Chapter 13, Problem 106SCQ

(a)

Interpretation Introduction

Interpretation: The molality, calculated freezing points and van’t Hoff factor (i) for the acids in the given table has to be determined

Concept introduction:

Freezing point depression: The freezing point of the solution varies with the solute concentration.

  $\Delta T_{fp}=K_{fp}{m}_{solute}$

  where,

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

van’t Hoff factor, i: it is the ration between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

  $van\text{'}tHofffactor,i=\frac{measured\Delta {\text{T}}_{fp}}{Calculated\Delta {\text{T}}_{fp}}$

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

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

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 106SCQ

Acid (1 mass %)	Molality (mol/kg ${\text{H}}_{\text{2}}\text{O}$)	$T_{measured}$ (${}^{\circ }C$)	$T_{calculated}$ (${}^{\circ }C$)	i
$HN{O}_3$	$\text{0}\text{.1603}\text{m}$	-0.56	$\text{-0}{\text{.2981}}^{\text{0}}\text{C}$	$1.8785$
$C{H}_{3}C{O}_{2}H$	$\text{0}\text{.1682}\text{m}$	-0.32	$\text{-0}{\text{.3128}}^{\text{0}}\text{C}$	$1.023$
$H_{2}S{O}_4$	$\text{0}\text{.1029}\text{m}$	-0.42	$\text{-0}{\text{.1914}}^{\text{0}}\text{C}$	$2.1943$
$H_2{C}_2{O}_4$	$\text{0}\text{.1122}\text{m}$	-0.30	$\text{-0}{\text{.2086}}^{\text{0}}\text{C}$	$1.4381$
$HC{O}_{2}H$	$\text{0}\text{.2194}\text{m}$	-0.42	$\text{-0}{\text{.408}}^{\text{0}}\text{C}$	$1.029$
$CC{l}_{3}C{O}_{2}H$	$\text{0}\text{.06182}\text{m}$	-0.21	$\text{-0}{\text{.1150}}^{\text{0}}\text{C}$	$1.826$

Explanation of Solution

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Molarity of any substance can be determined using the equation

,

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

Therefore,

For $HN{O}_3$,

In $100g$ of solution, $1g$ of $HN{O}_3$ is present (mass % = $1\%$ ) and mass of water is $99g$.

    $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{1g}}{63.01\text{g/mol}}\\ \text{=}\text{0}\text{.01587}\text{mol}\end{array}$

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.01587}\text{mol}}{\text{0}\text{.099}\text{kg}}\\ \text{=}\text{0}\text{.1603}\text{m}\end{array}$

Depression in freezing point is calculated by using the equation

  ${\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}$

Therefore,

For $HN{O}_3$,

  $\begin{array}{c}{\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}\\ \text{=}\left(\text{-1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.1603}\text{m}\right)\\ \text{=-0}{\text{.2981}}^{\text{0}}\text{C}\end{array}$

For $C{H}_{3}C{O}_{2}H$,

In $100g$ of solution, $1g$ of $C{H}_{3}C{O}_{2}H$ is present (mass % = 1%) and mass of water is $99g$.

    $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{1g}}{60.05\text{g/mol}}\\ \text{=}\text{0}\text{.01665}\text{mol}\end{array}$

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.01665}\text{mol}}{\text{0}\text{.099}\text{kg}}\\ \text{=}\text{0}\text{.1682}\text{m}\end{array}$

Depression in freezing point is calculated by using the equation

  ${\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}$

For $C{H}_{3}C{O}_{2}H$,

  $\begin{array}{c}{\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}\\ \text{=}\left(\text{-1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.1682}\text{m}\right)\\ \text{=-0}{\text{.3128}}^{\text{0}}\text{C}\end{array}$

For $H_{2}S{O}_4$,

In $100g$ of solution, $1g$ of $H_{2}S{O}_4$ is present (mass % = 1%) and mass of water is $99g$

    $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{1g}}{98.08\text{g/mol}}\\ \text{=}\text{0}\text{.01019}\text{mol}\end{array}$

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.01019}\text{mol}}{\text{0}\text{.099}\text{kg}}\\ \text{=}\text{0}\text{.1029}\text{m}\end{array}$

Depression in freezing point is calculated by using the equation

  ${\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}$

For $H_{2}S{O}_4$,

  $\begin{array}{c}{\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}\\ \text{=}\left(\text{-1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.1029}\text{m}\right)\\ \text{=-0}{\text{.1914}}^{\text{0}}\text{C}\end{array}$

For $H_2{C}_2{O}_4$,

In $100g$ of solution, $1g$ of $H_2{C}_2{O}_4$ is present (mass % = 1%) and mass of water is $99g$

