Chapter 10, Problem 40QAP

How many grams of the following nonelectrolytes would have to be mixed with 100.0 g of p-dichlorobenzene to increase the boiling point by 3.0°C? To decrease the freezing point by 2.0°C? (Use Table 10.2.)

(a) succinic acid (C₄H₆O₄)

(b) caffeine (C₈H₁₀N₄O₂)

Interpretation Introduction

(a)

Interpretation:

The mass of succinic acid mixed with 100.0 g of p-dichlorobenzene to increase the boiling point by $3.0{}^{\circ }\text{C}$ and freezing point by $2.0{}^{\circ }\text{C}$ needs to be determined.

Concept introduction:

Non-electrolytes are substances which do not dissociate into ions in the aqueous solution. Thus, they do not conduct electricity.

If a non−volatile solute is added to a volatile solvent, elevation in freezing point takes place. This can be determined using the following formula:

${\text{ΔT}}_{\text{b}}\text{=}{\text{k}}_{\text{b}}\text{×molality}\left(\text{m}\right)$

Here, ${\text{ΔT}}_{\text{b}}\text{=}$ Boiling point elevation

${\text{k}}_{\text{b}}=$ Molal boiling point elevation

Similarly, the depression in freezing point can be calculated as follows:

${\text{ΔT}}_f\text{=}{\text{k}}_f\text{×molality}\left(\text{m}\right)$

Here, ${\text{ΔT}}_f\text{=}$ Freezing point depression

And, ${\text{k}}_{\text{b}}=$ Molal freezing point depression

Molality is defined as number of moles of solute in 1 kg of the solvent.

$\text{Molality of the solution}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{kg}\text{of}\text{solvent}}$

Answer to Problem 40QAP

The mass of succinic acid mixed with 100.0 g of p-dichlorobenzene to increase the boiling point by $3.0{}^{\circ }\text{C}$ and freezing point by $2.0{}^{\circ }\text{C}$ is 5.67 g and 3.3 g respectively.

Explanation of Solution

The elevation in boiling point can be calculated using the following formula:

${\text{ΔT}}_{\text{b}}\text{=}{\text{k}}_{\text{b}}\text{×molality}\left(\text{m}\right)$

Here, ${\text{ΔT}}_{\text{b}}\text{=}$ Boiling point elevation

${\text{k}}_{\text{b}}=$ Molal boiling point elevation

Since,

$\begin{array}{l}{\text{k}}_{\text{b}}=6.2{}^{\text{0}}\frac{\text{C}}{\text{m}}\\ \Delta {\text{T}}_{\text{b}}=3^0\text{C}\end{array}$

So,

$\begin{array}{c}\text{Molality}\left(\text{m}\right)\text{=}\frac{{\text{ΔT}}_{\text{b}}}{{\text{k}}_{\text{b}}}\\ \text{=}\frac{\text{3}\bcancel{{}^{\text{0}}\text{C}}}{\text{6}\text{.2}\frac{\bcancel{{}^{\text{0}}\text{C}}}{\text{m}}}\\ \text{=}\text{0}\text{.48}\text{m}\end{array}$

Thus, to increase the boiling point by ${3.0}^0\text{C}$ the molality of the solution should be $0.48\text{m}$.

Now, molality is related to number of moles of solute as follows:

$\text{Molality of the solution}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Kg}\text{of}\text{solvent}}$

Put the values of mass solvent as $0.1000\text{kg}$ and molality $0.48\text{m}$

$\begin{array}{c}\text{moles of solute=}\text{molality}\left(\text{m}\right)\text{×mass }\text{solvent}\left(\text{kg}\right)\\ \text{=}\text{0}\text{.48}\text{m×0}\text{.1000}\text{kg}\\ \text{=}\text{0}\text{.048}\text{mol}\end{array}$

To calculate the freezing point depression, following formula is used.

${\text{ΔT}}_{\text{f}}\text{=}{\text{k}}_{\text{f}}\text{×molality}\left(\text{m}\right)$

Here ${\text{ΔT}}_{\text{f}}$ is freezing point lowering and ${\text{k}}_{\text{f}}$ is molal freezing point constant.

