Chapter 8.8, Problem 8.31P

Interpretation Introduction

(a)

Interpretation:

The boiling point of the solution formed upon the addition of $\text{2 mol}$ sucrose to one kilogram of water is to be determined.

Concept introduction:

The effect of the addition of a non-volatile component to a solvent is that there is an increase in the boiling point. This is called elevation in boiling point. Such an increase is observed since the surface of the solvent is saturated with solute particles. This results in the lesser escape of gaseous molecules and hence the vapor pressure of the solution gets lowered compared to the vapor pressure found above the pure solvent to which no solute has been added.

The temperature at which the vapor pressure of solution becomes equal to the atmospheric pressure is known as the boiling point. Due to the lesser vapor pressure, the solution needs to be heated to higher temperatures in order to boil it. Therefore, the boiling point of the solution is elevated.

One mole of the nonvolatile solute increases the boiling point of one kilogram of water by $0.51°\text{C}$.

The extent of elevation in boiling point is directly proportional to the amount of non-volatile solute added. For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$. Since there are two particles per mole of $\text{NaCl}$, it increases the boiling point calculated as follows:

  $\begin{array}{c}\text{Elevation in boiling point}\left(°\text{C}\right)=2\left(0.51°\text{C}\right)\\ =1.02°\text{C}\end{array}$

The formula to calculate the number of moles from the mass is given as follows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

The formula to calculate the molality is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

The formula to calculate the elevation in boiling point is as follows:

  $\Delta {\text{T}}_b={\text{K}}_b\text{m}$

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b{}^0-{\text{T}}_b$