Chapter 13, Problem 116QRT

Interpretation Introduction

Interpretation:

The boiling point, the freezing point and osmotic pressure of $\text{NaCl}$ solution have to be calculated.

Concept introduction

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$. That is water changes from liquid phase to gas phase at temperature ${\text{100}}^{\text{o}}\text{C}$.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}$.

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point.

  ${\text{T}}_{\text{b}}-$ Boiling point of the solution.

  ${\text{T}}_{\text{b}}^{\text{°}}-$ Boiling point of pure solvent.

TheBoiling point elevation${\text{(ΔT}}_{\text{b}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point.

  ${\text{K}}_{\text{b}}-$ Molal boiling point constant.

  $m-$ Molality of the solution.

Freezing point is the temperature at which liquid turns into solid.

TheFreezing point depression${\text{(ΔT}}_{\text{f}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

${\text{ΔT}}_{\text{f}}-$ Change in freezing point.

${\text{K}}_{\text{f}}-$ Molal freezing point constant.

$m-$ Molality of the solution.

Answer to Problem 116QRT

The Boiling point of the solution is $100.0957^{o}C.$ and the freezing point of the solution is $-0.348^{o}C$ and the osmotic pressure is $4.62atm.$

Explanation of Solution

Given data as follows:

Mole of $\text{NaCl}$$\text{= 0}\text{.100}\text{mole}$; molality of $\text{NaCl}$ solution $\text{= 0}\text{.100}\text{mol/kg}$;

‘i’ is $\text{= 1}\text{.87}$.

Mass of water $\text{= 1 kg}\text{.}$

Value of ${\text{K}}_{\text{b}}\text{=}\text{0}{\text{.512}}^{\text{o}}\text{C kg /mol}\text{.}$

Value of ${\text{K}}_f\text{=}-\text{1}{\text{.86}}^{\text{o}}\text{C kg/mol}\text{.}$

Calculate Freezing point of the solution:

Substituting the values of molality, $K_{f}and{K}_b$ the freezing point was calculated as shown below,

    $\begin{array}{c}{\text{ΔT}}_{\text{f }}{\text{= K}}_{\text{f}}\text{m =}\left(\text{-1}\text{.86°C kg/mol}\right)\left(\text{= 0}\text{.100}\text{mol/kg}\right)\left(1.87\right)\text{ = -0}{\text{.348}}^{\circ }\text{C}\\ \\ {\text{ΔT}}_{\text{f }}{\text{= T}}_{\text{f }}^{\text{°}}{\text{-T}}_{\text{f }}\\ \\ {\text{ -T}}_{\text{f }}={\text{ΔT}}_{\text{f }}-{\text{T}}_{\text{f }}^{\text{°}}.\end{array}$

    $\begin{array}{c}{\text{T}}_{\text{f }}={\text{T}}_{\text{f }}^{\text{°}}-{\text{ΔT}}_{\text{f }}\\ \\ =0\text{°C +}\left(\text{-0}{\text{.348}}^{\circ }\text{C}\right)\\ \\ =-0.348^{o}C.\end{array}$

Calculate Boiling point of the solution:

Substituting the values of molality, $K_{f}and{K}_b$ the boiling point was calculated as shown below,

  $\begin{array}{c}{\text{ΔT}}_{\text{b }}{\text{= K}}_{\text{b}}\text{m = }\left(\text{0}\text{.512°C kg/mol}\right)\left(0.100\text{m}\right)\left(1.87\right)\text{= 0}{\text{.0957}}^{\text{o}}\text{C}\\ \\ {\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}.\end{array}$

  $\begin{array}{c}{\text{T}}_{\text{b}}\left(solution\right)={\text{T}}_{\text{b}}^{\text{°}}\left(solvent\right)+{\text{ΔT}}_{\text{b}}\\ \\ ={100}^{\circ }\text{C }+\text{0}{\text{.0957}}^{\text{o}}\text{C}\\ \\ =100.0957^{o}C.\end{array}$

Therefore, the Boiling point of the solution is $100.0957^{o}C.$ and the freezing point of the solution is $-0.348^{o}C.$

Osmotic pressure calculation:

The molarity is calculated as follows:

  $\begin{array}{l}\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}\\ \\ \text{=}\left(\frac{\text{0}\text{.100}\text{mol NaCl}}{\text{1}\text{kg}\text{solvent}}\right)\left(\frac{\text{1}\text{kg}\text{solvent}}{\text{1}\text{kg}\text{solution}}\right)\left(\frac{\text{1}\text{.01 kg solution}}{\text{1}\text{L solution}}\right)\text{=}0.101M\text{NaCl}\text{.}\end{array}$

Hence, molarity obtained is $0.101M.$

Temperature ${\text{= 25}}^{\text{o}}\text{C}\text{+273}\text{=}\text{298}\text{K}\text{.}$

The formula of osmotic pressure is as shown below,

Osmotic pressure $\text{(π) = iMRT}\text{.}$

  $\begin{array}{l}\text{(π) = iMRT}\\ \\ \text{=}\left(\text{1}\text{.87}\right)\left(\text{0}\text{.101}\text{M}\right)\left(\text{0}\text{.08206}\frac{\text{L}\text{.atm}}{\text{mol}\text{.K}}\right)\left(\text{298K}\right)\\ \\ \text{=}4.62atm.\end{array}$

Therefore, the osmotic pressure of solution is $4.62atm.$