Since,

$\begin{array}{l}{\text{k}}_{\text{f}}=7.1{}^{\text{0}}\frac{\text{C}}{\text{m}}\\ \Delta {\text{T}}_{\text{f}}={2.0}^0\text{C}\end{array}$

So,

$\begin{array}{c}\text{Molality}\left(\text{m}\right)\text{=}\frac{{\text{ΔT}}_{\text{f}}}{{\text{k}}_{\text{f}}}\\ \text{=}\frac{2\bcancel{{}^{\text{0}}\text{C}}}{7.1\frac{\bcancel{{}^{\text{0}}\text{C}}}{\text{m}}}\\ \text{=}\text{0}\text{.28}\text{m}\end{array}$

To decrease the freezing point by $2^0\text{C}$ the molality of the solution should be $0.28\text{m}$.

Now,

$\text{Molality of the solution}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{kg}\text{of}\text{solvent}}$

Put the values of mass solvent as $0.1000\text{kg}$ and molality $0.28\text{m}$

$\begin{array}{c}\text{mole of}\text{solute=}\text{molality}\left(\text{m}\right)\text{×mass}\text{solvent}\left(\text{kg}\right)\\ \text{=}\text{0}\text{.28}\text{m×0}\text{.1000}\text{kg}\\ \text{=}\text{0}\text{.028}\text{mol}\end{array}$

Since, molar mass of succinic acid $\left({\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\right)$ is 118.08 thus, masses in both case can be calculated using the following relation:

$\mathrm{mass}=\text{number of moles×molar mass}$

For boiling point case:

$\begin{array}{c}\text{Mass}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{=}\text{0}\text{.048}\cancel{\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}}\text{×}\frac{\text{118}\text{.09}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}}{\text{1}\cancel{\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}}}\\ \text{=}\text{5}\text{.67}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\end{array}$

For freezing point case:

$\begin{array}{c}\text{Mass of}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{=}\text{0}\text{.028}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\text{×}\frac{\text{118}\text{.09}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}}{\text{1}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}}\\ =3.3\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The mass of caffeine mixed with 100.0 g of p-dichlorobenzene to increase the boiling point by $3.0{}^{\circ }\text{C}$ and freezing point by $2.0{}^{\circ }\text{C}$ needs to be determined.

Concept introduction:

Non-electrolytes are substances which do not dissociate into ions in the aqueous solution. Thus, they do not conduct electricity.

If a non−volatile solute is added to a volatile solvent, elevation in freezing point takes place. This can be determined using the following formula:

${\text{ΔT}}_{\text{b}}\text{=}{\text{k}}_{\text{b}}\text{×molality}\left(\text{m}\right)$

Here, ${\text{ΔT}}_{\text{b}}\text{=}$ Boiling point elevation

${\text{k}}_{\text{b}}=$ Molal boiling point elevation

Similarly, the depression in freezing point can be calculated as follows:

${\text{ΔT}}_f\text{=}{\text{k}}_f\text{×molality}\left(\text{m}\right)$

Here, ${\text{ΔT}}_f\text{=}$ Freezing point depression

And, ${\text{k}}_{\text{b}}=$ Molal freezing point depression

Molality is defined as number of moles of solute in 1 kg of the solvent.

$\text{Molality of the solution}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{kg}\text{of}\text{solvent}}$

Answer to Problem 40QAP

The mass of caffeine mixed with 100.0 g of p-dichlorobenzene to increase the boiling point by $3.0{}^{\circ }\text{C}$ and freezing point by $2.0{}^{\circ }\text{C}$ is 9.32 g and 5.4 g respectively.

