Chapter 16, Problem 16.26P

(a)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of $1.50m$ ethanol at ${25}^{\circ }C$ 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 16.26P

The vapor pressure lowering for the aqueous solution of ethanol is $0.834mbar.$

Explanation of Solution

Given that the molality of aqueous solution of ethanol is $1.5m.$  Ethanol is a weak electrolyte.  Therefore, the value of i for ethanol is one.  The colligative molality for aqueous solution of ethanol can be calculated as given below.

  $m_c=i\times m=1\times 1.5m=1.5m_c.$

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

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ Molesofethanol=1.5\frac{mol}{kg}\times 1kg=1.5mol.\end{array}$

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

  $\begin{array}{l}{\chi }_{ethanol}=\frac{n_{ethanol}}{n_{Water}+n_{ethanol}}\\ \\ {\chi }_{ethanol}=\frac{1.5}{\frac{Massofwater}{Molarmassofwater}+1.5}\\ \\ {\chi }_{ethanol}=\frac{1.5}{\frac{1000g}{18.02g/mol}+1.5}\\ \\ {\chi }_{ethanol}=0.0263\end{array}$

The vapor pressure of water at ${25}^{\circ }C$ is $31.7mbar.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.0263\right)\times \left(31.7mbar\right)=0.834mbar.$

Therefore, the vapor pressure lowering for the aqueous solution of ethanol is $0.834mbar.$

(b)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of $0.50m$$Tl{\left(Cl\right)}_3$ at ${25}^{\circ }C$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.26P

The vapor pressure lowering for the aqueous solution of $Tl{\left(Cl\right)}_3$ is $1.10mbar.$

Explanation of Solution

Given that the molality of aqueous solution of thallium (III) chloride is $0.50m.$$Tl{\left(Cl\right)}_3$ is a strong electrolyte and it dissociates completely into one mole of $T{l}^{3+}\left(aq\right)$ ion and three mole of $C{l}^{-}\left(aq\right)$ ion.  Therefore, the value of i for $Tl{\left(Cl\right)}_3$ is four.  The colligative molality for aqueous solution of $Tl{\left(Cl\right)}_3$ can be calculated as given below.

  $m_c=i\times m=4\times 0.50m=2.00m_c.$

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

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofTl{\left(Cl\right)}_3=2.00\frac{mol}{kg}\times 1kg=2.00mol.\end{array}$

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

  $\begin{array}{l}{\chi }_{Tl{\left(Cl\right)}_3}=\frac{n_{Tl{\left(Cl\right)}_3}}{n_{Water}+n_{Tl{\left(Cl\right)}_3}}\\ \\ {\chi }_{Tl{\left(Cl\right)}_3}=\frac{2.00}{\frac{Massofwater}{Molarmassofwater}+2.00}\\ \\ {\chi }_{Tl{\left(Cl\right)}_3}=\frac{2.00}{\frac{1000g}{18.02g/mol}+2.00}\\ \\ {\chi }_{Tl{\left(Cl\right)}_3}=0.035\end{array}$

The vapor pressure of water at ${25}^{\circ }C$ is $31.7mbar.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.035\right)\times \left(31.7mbar\right)=1.10mbar.$

Therefore, the vapor pressure lowering for the aqueous solution of thallium chloride is $1.10mbar.$

(c)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of $0.25m$$K_{2}S{O}_4$ at ${25}^{\circ }C$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.26P

The vapor pressure lowering for the aqueous solution of $K_{2}S{O}_4$ is $0.423mbar.$

Explanation of Solution

Given that the molality of aqueous solution of $K_{2}S{O}_4$ is $0.25m.$$K_{2}S{O}_4$ is a strong electrolyte and it dissociates completely into two moles of $K^{+}\left(aq\right)$ ions and one mole of $S{O}_4{}^{2-}\left(aq\right)$ ion.  Therefore, the value of i for $K_{2}S{O}_4$ is three.  The colligative molality for aqueous solution of $K_{2}S{O}_4$ can be calculated as given below.

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

The number of moles of $K_{2}S{O}_4$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ Molesof{K}_{2}S{O}_4=0.75\frac{mol}{kg}\times 1kg=0.75mol\end{array}$

The mole fraction of $K_{2}S{O}_4$ can be calculated as given below.

  $\begin{array}{l}{\chi }_{K_{2}S{O}_4}=\frac{n_{K_{2}S{O}_4}}{n_{Water}+n_{K_{2}S{O}_4}}\\ \\ {\chi }_{K_{2}S{O}_4}=\frac{0.75}{\frac{Massofwater}{Molarmassofwater}+0.75}\\ \\ {\chi }_{K_{2}S{O}_4}=\frac{0.75}{\frac{1000g}{18.02g/mol}+0.75}\\ \\ {\chi }_{K_{2}S{O}_4}=0.0133\end{array}$

The vapor pressure of water at ${25}^{\circ }C$ is $31.7mbar.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.0133\right)\times \left(31.7mbar\right)=0.423mbar.$

Therefore, the vapor pressure lowering for the aqueous solution of $K_{2}S{O}_4$ is $0.423mbar.$