Chapter 16, Problem 16.23P

(a)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution that is $0.25m$ in sodium chloride has to be calculated.

Concept Introduction:

Raoult’s law:

It states that the equilibrium vapor pressure of the solvent over a solution is directly proportional to the mole fraction of the solvent in the solution.

The expression for the vapor pressure lowering is given below.

  $\Delta P={\chi }_2{P}^{\circ }$

Where, $\Delta P$ is vapor pressure lowering, ${\chi }_2$ is the mole fraction of solute and $P^{\circ }$ is vapor pressure of the pure solvent.

Answer to Problem 16.23P

The vapor pressure lowering for the aqueous solution of sodium chloride is $0.1566Torr.$

Explanation of Solution

Given that the molality of aqueous solution of sodium chloride is $0.25m.$$NaCl$ is a strong electrolyte and it dissociates completely into one mole of $N{a}^{+}\left(aq\right)$ ion and one mole of $C{l}^{-}\left(aq\right)$ ion.  Therefore, the value of i for $NaCl$ is two.  The colligative molality for aqueous solution of $NaCl$ can be calculated as given below.

  $m_c=i\times m=2\times 0.25m=0.50m_c.$

The number of moles of sodium chloride can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofNaCl=0.50\frac{mol}{kg}\times 1kg=0.50mol.\end{array}$

The mole fraction of sodium chloride can be calculated as given below.

  $\begin{array}{l}{\chi }_{NaCl}=\frac{n_{NaCl}}{n_{Water}+n_{NaCl}}\\ \\ {\chi }_{NaCl}=\frac{0.50}{\frac{Massofwater}{Molarmassofwater}+0.50}\\ \\ {\chi }_{NaCl}=\frac{0.50}{\frac{1000g}{18.02g/mol}+0.50}\\ \\ {\chi }_{NaCl}=8.93\times 10^{-3}.\end{array}$

The vapor pressure of water at ${20}^{\circ }C$ is $17.54Torr.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(8.93\times {10}^{-3}\right)\times \left(17.54Torr\right)=0.1566Torr.$

Therefore, the vapor pressure lowering for the aqueous solution of sodium chloride is $0.1566Torr.$

(b)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution that is $0.25m$ in calcium chloride has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.23P

The vapor pressure lowering for the aqueous solution of calcium chloride is $0.234Torr.$

Explanation of Solution

Given that the molality of aqueous solution of calcium chloride is $0.25m.$$CaC{l}_2$ is a strong electrolyte and it dissociates completely into one mole of $C{a}^{2+}\left(aq\right)$ ion and two moles of $C{l}^{-}\left(aq\right)$ ion.  Therefore, the value of i for $CaC{l}_2$ is three.  The colligative molality for aqueous solution of $CaC{l}_2$ can be calculated as given below.

  $m_c=i\times m=3\times 0.25m=0.75m_c.$

The number of moles of calcium chloride can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofCaC{l}_2=0.75\frac{mol}{kg}\times 1kg=0.75mol.\end{array}$

The mole fraction of calcium chloride can be calculated as given below.

  $\begin{array}{l}{\chi }_{CaC{l}_2}=\frac{n_{CaC{l}_2}}{n_{Water}+n_{CaC{l}_2}}\\ \\ {\chi }_{CaC{l}_2}=\frac{0.75}{\frac{Massofwater}{Molarmassofwater}+0.75}\\ \\ {\chi }_{CaC{l}_2}=\frac{0.75}{\frac{1000g}{18.02g/mol}+0.75}\\ \\ {\chi }_{CaC{l}_2}=0.0133.\end{array}$

The vapor pressure of water at ${20}^{\circ }C$ is $17.54Torr.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.0133\right)\times \left(17.54Torr\right)=0.234Torr.$

Therefore, the vapor pressure lowering for the aqueous solution of calcium chloride is $0.234Torr.$

(c)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution that is $0.25m$ in sucrose has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.23P

The vapor pressure lowering for the aqueous solution of sucrose is $0.08Torr.$

Explanation of Solution

Given that the molality of aqueous solution of sucrose is $0.25m.$  Sucrose is a weak electrolyte.  Therefore, the value of i for sucrose is one.  The colligative molality for aqueous solution of sucrose can be calculated as given below.

  $m_c=i\times m=1\times 0.25m=0.25m_c.$

The number of moles of sucrose can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofSucrose=0.25\frac{mol}{kg}\times 1kg=0.25mol.\end{array}$

The mole fraction of sucrose can be calculated as given below.

  $\begin{array}{l}{\chi }_{Sucrose}=\frac{n_{Sucrose}}{n_{Water}+n_{sucrose}}\\ \\ {\chi }_{Sucrose}=\frac{0.25}{\frac{Massofwater}{Molarmassofwater}+0.25}\\ \\ {\chi }_{Sucrose}=\frac{0.25}{\frac{1000g}{18.02g/mol}+0.25}\\ \\ {\chi }_{Sucrose}=4.5\times 10^{-3}.\end{array}$

The vapor pressure of water at ${20}^{\circ }C$ is $17.54Torr.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(4.5\times {10}^{-3}\right)\times \left(17.54Torr\right)=0.08Torr.$

Therefore, the vapor pressure lowering for the aqueous solution of sucrose is $0.08Torr.$

(d)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution that is $0.25m$ in aluminum perchlorate has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.23P

The vapor pressure lowering for the aqueous solution of aluminum perchlorate is $0.3105Torr.$

Explanation of Solution

Given that the molality of aqueous solution of aluminum perchlorate is $0.25m.$$Al{\left(Cl{O}_4\right)}_3$ is a strong electrolyte and it dissociates completely into one mole of $A{l}^{3+}\left(aq\right)$ ion and three moles of $Cl{O}_4{}^{-}\left(aq\right)$ ion.  Therefore, the value of i for $Al{\left(Cl{O}_4\right)}_3$ is four.  The colligative molality for aqueous solution of $Al{\left(Cl{O}_4\right)}_3$ can be calculated as given below.

  $m_c=i\times m=4\times 0.25m=1.0m_c.$

The number of moles oaluminum perchlorate can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofAl{\left(Cl{O}_4\right)}_3=1.0\frac{mol}{kg}\times 1kg=1.0mol.\end{array}$

The mole fraction of aluminum perchlorate can be calculated as given below.

  $\begin{array}{l}{\chi }_{Al{\left(Cl{O}_4\right)}_3}=\frac{n_{Al{\left(Cl{O}_4\right)}_3}}{n_{Water}+n_{Al{\left(Cl{O}_4\right)}_3}}\\ \\ {\chi }_{Al{\left(Cl{O}_4\right)}_3}=\frac{1.0}{\frac{Massofwater}{Molarmassofwater}+1.0}\\ \\ {\chi }_{Al{\left(Cl{O}_4\right)}_3}=\frac{1.0}{\frac{1000g}{18.02g/mol}+1.0}\\ \\ {\chi }_{Al{\left(Cl{O}_4\right)}_3}=0.0177.\end{array}$

The vapor pressure of water at ${20}^{\circ }C$ is $17.54Torr.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.0177\right)\times \left(17.54Torr\right)=0.3105Torr.$

Therefore, the vapor pressure lowering for the aqueous solution of aluminum perchlorate is $0.3105Torr.$