    $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{1g}}{90.03\text{g/mol}}\\ \text{=}\text{0}\text{.0111}\text{mol}\end{array}$

$\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.0111}\text{mol}}{\text{0}\text{.099}\text{kg}}\\ \text{=}\text{0}\text{.1122}\text{m}\end{array}$

Depression in freezing point is calculated by using the equation

  ${\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}$

For $H_2{C}_2{O}_4$,

$\begin{array}{c}{\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}\\ \text{=}\left(\text{-1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.1122}\text{m}\right)\\ \text{=-0}{\text{.2086}}^{\text{0}}\text{C}\end{array}$

For $HC{O}_{2}H$,

In $100g$ of solution, $1g$ of $HC{O}_{2}H$ is present (mass % = 1%) and mass of water is $99g$

    $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{1g}}{\text{46}\text{.03g/mol}}\\ \text{=}\text{0}\text{.02172}\text{mol}\end{array}$

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.02172}\text{mol}}{\text{0}\text{.099}\text{kg}}\\ \text{=}\text{0}\text{.2194}\text{m}\end{array}$

Depression in freezing point is calculated by using the equation

  ${\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}$

For $HC{O}_{2}H$,

  $\begin{array}{c}{\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}\\ \text{=}\left(\text{-1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.2194}\text{m}\right)\\ \text{=-0}{\text{.408}}^{\text{0}}\text{C}\end{array}$

For $CC{l}_{3}C{O}_{2}H$,

In $100g$ of solution, $1g$ of $CC{l}_{3}C{O}_{2}H$ is present (mass % = 1%) and mass of water is $99g$.

    $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{1g}}{163.378\text{g/mol}}\\ \text{=}\text{0}\text{.00612}\text{mol}\end{array}$

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.00612}\text{mol}}{\text{0}\text{.099}\text{kg}}\\ \text{=}\text{0}\text{.06182}\text{m}\end{array}$

Depression in freezing point is calculated by using the equation

  ${\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}$

For $CC{l}_{3}C{O}_{2}H$,

  $\begin{array}{c}{\text{ΔT}}_{\text{fp}}{\text{= K}}_{\text{fp}}{\text{. m}}_{\text{solute}}\\ \text{=}\left(\text{-1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.06182}\text{m}\right)\\ \text{=-0}{\text{.1150}}^{\text{0}}\text{C}\end{array}$

Van’t Hoff factor can be find out using the equation,

  $van\text{'}tHofffactor,i=\frac{measured\Delta {\text{T}}_{fp}}{Calculated\Delta {\text{T}}_{fp}}$

For $HN{O}_3$,

  $\begin{array}{c}i=\frac{-{0.56}^0\text{C}}{-{0.2981}^0\text{C}}\\ =1.8785\end{array}$

For $C{H}_{3}C{O}_{2}H$,

  $\begin{array}{c}i=\frac{-{0.32}^0\text{C}}{-{0.3128}^0\text{C}}\\ =1.023\end{array}$

For $H_{2}S{O}_4$,

  $\begin{array}{c}i=\frac{-{0.42}^0\text{C}}{-{0.1914}^0\text{C}}\\ =2.1943\end{array}$

For $H_2{C}_2{O}_4$,

  $\begin{array}{c}i=\frac{-{0.30}^0\text{C}}{-{0.2086}^0\text{C}}\\ =1.4381\end{array}$

For $HC{O}_{2}H$,

  $\begin{array}{c}i=\frac{-{0.42}^0\text{C}}{-{0.408}^0\text{C}}\\ =1.029\end{array}$

For $CC{l}_{3}C{O}_{2}H$,

  $\begin{array}{c}i=\frac{-{0.21}^0\text{C}}{-{0.1150}^0\text{C}}\\ =1.826\end{array}$

(b)

Interpretation Introduction

Interpretation: The relation of acidic strength with the i value has to be given.

Concept introduction:

van’t Hoff factor, i: it is the ration between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

  $van\text{'}tHofffactor,i=\frac{measured\Delta {\text{T}}_{fp}}{Calculated\Delta {\text{T}}_{fp}}$

Answer to Problem 106SCQ

The values of ‘i’ increases in the increasing order of strength of acids.

Explanation of Solution

The increasing order of acidic strength for the given acids is,

    $C{H}_{3}C{O}_{2}H<HC{O}_{2}H<H_2{C}_2{O}_4<CC{l}_{3}C{O}_{2}H<HN{O}_3<H_{2}S{O}_4$

The increasing order of ‘i’ value for the given acids is,

  $C{H}_{3}C{O}_{2}H<HC{O}_{2}H<H_2{C}_2{O}_4<CC{l}_{3}C{O}_{2}H<HN{O}_3<H_{2}S{O}_4$

Hence,

The values of ‘i’ increases in the increasing order of strength of acids.