Chapter 12, Problem 12.144QP

Interpretation Introduction

Interpretation:

The general misconception that is adding salt in water helps to cook spaghetti faster should be explained. The boiling point of salted water should be calculated and the does this temperature increase will make more difference in cooking time should be identified.

Concept Introduction:

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$. That is water changes from liquid phase to gas phase at temperature ${\text{100}}^{\text{o}}\text{C}$.

${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}$

Where,

${\text{ΔT}}_{\text{b}}-$ Change in boiling point

${\text{T}}_{\text{b}}-$ Boiling point of the solution

${\text{T}}_{\text{b}}^{\text{°}}-$ Boiling point of pure solvent

The Boiling point elevation${\text{(ΔT}}_{\text{b}}\text{)}$ is indicated as follows,

${\text{ΔT}}_{\text{b}}{\text{= iK}}_{\text{b}}\text{m}$

Where,

${\text{ΔT}}_{\text{b}}-$ Change in boiling point

${\text{K}}_{\text{b}}-$ Molal boiling point constant

$m-$ Molality of the solution

$i-$ Von’t Hoff’s factor

Answer to Problem 12.144QP

The boiling point increase was calculated as $0.12{}^{\text{o}}\text{C}$. This very small temperature increase in boiling point will not make more effect over cooking time.

Explanation of Solution

The increase in boiling point of salted solution is calculated by first determining the concentration of salt in the salted water used to cook spaghetti.

Calculate$NaCl$moles:

$\begin{array}{c}\text{=}\cancel{\text{1}\text{Tbsp}\text{salt}}\text{×}\frac{\text{15}\cancel{\text{mL}\text{salt}}}{\cancel{\text{1}\text{Tbsp}\text{salt}}}\text{×}\frac{\cancel{\text{1}{\text{cm}}^{\text{3}}\text{salt}}}{\cancel{\text{1mL}\text{salt}}}\text{×}\frac{\text{2}\text{.2}\cancel{\text{g}}\text{NaCl}}{\cancel{\text{1}{\text{cm}}^{\text{3}}\text{salt}}}\text{×}\frac{\text{1}\text{mol}\text{NaCl}}{\begin{array}{l}\text{58}\text{.44}\cancel{\text{g}}\text{NaCl}\\ \end{array}}\\ \text{= 0}\text{.6}\text{mol}\text{NaCl}\end{array}$

Assuming that one table spoon of salt contains $\text{15}\text{mL}$ salt which then converted into moles.

Calculate mass of water:

$\begin{array}{l}=5\cancel{L}\times \frac{1000\cancel{mL}}{1\cancel{L}}\times \frac{1\cancel{g}water}{1\cancel{mL}}\times \frac{1kgwater}{1000\cancel{g}water}\\ =5kg\end{array}$

The mass of water determined by assuming one table spoon of salt added to 5 quarts of water.

Calculate molality of the salted solution:

$\begin{array}{l}\text{=}\frac{\text{0}\text{.6}\text{mol}\text{NaCl}}{\text{5}\text{kg}\text{water}}\\ \text{=}\text{0}\text{.12}\text{m}\end{array}$

The molality of solution was calculated by dividing solute moles and mass of solvent.

Calculate the increase in boiling point:

Generally, i is the value that denotes the number of ions, those are obtained after dissociation of electrolyte.

For example the i value for $NaCl$ is 2 since dissociation of $NaCl$ yields two ions ${\text{Na}}^{\text{+}}\text{and}{\text{Cl}}^{\text{-}}$.

$\begin{array}{c}{\text{ΔT}}_{\text{b}}{\text{= iK}}_{\text{b}}\text{m}\\ \text{=2×0}\text{.52}\frac{{}^{\text{o}}\text{C}}{\cancel{\text{m}}}\text{×0}\text{.12}\cancel{\text{m}}\\ =0.12{}^{\text{o}}\text{C}\\ \text{Where}{\text{K}}_{\text{b}}=\text{0}\text{.52}\raisebox{1ex}{${}^{\text{o}}\text{C}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.\\ \text{i}=2\end{array}$

The boiling point increase was calculated by multiplying the Von’t Hoff’s factor of $\text{NaCl}$ with molality and the Molal boiling point constant.

Conclusion

The general misconception that is adding salt in water helps to cook spaghetti faster was explained. The boiling point of salted water was calculated and this temperature increase does not make more difference in cooking time.