Chapter 13, Problem 37PS

(a)

Interpretation Introduction

Interpretation: The freezing point of the given solution has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Change in freezing point is calculated by using the equation,

  $\Delta T_{fp}=K_{fp}{m}_{solute}$

  where,

  $K_{fp}$ is the molal freezing point elevation constant.

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 37PS

1. a) The freezing point of the solution is $-\text{0}{\text{.348}}^{\text{0}}C$

Explanation of Solution

Given,

Mass of phenylalanine is $\text{3\%}\text{=}\frac{\text{3}}{\text{100}}\text{×}\text{100}\text{g}\text{=}\text{3}\text{g}$

Mass of solvent is $\text{100}\text{g}-\text{3}\text{g}\text{=}\text{97}\text{g}\text{=}\text{0}\text{.097}\text{kg}$

Molar mass of phenylalanine is $165.19\text{g/mol}$

Molal freezing point elevation constant of water is $-\text{1}{\text{.86}}^{\text{0}}\text{C}{\text{m}}^{-1}$

  $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{3}\text{g}\text{phenylalanine}}{\text{165}\text{.19}\text{g/mol}}\\ =0.018mol\end{array}$

Molality of the solute is calculated,

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}\\ \text{=}\frac{\text{0}\text{.01816}}{\text{0}\text{.097}\text{kg}}\\ \text{=}\text{0}\text{.187}\text{m}\end{array}$

Hence,

Change in freezing point is,

  $\begin{array}{c}{\text{ΔT}}_{\text{fp}}\text{=}{\text{K}}_{\text{fp}}{\text{m}}_{\text{solute}}\\ \text{=}-\left(\text{1}{\text{.86}}^{\text{0}}\text{C}{\text{m}}^{-1}\right)\text{×}\text{0}\text{.187}\text{m}\\ \text{=}-\text{0}{\text{.3482}}^{\text{0}}C\end{array}$

Therefore,

The freezing point of the solution is $-\text{0}{\text{.348}}^{\text{0}}C$

(b)

Interpretation Introduction

Interpretation: The boiling point of the given solution has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Change in boiling point is calculated by using the equation,

  $\Delta T_{bp}=K_{bp}{m}_{solute}$

  where,

  $K_{bp}$ is the molal boiling point elevation constant.

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 37PS

1. b) The boiling point of the solution is $100.{\text{0959}}^{\text{0}}C$

Explanation of Solution

Given,

Mass of phenylalanine is $\text{3\%}\text{=}\frac{\text{3}}{\text{100}}\text{×}\text{100}\text{g}\text{=}\text{3}\text{g}$

Mass of solvent is $\text{100}\text{g}-\text{3}\text{g}\text{=}\text{97}\text{g}\text{=}\text{0}\text{.097}\text{kg}$

Molar mass of phenylalanine is $165.19\text{g/mol}$

Molal boiling point elevation constant of water is ${0.5121}^{\text{0}}\text{C}{\text{m}}^{-1}$

  $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{3}\text{g}\text{phenylalanine}}{\text{165}\text{.19}\text{g/mol}}\\ =0.018mol\end{array}$

Molality of the solute is calculated,

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}\\ \text{=}\frac{\text{0}\text{.01816}}{\text{0}\text{.097}\text{kg}}\\ \text{=}\text{0}\text{.187}\text{m}\end{array}$

Hence,

Change in boiling point is,

  $\begin{array}{c}{\text{ΔT}}_{\text{bp}}\text{=}{\text{K}}_{\text{bp}}{\text{m}}_{\text{solute}}\\ \text{=}{0.5121}^{\text{0}}\text{C}{\text{m}}^{-1}\text{×}\text{0}\text{.187}\text{m}\\ \text{=}\text{0}{\text{.09588}}^{\text{0}}C\end{array}$

Therefore,

The boiling point of the solution is,

  $\begin{array}{c}bp={100}^{0}C-\text{0}{\text{.0959}}^{\text{0}}C\\ =100.{\text{0959}}^{\text{0}}C\end{array}$

The boiling point of the solution is $100.{\text{0959}}^{\text{0}}C$

(c)

Interpretation Introduction

Interpretation: The osmotic pressure of the given solution has to be determined. The values that are easily measureable in laboratory have to be identified among these values.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Osmotic pressure is calculated by using the equation,

  $\pi =cRT$

  where,

  c is the molar concentration.

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 37PS

1. c) The osmotic pressure of the solution is $4.58atm$

It is found that osmotic pressure can be easily measured experimentally in laboratories.

Explanation of Solution

Given,

Mass of phenylalanine is $\text{3\%}\text{=}\frac{\text{3}}{\text{100}}\text{×}\text{100}\text{g}\text{=}\text{3}\text{g}$

Mass of solvent is $\text{100}\text{g}-\text{3}\text{g}\text{=}\text{97}\text{g}\text{=}\text{0}\text{.097}\text{kg}$

Molar mass of phenylalanine is $165.19\text{g/mol}$

$\begin{array}{c}\text{R}\text{=}\text{0}\text{.08206}\text{L}{\text{.atm K}}^{-1}{\text{mol}}^{-1}\\ T={25}^{0}C=\left(25+273.15\right)K=298.15K\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{3}\text{g}\text{phenylalanine}}{\text{165}\text{.19}\text{g/mol}}\\ =0.018mol\end{array}$

Molality of the solute is calculated,

  $\begin{array}{c}\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}\\ \text{=}\frac{\text{0}\text{.01816}}{\text{0}\text{.1L}}\\ \text{=}\text{0}\text{.1816}\text{M}\end{array}$

Therefore,

The osmotic pressure of the solution is,

  $\begin{array}{c}\pi =cRT\\ =\text{0}\text{.1816}\text{M}\times \text{0}\text{.08206}\text{L}{\text{.atm K}}^{-1}{\text{mol}}^{-1}\times 298.15K\\ =4.58atm\end{array}$

The osmotic pressure of the solution is $4.58atm$

• Comparing to other colligative properties, osmotic pressure is the measurement with large values. Therefore, it can be easily measured experimentally in laboratories with less error.