Explanation of Solution

The elevation in boiling point can be calculated using the following formula:

${\text{ΔT}}_{\text{b}}\text{=}{\text{k}}_{\text{b}}\text{×molality}\left(\text{m}\right)$

Here, ${\text{ΔT}}_{\text{b}}\text{=}$ Boiling point elevation

${\text{k}}_{\text{b}}=$ Molal boiling point elevation

Since,

$\begin{array}{l}{\text{k}}_{\text{b}}=6.2{}^{\text{0}}\frac{\text{C}}{\text{m}}\\ \Delta {\text{T}}_{\text{b}}=3^0\text{C}\end{array}$

So,

$\begin{array}{c}\text{Molality}\left(\text{m}\right)\text{=}\frac{{\text{ΔT}}_{\text{b}}}{{\text{k}}_{\text{b}}}\\ \text{=}\frac{\text{3}\bcancel{{}^{\text{0}}\text{C}}}{\text{6}\text{.2}\frac{\bcancel{{}^{\text{0}}\text{C}}}{\text{m}}}\\ \text{=}\text{0}\text{.48}\text{m}\end{array}$

Thus, to increase the boiling point by ${3.0}^0\text{C}$ the molality of the solution should be $0.48\text{m}$.

Now, molality is related to number of moles of solute as follows:

$\text{Molality of the solution}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{kg}\text{of}\text{solvent}}$

Put the values of mass solvent as $0.1000\text{kg}$ and molality $0.48\text{m}$

$\begin{array}{c}\text{moles of solute=}\text{molality}\left(\text{m}\right)\text{×mass }\text{solvent}\left(\text{kg}\right)\\ \text{=}\text{0}\text{.48}\text{m×0}\text{.1000}\text{kg}\\ \text{=}\text{0}\text{.048}\text{mol}\end{array}$

To calculate the freezing point depression, following formula is used.

${\text{ΔT}}_{\text{f}}\text{=}{\text{k}}_{\text{f}}\text{×molality}\left(\text{m}\right)$

Here ${\text{ΔT}}_{\text{f}}$ is freezing point lowering and ${\text{k}}_{\text{f}}$ is molal freezing point constant.

Since,

$\begin{array}{l}{\text{k}}_{\text{f}}=7.1{}^{\text{0}}\frac{\text{C}}{\text{m}}\\ \Delta {\text{T}}_{\text{f}}={2.0}^0\text{C}\end{array}$

So,

$\begin{array}{c}\text{Molality}\left(\text{m}\right)\text{=}\frac{{\text{ΔT}}_{\text{f}}}{{\text{k}}_{\text{f}}}\\ \text{=}\frac{2\bcancel{{}^{\text{0}}\text{C}}}{7.1\frac{\bcancel{{}^{\text{0}}\text{C}}}{\text{m}}}\\ \text{=}\text{0}\text{.28}\text{m}\end{array}$

To decrease the freezing point by $2^0\text{C}$ the molality of the solution should be $0.28\text{m}$.

Now,

$\text{Molality of the solution}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Kg}\text{of}\text{solvent}}$

Put the values of mass solvent as $0.1000\text{kg}$ and molality $0.28\text{m}$

$\begin{array}{c}\text{mole of}\text{solute=}\text{molality}\left(\text{m}\right)\text{×mass}\text{solvent}\left(\text{kg}\right)\\ \text{=}\text{0}\text{.28}\text{m×0}\text{.1000}\text{kg}\\ \text{=}\text{0}\text{.028}\text{mol}\end{array}$

Since, molar mass of caffeine $\left({\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\right)$ is 194.20 g/mol thus, mass can be calculated as follows:

$\mathrm{mass}=\text{number of moles×molar mass}$

For boiling point case:

$\begin{array}{c}\text{Mass}\text{of}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\text{=}\text{0}\text{.048}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\text{×}\frac{\text{194}\text{.20}\text{g}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\\ =9.32\text{g}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\end{array}$

For freezing point case:

$\begin{array}{c}\text{Mass}\text{of}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\text{=}\text{0}\text{.028}\cancel{\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\text{×}\frac{\text{194}\text{.20}\text{g}}{\text{1}\cancel{\text{mol}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}}\\ =5.4\text{g}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\end{array}$