Chapter 16, Problem 16.30P

(a)

Interpretation Introduction

Interpretation:

The molality of potassium iodide in the aqueous solution 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 16.30P

The molality of $KI$ in the solution is $3.58m.$

Explanation of Solution

Given that, the vapor pressure lowering of an aqueous solution of potassium iodide is $2.00Torrat{20}^{\circ }C.$

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

Then,

  $\begin{array}{l}\Delta P={\chi }_2{P}^{\circ }={\chi }_2\times \left(17.54Torr\right)\\ \\ {\chi }_2=\frac{\Delta P}{17.54Torr}=\frac{2.00Torr}{17.54Torr}=0.114\end{array}$

Suppose the molality of aqueous solution of $KI$ is $m.$$KI$ is a strong electrolyte and it dissociates completely into one mole of $K^{+}\left(aq\right)$ ion and one mole of $I^{-}\left(aq\right)$ ion.  Therefore, the value of i for $KI$ is two.  The colligative molality for aqueous solution of $KI$ can be calculated as given below.

  $m_c=i\times m=2\times m=2m_c.$

The number of moles of $KI$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofKI=2m_c\frac{mol}{kg}\times 1kg=2mmol.\end{array}$

The mole fraction of $KI$ can be calculated as given below.

  $\begin{array}{l}{\chi }_{KI}=\frac{n_{KI}}{n_{Water}+n_{KI}}\\ \\ 0.114=\frac{2m}{\frac{Massofwater}{Molarmassofwater}+2m}\\ \\ 0.114=\frac{2m}{\frac{1000g}{18.02g/mol}+2m}\\ \\ 0.114=\frac{2m}{55.5+2m}\\ \\ \left(0.114\right)\times \left(55.5+2m\right)=2m\\ \\ 2m=6.327+0.228m\\ \\ m=3.58.\end{array}$

Now, molality of aqueous solution of $KI$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molality=\frac{3.58mol}{1kg}=3.58m.\end{array}$

Therefore, the molality of $KI$ in the solution is $3.58m.$

(b)

Interpretation Introduction

Interpretation:

The molality of strontium chloride in the aqueous solution 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 16.30P

The molality of $SrC{l}_2$ in the solution is $2.38m.$

Explanation of Solution

Given that, the vapor pressure lowering of an aqueous solution of $SrC{l}_2$ is $2.00Torrat{20}^{\circ }C.$

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

Then,

  $\begin{array}{l}\Delta P={\chi }_2{P}^{\circ }={\chi }_2\times \left(17.54Torr\right)\\ \\ {\chi }_2=\frac{\Delta P}{17.54Torr}=\frac{2.00Torr}{17.54Torr}=0.114\end{array}$

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

  $m_c=i\times m=3\times m=3m_c.$

The number of moles of $SrC{l}_2$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofSrC{l}_2=3m_c\frac{mol}{kg}\times 1kg=3mmol.\end{array}$

The mole fraction of $SrC{l}_2$ can be calculated as given below.

  $\begin{array}{l}{\chi }_{SrC{l}_2}=\frac{n_{SrC{l}_2}}{n_{Water}+n_{SrC{l}_2}}\\ \\ 0.114=\frac{3m}{\frac{Massofwater}{Molarmassofwater}+3m}\\ \\ 0.114=\frac{3m}{\frac{1000g}{18.02g/mol}+3m}\\ \\ 0.114=\frac{3m}{55.5+3m}\\ \\ \left(0.114\right)\times \left(55.5+3m\right)=3m\\ \\ 3m=6.327+0.342m\\ \\ m=2.38.\end{array}$

Now, molality of aqueous solution of $SrC{l}_2$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molality=\frac{2.38mol}{1kg}=2.38m.\end{array}$

Therefore, the molality of $SrC{l}_2$ in the solution is $2.38m.$

(c)

Interpretation Introduction

Interpretation:

The molality of ammonium sulfate in the aqueous solution 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 16.30P

The molality of ${\left(N{H}_4\right)}_{2}S{O}_4$ in the solution is $2.38m.$

Explanation of Solution

Given that, the vapor pressure lowering of an aqueous solution of ${\left(N{H}_4\right)}_{2}S{O}_4$ is $2.00Torrat{20}^{\circ }C.$

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

Then,

  $\begin{array}{l}\Delta P={\chi }_2{P}^{\circ }={\chi }_2\times \left(17.54Torr\right)\\ \\ {\chi }_2=\frac{\Delta P}{17.54Torr}=\frac{2.00Torr}{17.54Torr}=0.114\end{array}$

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

  $m_c=i\times m=3\times m=3m_c.$

The number of moles of ${\left(N{H}_4\right)}_{2}S{O}_4$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ Molesof{\left(N{H}_4\right)}_{2}S{O}_4=3m_c\frac{mol}{kg}\times 1kg=3mmol.\end{array}$

The mole fraction of ${\left(N{H}_4\right)}_{2}S{O}_4$ can be calculated as given below.

  $\begin{array}{l}{\chi }_{{\left(N{H}_4\right)}_{2}S{O}_4}=\frac{n_{{\left(N{H}_4\right)}_{2}S{O}_4}}{n_{Water}+n_{{\left(N{H}_4\right)}_{2}S{O}_4}}\\ \\ 0.114=\frac{3m}{\frac{Massofwater}{Molarmassofwater}+3m}\\ \\ 0.114=\frac{3m}{\frac{1000g}{18.02g/mol}+3m}\\ \\ 0.114=\frac{3m}{55.5+3m}\\ \\ \left(0.114\right)\times \left(55.5+3m\right)=3m\\ \\ 3m=6.327+0.342m\\ \\ m=2.38.\end{array}$

Now, molality of aqueous solution of ${\left(N{H}_4\right)}_{2}S{O}_4$ can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molality=\frac{2.38mol}{1kg}=2.38m.\end{array}$

Therefore, the molality of ${\left(N{H}_4\right)}_{2}S{O}_4$ in the solution is $2.38m.$