Textbook Problem

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What is the boiling point of a solution containing 0.200 mol of a nonvolatile solute in 125 g of benzene (C₆H₆)?

Interpretation Introduction

Interpretation: The Boiling point of solution (Benzene) has to be identified.

Concept introduction:

Boiling point elevation: The boiling point of the solution varies with the solute concentration.

$\begin{array}{l}\text{Elevation}\text{ in boiling point = }\Delta {\text{T}}_{bp}{\text{= K}}_{bp}{\text{. m}}_{solute}\text{,}\\ \text{where,}{\text{K}}_{bp}\text{is}\text{molal boiling point elevation constant,}\\ {\text{m}}_{solute}\text{is molality of solute}\text{. }\end{array}$

Explanation of Solution

Given data:

$\begin{array}{l}\text{Mole of solute = 0}\text{.200 mol;}\\ \text{Mass of Benzene =}\text{125 g = 0}\text{.125 kg }\\ \text{Boiling}\text{point}\text{of}\text{solution}\text{= ?}\end{array}$

Known data:

$\begin{array}{l}\text{Boiling point of Benzene = 80}{\text{.10}}^{\text{o}}\text{C}\\ {\text{K}}_{\text{bp}}\text{of}\text{benzene=}\text{+2}{\text{.53}}^{\text{o}}\text{C/m}\end{array}$

Molality of a solute is defined as the amount of solute dissolves in kilogram of solvent.

  $\text{Molality}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{kg}}$

$\text{molality = }\frac{\text{Moles of solute (mol)}}{\text{Amount of solvent}\left(\text{in kg}\right)}\text{ = }\frac{\left(\text{0}\text{.200 mol}\right)}{\text{0}\text{.125 kg}}\text{= 1}\text{.6}\text{m}$

$\begin{array}{l}\underset{\_}{\text{Boiling point elevation:}}\\ \text{Elevation}{\text{ in boiling point = ΔT}}_{\text{bp}}{\text{= K}}_{\text{bp}}{}_{}\end{array}$