}}$

Since, the two solutions have the same boiling point.

$\begin{array}{c}\\\Delta {{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{K}}_{\rm{b}}} \times {\rm{m}}\\\\{{\rm{T}}_{\rm{b}}}\left( {{\rm{solution}}} \right) - {{\rm{T}}_{\rm{b}}}\left( {{\rm{Solvent}}} \right){\rm{ = }}{{\rm{K}}_{\rm{b}}} \times {\rm{m}}\\\\{{\rm{T}}_{\rm{b}}}\left( {{\rm{solution}}} \right) = {{\rm{T}}_{\rm{b}}}\left( {{\rm{Solvent}}} \right) + \left( {{{\rm{K}}_{\rm{b}}} \times {\rm{m}}} \right)\\\end{array}$

${\rm{Since, the boiling point of two solutions is same}}$

$\begin{array}{c}\\\left[ {{{\rm{T}}_{\rm{b}}}\left( {{\rm{Solvent}}} \right) + \left( {{{\rm{K}}_{\rm{b}}} \times {\rm{m}}} \right)} \right]\left( {{\rm{Benzene}}} \right) = \,\left[ {{{\rm{T}}_{\rm{b}}}\left( {{\rm{Solvent}}} \right) + \left( {{{\rm{K}}_{\rm{b}}} \times {\rm{m}}} \right)} \right]\left( {{\rm{CC}}{{\rm{l}}_4}} \right)\\\\{80.1^{\rm{o}}}{\rm{C + }}\left( {{\rm{2}}{\rm{.5}}{{\rm{3}}^{\rm{o}}}{\rm{C/m}}} \right) \times {\rm{m = 76}}{\rm{.}}{{\rm{8}}^{\rm{o}}}{\rm{C + }}\left( {{{5.03}^{\rm{o}}}{\rm{C/m}}} \right) \times {\rm{m}}\\\\{\rm{3}}{\rm{.}}{{\rm{3}}^{\rm{o}}}{\rm{C = }}\left( {{\rm{2}}{\rm{.}}{{\rm{5}}^{\rm{o}}}{\rm{C/m}}} \right) \times {\rm{m}}\\\\{\rm{Molality = }}\frac{{{\rm{3}}{\rm{.}}{{\rm{3}}^{\rm{o}}}{\rm{C }}}}{{\left( {{\rm{2}}{\rm{.}}{{\rm{5}}^{\rm{o}}}{\rm{C/m}}} \right)}}\\\\ = 1.32{\rm{m}}\\\end{array}$

The boiling point is calculated below:

$\begin{array}{c}\\{{\rm{T}}_{\rm{b}}}\left( {{\rm{solution}}} \right) - {{\rm{T}}_{\rm{b}}}\left( {{\rm{Solvent}}} \right){\rm{ = }}{{\rm{K}}_{\rm{b}}} \times {\rm{m}}\\\\{{\rm{T}}_{\rm{b}}}\left( {{\rm{solution}}} \right){\rm{ = 80}}{\rm{.}}{{\rm{1}}^{\rm{o}}}{\rm{C + }}\left( {{\rm{2}}{\rm{.5}}{{\rm{3}}^{\rm{o}}}{\rm{C/m}}} \right) \times {\rm{1}}{\rm{.32m}}\\\\{\rm{ = 83}}{\rm{.439}}{{\rm{6}}^{\rm{o}}}{\rm{C}}\\\end{array}$

Ans:

The molal concentration, m (or b), and boiling point, Tb is determined below:

$\begin{array}{c}\\{\rm{The molal concentration}}\left( {\rm{m}} \right){\rm{ = 1}}{\rm{.32m}}\\\\{\rm{The boiling point}}\left( {{{\rm{T}}_{\rm{b}}}} \right) = {83.4396^{\rm{o}}}{\rm{C}}\\\end{array}$