Chapter 16, Problem 16.25P

(a)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of $2.00m$ sucrose 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 16.25P

The vapor pressure lowering for the aqueous solution of sucrose is $1.11\times 10^{-3}bar.$

Explanation of Solution

Given that the molality of aqueous solution of sucrose is $2.00m.$  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.

  $m_c=i\times m=1\times 2.00m=2.00m_c.$

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

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofSucrose=2.00\frac{mol}{kg}\times 1kg=2.00mol.\end{array}$

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

  $\begin{array}{l}{\chi }_{Sucrose}=\frac{n_{Sucrose}}{n_{Water}+n_{sucrose}}\\ \\ {\chi }_{Sucrose}=\frac{2.00}{\frac{Massofwater}{Molarmassofwater}+2.00}\\ \\ {\chi }_{Sucrose}=\frac{2.00}{\frac{1000g}{18.02g/mol}+2.00}\\ \\ {\chi }_{Sucrose}=0.035.\end{array}$

The vapor pressure of water at ${25}^{\circ }C$ is $0.0317bar.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.035\right)\times \left(0.0317bar\right)=1.11\times 10^{-3}bar.$

Therefore, the vapor pressure lowering for the aqueous solution of sucrose is $1.11\times 10^{-3}bar.$

(b)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of $2.00m$$NaCl$ at ${25}^{\circ }C$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.25P

The vapor pressure lowering for the aqueous solution of sodium chloride is $2.1313\times 10^{-3}bar.$

Explanation of Solution

Given that the molality of aqueous solution of sodium chloride is $2.00m.$$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.

  $m_c=i\times m=2\times 2.00m=4.00m_c.$

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

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofNaCl=4.00\frac{mol}{kg}\times 1kg=4.00mol.\end{array}$

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

  $\begin{array}{l}{\chi }_{NaCl}=\frac{n_{NaCl}}{n_{Water}+n_{NaCl}}\\ \\ {\chi }_{NaCl}=\frac{4.00}{\frac{Massofwater}{Molarmassofwater}+4.00}\\ \\ {\chi }_{NaCl}=\frac{4.00}{\frac{1000g}{18.02g/mol}+4.00}\\ \\ {\chi }_{NaCl}=0.0672.\end{array}$

The vapor pressure of water at ${25}^{\circ }C$ is $0.0317bar.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.0672\right)\times \left(0.0317bar\right)=2.1313\times 10^{-3}bar.$

Therefore, the vapor pressure lowering for the aqueous solution of sodium chloride is $2.1313\times 10^{-3}bar.$

(c)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of $2.00m$$CaC{l}_2$ at ${25}^{\circ }C$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 16.25P

The vapor pressure lowering for the aqueous solution of $CaC{l}_2$ is $3.01\times 10^{-3}bar.$

Explanation of Solution

Given that the molality of aqueous solution of calcium chloride is $2.00m.$$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.

  $m_c=i\times m=3\times 2.00m=6.00m_c.$

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

  $\begin{array}{l}Molality=\frac{Molesofsolute}{kgofsolvent}\\ \\ Molesofsolute=Molality\times kgofsolvent\\ \\ MolesofCaC{l}_2=6.00\frac{mol}{kg}\times 1kg=6.00mol\end{array}$

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

  $\begin{array}{l}{\chi }_{CaC{l}_2}=\frac{n_{CaC{l}_2}}{n_{Water}+n_{CaC{l}_2}}\\ \\ {\chi }_{CaC{l}_2}=\frac{6.00}{\frac{Massofwater}{Molarmassofwater}+6.00}\\ \\ {\chi }_{CaC{l}_2}=\frac{6.00}{\frac{1000g}{18.02g/mol}+6.00}\\ \\ {\chi }_{CaC{l}_2}=0.0975.\end{array}$

The vapor pressure of water at ${25}^{\circ }C$ is $0.0317bar.$

The vapor pressure lowering can be calculated as given below.

  $\Delta P={\chi }_2{P}^{\circ }=\left(0.0975\right)\times \left(0.0317bar\right)=3.01\times 10^{-3}bar.$

Therefore, the vapor pressure lowering for the aqueous solution of calcium chloride is $3.01\times 10^{-3}bar.$