Chapter 16, Problem 16.76P

(a)

Interpretation Introduction

Interpretation:

The molality, colligative molality, freezing point, and the boiling point of $5.0gof{K}_{2}S{O}_4\left(s\right)in0.250kg$ of water have to be calculated.

Concept Introduction:

Molarity can be defined as the number of moles of solute per liter of solution.  Molality can be defined as the number of moles of solute per kilogram of solvent.

Boiling point elevation:

The boiling point of a solution is higher than the boiling point of the pure solvent.  The amount by which the boiling point of solution exceeds the boiling point of the pure liquid that is called the boiling point elevation.  The mathematical relationship is given below.

  $\Delta T_b=T_b-T_b{}^{\circ }=K_b{m}_c$

Where, $\Delta T_b$ is boiling point elevation, $T_b$ is the boiling point of solution, $T_b{}^{\circ }$ is the boiling point of pure solvent, $K_b$ is the boiling point elevation constant and $m_c$ is the colligative molality.

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent.  This effect is called as the freezing point depression.  The mathematical relationship is given below.

  $\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c$

Where, $\Delta T_f$ is freezing point depression, $T_f$ is the freezing point of solution, $T_f{}^{\circ }$ is the freezing point of pure solvent, $K_f$ is the freezing point depression constant and $m_c$ is the colligative molality.

Explanation of Solution

Given that $5.0gof{K}_{2}S{O}_4\left(s\right)$ is dissolved in $0.250kg$ of water.

$K_{2}S{O}_4\left(s\right)$ is a strong electrolyte and it completely dissociates into two moles of $K^{+}\left(aq\right)$ ions and one mole of $S{O}_4{}^{2-}\left(aq\right)$ ion.

The number of moles of $K_{2}S{O}_4\left(s\right)$ can be calculated as shown below.

  $Moles=\frac{Mass}{Molarmass}=\frac{5.0g}{174.26g.mo{l}^{-1}}=0.0287mol.$

The molality and colligative molality of the solution can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Moles}{kgofsolvent}=\frac{0.0287mol}{0.250kg}=0.115m.\\ \\ M_c=i\times m=3\times 0.115m=0.345m_c.\end{array}$

The value of $K_{f}and{K}_b$ of water are $0.513K.m_c{}^{-1}and1.86K.m_c{}^{-1}$ respectively.  The freezing and boiling point of water are $0^{\circ }Cand{100}^{\circ }C$ respectively.

The freezing point of the solution can be calculated as given below.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=\left(1.86K.m_c{}^{-1}\right)\times \left(0.345m_c\right)=0.64K\\ \\ T_f{}^{\circ }-T_f=0.64K\\ \\ T_f=T_f{}^{\circ }-0.64K=0^{\circ }C-{0.64}^{\circ }C=-{0.64}^{\circ }C\\ \\ T_f=-0.64^{o}C.\end{array}$

Therefore, the freezing point of the solution is $-0.64^{o}C.$

The boiling point of the solution can be calculated as given below.

  $\begin{array}{l}\Delta T_b=T_b-T_b{}^{\circ }=K_b{m}_c=\left(0.513K.m_c{}^{-1}\right)\times \left(0.345m_c\right)=0.18K\\ \\ T_b-T_b{}^{\circ }=0.18K\\ \\ T_b=0.18K+T_b{}^{\circ }={0.18}^{\circ }C+{100}^{\circ }C={100.18}^{\circ }C\\ \\ T_b=100.18^{o}C.\end{array}$

Therefore, the boiling point of the solution is $100.18^{o}C.$

(b)

Interpretation Introduction

Interpretation:

The molality, colligative molality, freezing point, and the boiling point of $5.0gofethanolin0.250kg$ of water have to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given that $5.0gof$ ethanol is dissolved in $0.250kg$ of water.

Ethanol is a weak electrolyte.  Therefore, its colligative molality and molality are same.

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

  $Moles=\frac{Mass}{Molarmass}=\frac{5.0g}{46.07g.mo{l}^{-1}}=0.1085mol.$

The molality and colligative molality of the solution can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Moles}{kgofsolvent}=\frac{0.1085mol}{0.250kg}=0.434m.\\ \\ M_c=i\times m=1\times 0.434m=0.434m_c.\end{array}$

The value of $K_{f}and{K}_b$ of water are $0.513K.m_c{}^{-1}and1.86K.m_c{}^{-1}$ respectively.  The freezing and boiling point of water are $0^{\circ }Cand{100}^{\circ }C$ respectively.

The freezing point of the solution can be calculated as given below.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=\left(1.86K.m_c{}^{-1}\right)\times \left(0.434m_c\right)=0.81K\\ \\ T_f{}^{\circ }-T_f=0.81K\\ \\ T_f=T_f{}^{\circ }-0.81K=0^{\circ }C-{0.81}^{\circ }C=-{0.81}^{\circ }C\\ \\ T_f=-0.81^{o}C.\end{array}$

Therefore, the freezing point of the solution is $-0.81^{o}C.$

The boiling point of the solution can be calculated as given below.

  $\begin{array}{l}\Delta T_b=T_b-T_b{}^{\circ }=K_b{m}_c=\left(0.513K.m_c{}^{-1}\right)\times \left(0.434m_c\right)=0.22K\\ \\ T_b-T_b{}^{\circ }=0.22K\\ \\ T_b=0.22K+T_b{}^{\circ }={0.22}^{\circ }C+{100}^{\circ }C={100.22}^{\circ }C\\ \\ T_b=100.22^{o}C.\end{array}$

Therefore, the boiling point of the solution is $100.22^{o}C.$